Answer to the Friday Puzzle….

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First, the tickets for the next event at the Edinburgh Secret Society go on sale at 10 a.m. today!

On Friday I posted this puzzle….

A lovely puzzle this week.

I have a box and inside the box are four stones. One stone is white, another is yellow, the third is blue and the fourth stone is also blue. I put my hand into the box and pick up two stones. I bring my hand out in a fist, look inside my fist and remove a blue stone. What are the chances of the other stone in my hand also being blue?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

There are 4 stones: W, Y, B1 and B2.

There are the following 6 possibilities for drawing 2 stones:

W-Y

w-B1

W-B2

Y-B1

Y-B2

B1-B2

W-Y is eliminated, but all of the other possibilities involve at least 1 blue stone. However, only one of them involves 2 blue stones so the answer is 1 in 5. Did you solve it? Any other answers?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

314 comments on “Answer to the Friday Puzzle….

  1. Roland says:

    Got it. Sort of Baysian statistics, I believe

  2. Michael Sternberg says:

    Yes. Formally, 20% is the conditional probability of finding the second stone to be blue, under the condition that the first was found blue, for the reason stated.

    However, the conditional probability is awkward to formulate colloquially. No statement is made to the effect that a draw where the first stone comes out as not blue is disregarded. Instead we read “look inside my fist and remove a blue stone.” This would indicate that the protocol involves choosing. Even here, no statement is made should the draw contain two non-blue stones. Moreover, the choosing process will mean that, as far as the chooser is concerned, all uncertainty is removed, and the probability for the second stone also being blue is 0% or 100%, under the condition that is was or was not seen as blue, respectively. This may sound tautological, but conditional probability is not to be trifled with.

    • thequiet1 says:

      “No statement is made to the effect that a draw where the first stone comes out as not blue is disregarded.”

      Given that the first stone in this trial is blue, we need not consider that possibility. It isn’t asking for the probability of any event outside of this one trial.

    • Mike Torr says:

      Thank you Michael, for pointing that out. I have spent some time trying to point out to people here about the 0 and 1 (0% and 100%) argument, and I’ve just gone back and read your post: it appears you were first :)

  3. Gib says:

    I got 1 out of 3. Because if you selected B1 first, then you have a 1 in 3 change of getting B2 second. If you selected B2 first, then you have a 1 in 3 change of getting B1 second.

    So, either way, it’s 1 in 3.

    Richard, I think if your question had a friend looking into your hand, and telling you whether you had at least one blue rock, then your answer might be correct. But your question had you only looking at the first rock you pulled out.

    • Eddie says:

      Totally agree, Gib! I got 1 in 3 for exactly the same reason, Just as Richard’s questions can sometimes be vague, his answers can sometimes be wrong!

    • Michael Sternberg says:

      “But your question had you only looking at the first rock you pulled out.”

      The question does not say that. It says “look inside my fist”. The issue is wether that means one or two stones being seen, and hence choosing being done or not. Equivalently, wether the stones being pulled out are an ordered pair (1 in 3) or not (1 in 5).

    • Eddie (@edzeteito) says:

      I just assumed (fairly reasonably I think) that ‘looking inside his fist’ meant unfurling/opening his forefinger and thumb slightly to reveal the first ball with the one behind it obscured. Thus the question I answered is what is the probability of the ball behind the revealed ball being blue and the answer is 1/3. Richard’s answer suggests ‘x-ray’/colour vision.

    • Gib says:

      Michael, you’re right. Looking at the question again it does seem he looked at both stones. So, I sit corrected. 1 in 5 it is.

  4. Hannes Naude says:

    Nope. 1 in 3.

  5. John Loony says:

    The answer is either 1/3 or 1/5, depending on how you interpret the question.

    My first interpretation of the question was that “I look inside my fist” means that I peek into the fist in such a way that I can only see one stone, and I do not know the colour of the other stone (in other words, I have no choice over which stone presents itself to me). In this case, there are 6 possible instances of a blue stone being visible, 2 of which have a blue stone as the second (still hidden) stone in the fist. The probability is therefore 2 out of 6, or 1/3.

    My second interpretation of the question is that “I look inside my fist” means that I open my fist in such a way that I can see both stones, and can therefore choose either one, but keep it hidden from the neutral observer. The person holding the stones is thus transformed into a version of the game show presenter in the “Monty Hall” situation, where the TV presenter (the person holding the stones) has more knowledge than the neutral observer, and can use this knowledge to make a selection and thus influence the probability. In this case, it is 1/5 (as stated above).

    Having followed these Friday puzzles for a few weeks now, I am gradually coming to the view that the wording of the puzzle is deliberately ambiguous and deliberately open to multiple interpretations and different answers. Thus Comrade Wiseperson is being naughty and mischievous in pretending that his answer of 1/5 is somehow “the” answer, or “the” “correct” answer.

    • Hannes Naude says:

      Very nice explanation. I would only add that it is also necessary to know that the chooser is following a strategy of showing a blue stone if one is available. If the chooser is still choosing at random (or following some other strategy) the answer remains 1 in 3 (or some other strategy dependent answer).

    • Michael Sternberg says:

      Cogent explanation. However, the question stated “look inside my fist and remove a blue stone.” This would favor the second interpretation (choosing), since seeing only the stone that will be removed is redundant – the outcome is the same as if the first stone is simply removed blindly and found outside to be blue.

    • thequiet1 says:

      You are looking at the problem as though you need to know the probability before you pick up the stones. We are only concerned with the probability after we have seen 1 blue stone. It doesn’t matter if it was chosen after looking at both stones or if it was chosen from the fist blind – the probability is the same.

      The question was completely unambiguous.

    • Hannes Naude says:

      It absolutely does matter whether the chooser saw one or two stones AND how he chose which one to show. An intuitive way to see this is to consider that the chooser saw both stones and follows the following strategy to choose which one to show.

      Show a white stone if it is available,
      If not, show a yellow stone if it is available,
      If not, show a blue stone.

      In this case the answer would be 100% since seeing a blue stone being shown implies with 100% certainty that there are no yellow or white stones in the hand. The 1 in 5 answer only holds if we know that the chooser’s strategy started with

      Show a blue stone if it is available.

      There is absolutely nothing in the wording of the problem to suggest that this is the case.

  6. John Loony says:

    P.S. The answer given is also rubbish because, at some point in time during the weekend, the staones seem to have Schrödingerily transsubstantiated themselves into “marbles”. This metaphysical abomination is overwhelmingly less likely than either of the two possible answers of 1/3 and 1/5.

  7. safc4ever says:

    It is 1 in 3 (1/3):

    The chances of the first one you look at being blue are 2/4 (or 1/2); Taking out the stone leaves you with 3 stones to choose from, only 1 of which is blue which has a probability of 1/3.

    To check, multiply the probabilities of the first and the second being blue to find out the probability of them both being blue: 1/2 x 1/3 = 1/6 If the answer really was 1/5, the probability of selecting both blue stones would be 1/2 x 1/5 = 1/10 which is obviously not the case as there are only 6 combinations as shown in the ‘Answer’.

    • Except…

      Think of it this way. There are six possible pairs of stones. Five of them have a single blue stone in the pair. What the whole “I reveal that there is one blue stone” clue shows is that the single pair that does not contain a single blue stone (white-yellow) is eliminated, making a total of five possible pairs of stones. There is only one pair of stones that has both blue stones. Therefore, the answer is 1 pair out of a possible five pairs.

    • Tomas Blomberg says:

      thequixoticman, you should also be removing all those cases where a White and Blue was pulled and a White was shown, just as you are removing all the cases where a white and yellows was pulled and a white or yellow was shown.

      Didn’t you guys see the output from the simulation? That is how to estimate a conditional probability by simulation.

      Here is the pseudo code if you want to try it out. It will estimate the probability for the cases

      A: No apriori knowledge of the preference of any color or aversion to any color, so White is shown about half the time when the pair is White/Blue. Target group is when a Blue is shown.

      B: The question “Do you have at least one Blue?” is asked. The target group is when the answer is “Yes.”

      Hit = 0
      A = 0
      B = 0
      Repeat until satisfied
      ..Pull two balls at random
      ..If the one shown is Blue, ++A
      ..If at least one Blue, ++B
      ..If both are Blue, ++Hit
      ..Print Hit/A
      ..Print Hit/B

      It is not reasonable to assume that there is the strategy “Always show a Blue if there is one” any more than you can assume “He hates Blue so he tries to never show it.” That is a _huge_ assumption to take one of these strategies over the other. Without apriori knowledge of any strategy we must assume that there is none. In other words, case A above.

      Or Wiseman could have just phrased the puzzle as it should be phrased (case B) and the answer would have been 1/5 without anybody questioning it. B is how these puzzles are phrased by people who actually understands how important the phrasing is.

    • Steve Jones says:

      I fear the 1 in 5 is the only answer crowd will win out due to the simplistic nature of their argument. Those of use who will argue the ambiguous nature of question and therefore there are a range of answers will lose out.

      However, I too have written a simulator, and it does indeed come up with the numbers (within expected statistical variation) that bayes theorem would indicate.

      As it is, this question would have been better placed as one of game theory.

      nb. another little issue which is missed by many is the time at which the question is posed. If it is formulated only after the first blue stone is revealed, then that adds yet another complexity which tells us this problem is more related to game theory – that is about optimal strategies (albeit we don’t have the required information to tell us what the aims of such a game might be – perhaps wrong-footing the guesser).

    • Tomas Blomberg says:

      Steve Jones, in your simulator, which ball does the person show when there is a White and Blue in his hand?

      If you say “Always Blue since he loves Blue” then that is a huge assumption of the person’s strategy.

      If you say “Only Blue when he has to, since he hates Blue” that is an equally as huge assumption of the person’s strategy.

      If you say “White about half the time and Blue about half the time.” you’d have a good basis for simulating the problem as posed.

      Again, the way the problem was posed by Wiseman is _not_ how you would phrase it if you understand it and want 1/5 to be the answer.

    • Steve Jones says:

      I simulated three scenarios with the “fists” being generated randomly. Scenario 1 is when the picker always shows a blue stone if there is one in his fist, scenario 2 where he shows the first stone in his fist (equivalent to a random selection) and scenario 3 where he only shows a blue stone when both stones are that colour.

      These all act to filter different proportions of “fists” with just one blue stone. Scenario 1 (equivalent to Richard’s solution) filters out none of the single blue stone combinations and hence produces a 1 in 5 probability. Scenario 2 filters out half the “single blue stone” combinations and yields a 1 in 3 probability. Scenario 3 filters out all the single blue stone combinations and yields a probability of 1.

      In fact it is possible to come up with an infinite number of policies by replacing the hard rules over the preference to show a blue stone with a probability.

      I make no claim whatsoever that scenarios (1), (2) or (3) are the correct ones. I’m just pointing out that the wording of the question does not tell us what policy is chosen by the picker. As such it is impossible to be definitive, although I’d guessed that scenario (1) was the most likely one that RIchard will have intended

      I’d also point out that the question is only posed after the blue stone is revealed. Given that by that time the picker will have seen both stones in his hand, he might ask the question (depending on motive) which is most likely to deceive. For instance, if he knew the second stone was also a blue, he might ask what is the probability of the second one being blue knowing that on a purely random choice, it is less likely. So any question based on privileged information makes a definitive answer near impossible. This is a question about game theory, and not random chance alone.

      So – to repeat, I am merely claiming that the motives and policies of the stone picker play a critical part in the probability, and in this particular case it’s impossible to be definitive. The scenarios are simply examples of how this affects probabilities.

    • Tomas Blomberg says:

      And when you don’t have knowledge of the strategy, the correct assumption is to assume none. That is because if you happened to be presented by the same problem by different people during your lifetime, the strategies they have (if any) would vary, making the assumtion of a lack of strategy the best one.

      Again the correct way to state such a puzzle is that someone asks if a certain color is present, and the person holding the balls looks into his hand to confirm or deny it. No strategy or lack of strategy needs to be assumed then.

    • Steve Jones says:

      Assuming lack of strategy is still an assumption. In any event, it’s quite a bit assumption that there is no strategy given that we were explicitly told he looked. If that wasn’t relevant, why were we told?

      As I say, it’s impossible to be definitive about the probability in this case.

    • Tomas Blomberg says:

      Exactly. Let’s say that him looking into his hand is supposed to give us a hint that he hates Blue. It could also be that he peeked into his hand because he loves Yellow and would like to show you one, but in this case he couldn’t find a Yellow. It could also be that he peeked inside the hand because he would love to show you a White ball, but in this instance he can’t.

      This reasoning should show anyone that it is absurd to assume any strategy but a squarely distributed random strategy. That is what we can be allowed to do, since it’s easy to assume that even if a strategy/love/aversion to certain colors exist, in the long run it would vary from person to person, still making the random choice the only reasonable assumption to calculate this probability.

      Did I mention that the problem is poorly phrased? :)

    • it’s not 2 step process. You pick 2 stones. What are the chances that both are blue? 1 in Six. Given that atleast one is blue, what are the chances that both are blue? 1 in 5. I was sceptical earlier, as I too got the wrong answer, but re-read the question, and it clear now

    • Tomas Blomberg says:

      “Given that atleast one is blue”

      Is it given that you in all cases always will know when at least one is blue? That is not stated in this problem.

      Knowing that at least one is blue is _not_ the same as always knowing when at least one is blue in all possible cases.

      It’s hard to explain it clearer.

  8. Steve Jones says:

    The answer given is one of the three I came up with and is true if the picker always shows a blue stone if available. However, if the picker chooses a random stone of the two, it’s 1 in 3 whilst if the picker only chooses a blue stone if he has no choice (i.e. both stones are blue) then the probability of a blue is 1.

    This may seem counter-intuitive, but the reason is quite simple. It comes down to which of the 6 possible original configurations (1 of which has no blue stones) are eliminated before they get to the second stage.

    As Richard states, there are 5 possible combinations of the two stones. In scenario (1) where the picker chooses a blue stone if available, then there are 5 equally likely hands which will contain one blue stone which we will always get to see first. However, as only one of those 5 combinations will contain a second blue stone, the probability of the second stone being blue is 1 in 5 or 0.2.

    However, consider the opposite scenario which I’ll call (2). Here the picker only chooses a blue stone if he has no choice. That is if both stones are blue. Then there is only one combination where we will see the first stone as blue. The other 5 are eliminated as the policy effectively screens these out of this question. With only 1 configuration where the second stone is always blue, the probability of the second stone being blue is 1.

    In scenario (3), the blue stone was picked at random. As there are 4 configurations containing just one blue stone and 1 where both are blue, we have to treat this differently. In the “single blue” combination, the random choice will reveal a blue stone on only half the occasions whilst with the two blue stone combinations will always, of course, reveal a blue stone. This means that, on average, out of the six possible combinations a blue stone will be revealed first on 3 of them. Of these 3, only 1 has the second blue stone so the probability in that case is 1 in 3.

    To check this, I wrote a simulator which confirms the results. There is a serious point here – if somebody is making a decision based on privileged information to which you are not a party, then you are at a considerable disadvantage and it will often be impossible to work out the probabilities. In this case, you need to know the picker would always choose a blue stone if available.

    nb. my bet was that scenario (1) would be the answer, but the other two are valid. Indeed there’s an infinite range – there are intermediate policies which can be chosen where lone blue stones (if available) are picked with a probability P.

    • thequiet1 says:

      The answer for option 3 is the same as for option 1; 1 in 5. Have a good think about it.

    • Steve Jones says:

      @thequiet1

      The answer to option 3 is not the same as option 1. If the picker only ever chooses a blue stone if he has two already then the probability is 1. Given that policy, the guesser will only ever see a blue stone when the picker has two blue stones in his hand. In effect, 5 out of the 6 original combinations will never be represented by a single blue stone.

      As it happens this is really more like a game theory problem save that we don’t know what the full rules are. If the purpose is to make the guesser get it wrong most of the time, then the stone picked and the question asked will be critical.

    • Tort says:

      This is a commonly misunderstood part of probabilities. You are including in your probabilities calculations that determine whether the person is going to find a blue stone. We already know they found a blue stone the question is given that what is the probability that there is another blue stone. Working it out correctly the first and third options give the same answer.

    • Steve Jones says:

      I’d got the order of my scenarios wrong (the 100% was scenario (2).

      Anyway, it doesn’t matter in that scenario (1) & (3) do not lead to the same probability. I suggest you go and read about the “Tuesday Boy” puzzle to find out why.

      This is one discussion (of many)

      http://letterstonature.wordpress.com/2010/06/11/what-is-tuesday-boy-telling-us/

  9. M says:

    IT IS 1/5 !

    That is the only correct answer and Richard explained it correctly.

    The question was very, very clear, and everbody who states it is 1/3 is wrong!

    He looks in his fist, in order to look for a blue marble. If only one of them is blue, he HAS to pick that one. If both are blue, he can choose.

    If he has the Y-W case, he is unable to pick out a blue stone, so he has to retry. That leaves 5 valid scenario’s, and only one of them is B-B.

    • John Loony says:

      You are wrong.

      There are four possible handfuls with one blue, and there are two possible blues in the one handful with two blues. Therefore there are six possible blues which can be picked, and two of those have a blue as the second one.

    • Yat says:

      “He looks in his fist, in order to look for a blue marble. If only one of them is blue, he HAS to pick that one. If both are blue, he can choose.”…

      Woaw ! Indeed, the question is very clear if you read in Richard’s mind and extract sentenses like this one. Nothing like this is told in the original problem. It says :

      “I bring my hand out in a fist, look inside my fist and remove a blue stone.”.

      So If you can translate that into the first quoted sentence, then I could as well translate it into the following one :

      “He looks in his fist, in order to randomly pick one of the two stones. He happens to pick a blue one”

      or even

      ‘He looks in his fist, in order to look for a non-blue marble. If both marbles are blue he HAS to pick one. If only one was blue then he would have picked the other”.

      It involves assumptions on the real meaning of the original sentence, but not more than your translation.

    • M says:

      It says
      “I … look inside my fist and remove a blue stone”

      That means there is no random. If it was not intentional to pick out a blue one, then why does he say he looks in his fist? It is clearly worded.

      In 5 out of 6 cases he is able to do this. And this sentence includes that the 6th case did not happen.

      Only in one of the 5 remaining cases he has another blue stone. And in that case it does not matter which one he picks. So @John: that case does NOT count double!

    • Yat says:

      M, even if I twist my mind to pretend I agree that the fact that he looks in his fist means that he does not chose randomly, you are still wrong :

      Okay, let’s assume he looks in his hand for a good reason. Why would it be more valid to assume that the reason is that he wants to remove a blue stone ? The assumption that he wants to avoid removing a blue stone has exactly the same validity. In that case, he will remove a blue stone only when he does not have a choice, so the probability that the remaining stone is blue is 100%.

  10. Peter O says:

    To those people sating the odds were 1/3 or 1/6 I’m afraid you missed the vital clue in the question. The whole point of looking in the fist and removing a blue stone was to discount the possibility of W/Y being an option. Therefore the answer had to be ?/5 – not ?/6. My first answer was 1/6 but then I read the question again because (and thinking logically this time) it wouldn’t be a brain teaser if it were that straight forward!

    • Michael Sternberg says:

      1 in 6 indeed out entirely. 1 in 3 however, is reasonable to consider, but not for the reason you dismissed.

    • I was V confident and happy with my 1/5 answer and delighted to have it confirmed by Mr Wiseman. However; I’ve been mulling it over in the shower and now think there’s a couple of flaws. Firstly the odds of this event change with time. Secondly the 1/5 answer ASSUMES that if the subject picks W/Y he gets another chance for free. This is NOT stated in the question. Therefore the three possible answers apply to 3 very specific questions:
      >> Before picking any stones the odds of selecting 2xBlue = 1/6
      >> Assuming a free go if you pick W/Y then odds at start =1/5
      >> Regardless of the above 2 scenarios the odds of picking a second Blue after revealing the first = 1/3 .

      AND HERE’S THE CRUX –
      The question is asked AFTER the first Blue is revealed. Therefore; AT THAT POINT IN TIME in time the ONLY answer must be 1/3.

      Unfortunately for me this is the one answer I didn’t choose on Friday.
      Mr Wiseman – please can you re-consider your answer?
      Cheers all and let the debate continue!

    • John Loony says:

      Nobody said it was 1/6.

      The answer is 1/3 because it’s 2/6. There are 6 possible blues in 5 possible fistfuls (1 fistful with 2 blues and 4 fistfuls with 1 blue), and 2 of those blues are paired with a second blue.

    • Some people did say 1/6 – I was one of them! And if the question was “What are the odds of picking up two blue stones” at the start the answer is still 1/6. ie 2/4 x 1/3.

      As I tried to explain above I still think the crux to this problem is about WHEN is the question being asked.

      Regardless of looking inside your fist; regardless of throwing away a pair of non blues; at the point in the problem when the question is asked the subject has already revealed one blue stone and at that point in time there are only 3 stones left of which 1 is blue. The answer MUST be 1/3

      I’m glad this has provoked such debate but still feel this is due to a potentially ambiguous problem.

      ps Please not that my first answer on Friday was 1/6, I then changed it to 1/5 but now – after 1/5 has been declared correct I have changed my mind and think it should be 1/3 – I am either more stupid or more clever than I think – but I don’t know which!

  11. Mike Torr says:

    1/5 – yep, that’s what I got. I’ll state again what I said on the question page: the policy in the mind of the selector when peeking inside the fist is not relevant when calculating the probability from the point of view of the observer. All that matters is that a blue stone is produced from the fist, because that’s that the observer sees.

    • Bleach says:

      The policy of the selector is relevant. If the selector would prefer to pick a non-blue stone, but still produces a blue one, then chances of the second stone being blue increase to a 100%.

    • Mike Torr says:

      No: the policy of the selector is not part of the set of information available to the observer, and therefore is not part of the probability calculation. Probabilities are relative to a given set of known facts: they are not inherent constants that are unchanging. If they were, they would always be 0 or 1.

    • Yat says:

      So it is a known fact that the presenter was actually looking for a blue stone when he opened his hand ? Can you tell me where it is writtent ?

    • Mike Torr says:

      I didn’t say anything of the kind. Please read what I wrote. The only “known fact” I’m writing about is the fact that the stone pulled out is blue.

    • Yat says:

      Yes, that is the only fact that we have. But when you make your calculations, you actually consider that the presenter was looking for a blue stone. If you did not, you would not be able to get 1/5. Is it this part that you don’t understand ? You don’t actually see that 1/5 is the result of the assumption that the host was looking for a blue stone ?

    • Mike Torr says:

      Maybe there is another way to get 1/5, without making that assumption. Perhaps making that assumption does give you 1/5 the way you calculate it, but that doesn’t mean you can’t get 1/5 via a different approach.

      Note that in my way of calculating probability, it makes no sense to say one is (or is not) making an assumption about the selection process, because the selection process is not part of the calculation.

    • Yat says:

      No, there is no other way. The only way that the final solution is 1/5 is that whenever there is a blue stone in the hand, the host removes this one. Someone here even gave the formula to get the final probability from the probability that the blue stone is removed when there is one blue stone only. The 1/5 solution just assumes this probability is 1.

      You can even see it with the basic nearly hand-waving solution which consists of enumerating the different options. You can see that, when “1in5ers” apply this method, they consider that every combination where there is a blue stone is considered to be a valid option. This is equivalent to saying that whenever there is a blue stone, it will be the one to be removed.

  12. Anataboga says:

    1 in 5 was my answer too. I know there was much discussion but I really didn’t see why, the wording of the puzzle was pretty clear for once.

  13. Nick Sharrratt says:

    Since there is nothing in the question which states that the 2 stones selected are done blindly/randomly, it’s possible that the odds of the 2nd stone in the fist at the end is either 0% (a blue and non-blue chosen deliberately) or 100% (2 blues chosen deliberately). There is also nothing in the question to indicate that if repeated, a blue would always be removed, that choice could have been randomly made or it could have been done as in the Monty Hall problem of always revealing a goat behind one door, or it could have been a choice to only show a blue if both were blue and nothing otherwise.

    But, since the only way to have a calculable solution is to assume the initial choice is un-directed and that a blue will be removed if one is present, I took it as having 12 options in the selection (depending on stone choice order), 2 combinations of which have been shown to be impossible once a blue stone is present, leaving only 2 options out of the remaining 10 which has a 2nd blue stone in the fist or 1 in 5.

  14. Anonymous says:

    So where are the sodding flagpoles?

  15. thequiet1 says:

    Mike Torr, I agree completely. Not many people seem to get it.

    The only thing we need to assume is that he is not deliberately deceiving us by showing a blue stone only when their are two blue stones as that will result in a probability of 1. I think it is reasonable to discount this possibility as it much as the possibility he is switching stones from his pocket when we aren’t looking; they both offer equally unsatisfying and silly answers to the puzzle.

    But otherwise, whether he is picking the stones randomly or preferentially picking a blue stone when at least one is available the results are the same.

    We are not in any way concerned with how the first stone was chosen. Get that our of your head.

    All we are concerned with is this:
    given that 1 stone is blue, how many initial configurations could their be to satisfy this condition? 5

    How many of those configurations will give us a second blue stone? 1

    If he randomly chooses the white or yellow stone first, we wouldn’t be asking the question, so it can’t effect the probability! Why on earth are people so hung up on that?

    • Yat says:

      “But otherwise, whether he is picking the stones randomly or preferentially picking a blue stone when at least one is available the results are the same.”

      No, they are not. You should do at least one of these 3 things :

      Read Steve Jones’ comment
      Do the maths
      try it for real

    • Mike Torr says:

      Trying this one “for real” is actually problematic. If you look at a simple probability question involving coin tosses are playing cards, it’s possibly to generate a load of random outcomes in a spreadsheet and verify the answer; but in this case, part of the unknown information involves a person’s thought processes. How will you simulate these?

    • Mike Torr says:

      Oops. I meant “OR playing cards”, of course..

    • Yat says:

      Mike Torr,
      It is very interresting that you fully admit that we are missing information to solve the problem, and at the same time say that the only right solution is 1/5. I think you don’t realise that when you assume the host removes the blue stone whenever there is at least one, you make exactly the same type of assumption that when one assumes the host just removes a random stone.

    • Mike Torr says:

      Yat, I knew you would say that. :-|

      The fact that we are missing information does not give us the right to start giving multiple answers. On the contrary, it is part of the reason that we arrive at the one correct answer of 1/5. If we knew more about the decision process in picking the blue stone, the probability would change: there would still be only one answer, and it would be different.

      Once you allow yourself the luxury of speculating about unknowns, you can add more and more possibilities (perhaps the selector has a favourite colour; perhaps they decide to select a blue stone when the other one is not blue, but only if it’s a Thursday…) until in theory you could claim that there were octillions of answers, broken down by all the quantum events in the local region of space time. Each one of these answers would be 0 (if the the second stone turned out not to be blue) or 1 (if it turned out to be blue). In other words, you would have completely analysed every possibility afforded by the universe. This is clearly silly and impossible – but in fact it’s just an extreme extrapolation of any claim that the unknown decision process of the selector has an influence on the probability, or that there is more than one possible answer.

      Does that help to clarify?

    • Steve Jones says:

      @Mike Torr

      You appear to have completely inverted the logic. It’s when we don’t have complete information that we have to give a spread of possibilities.

      Note this is not to say that it’s not reasonable to make assumptions, but these have to be stated and, indeed, I’d identified the 1 in 5 solution as the most likely. However, it’s unreasonable to claim this is definitively the only possible answer as the wording allows for other interpretations. Indeed the very question that is asked matters – perhaps the picker only raises the question put to mislead the guesser. We are into the whole ara of game theory at this point.

      Perhaps you missed the very last sentence of Richard Wiseman’s solution. “Any other answers?”. In other words he does not rule out the possibility that there are other valid answers, a step you appear to have taken. Indeed some might think the most interesting part of this problem is not the fairly trivial puzzle which can easily be solved when the ambiguities are removed, but the debate about how important these unstated assumptions are in coming to an answer.

    • Yat says:

      You allow yourself the luxury of speculating about unknowns when you assume the host was specifically looking for a blue stone to remove it. This is assumption, speculation or whatever you want to call it. The problem does not contain the information, so there are several possible answers. Exactly like if I said “what is the color of Robert’s car ?” Since you absolutely don’t know about which robert I am talking about, you can’t have the answer. And you can certainly not say it is green because you know a Robert with a green car.

      The point is that to chose between 1/5 and 1/3 (and actually, any other answer between 1/5 and 1), you _have_ to make an assumption about the protocol. And you actually are : you assume the host is looking for a bnlue stone to remove. A random selection is not more of an assumption, so 1/5 is not more valid than 1/3

      Thanks for the condescendence, let me give it a try too :
      Does that help you to understand why you are wrong ?

    • Mike Torr says:

      @Steve Jones – I agree that the debate is far more interesting than the puzzle. To me, the question seems unambiguous because it’s about conditional probabilities, and the whole point of these is to arrive at an answer based on the known facts. If the question were about something else – perhaps the way to cut up a square so that it could be re-assembled into some other shape – then there could be several answers.

      @Yat – Which part of what I wrote was condescending? That was not my intention and if it came across that way I apologise. I am just trying to clarify my argument.
      Regarding what you say about my assumptions, I have nothing to add except that I am not assuming anything – and that is the whole point of my approach. It is by avoiding all assumptions that I arrive at my (unique) answer. I am still confused about why you think I am assuming that the host is looking for a blue stone to remove. I have not assumed that at all: I have only observed that a blue stone was removed.

    • Yat says:

      the “Does that help to clarify?” was condescending. I don’t need clarification about the answer or why you stick to your point of view. I just need you to stop ignoring my arguments. I answer to everything you say, and you keep wording the same refuted statements differently.

      The only thing I don’t understand is how you can keep saying that you don’t make assumptions whereas you obviously are. Please help me understand that by giving me an answer to my problem with two bags with 1000 marbles. I think it would be much easier to understand your point of view with it.

  16. Edgar 2 says:

    I initially had 1/5. I then read the comments and added 1/3 as an alternative; I was initially persuaded of the apparent ambiguity in the question. But on reflection I have to say that in my view all this stuff about choice strategies is a red herring and the only admissible answer is 1/5. Nothing in the question suggests that “I” have a strategy when “I remove” (note the use of the neutral word “remove” rather than the more active “select” or choose”) a blue stone and to deduce otherwise from the words used stretches the ordinary and natural meaning of the words beyond breaking point.

    In any event, because (as SJ has pointed out) there are other strategies, the question becomes meaningless if you admit the concept of a choosing strategy, and thus insoluble. If we are to keep our feet on the ground we have to assume that we are being set a question which is meaningful (ie admits of a single answer, or determinable set of answers) and not a set of answers each of which depends upon a different assumption none of which is capable of being determined from the original question. We must therefore assume that the “chooser” is not in fact exercising choice – because we aren’t told what strategy s/he uses to make the choice – and therefore that the removal of the first stone is random. And if it is then the answer is 1/5.

    • Yat says:

      You must have missed something. If the guy choses randomly, then the answer is 1/3. The 1/5 is only if the guy specifically wants to pick a blue stone if there is one.

    • Edgar 2 says:

      Well, not missed exactly – in fact I see now that in my post I simply got the two probabilities transposed. My position, FWIW, is that the answer is 1/3 because to get to 1/5 you have (without there being any suggestion in the wording of the puzzle) to take into account a choosing strategy about which we have no information. There are a range of such strategies which will give differing results and thus if such strategies are to be taken into account we have a meaningless question. So, such strategies are *not* to be taken into account, and thus the answer is 1/3.

  17. jh says:

    I got 1 in 5 myself, and did not find the wording ambiguous at all. I understood that you open your fist and removed a blue stone.

    I looked at the problem in this perspective: when you pick two stones from the box there are twelve equally likely possibilities:

    a) W-Y
    b) W-B1
    c) W-B2
    d) Y-W
    e) Y-B1
    f) Y-B2
    g) B1-W
    h) B1-Y
    i) B1-B2
    j) B2-W
    k) B2-Y
    l) B2-B1

    If you completely open your fist and remove one blue stone, you know that there was at least one blue stone, so there are 2 possible combinations that can be dismissed (a and d). The chance of the other stone being blue is 2 in 10, or 1 in 5.

    If you assume you are peeking into the fist and see only one stone, being it blue and removing it, this information tells you that you must dismiss all the possibilities from a to f, and you are left with 6 possible combinations. Of these combinations, two (i and l) have the second stone also blue, therefore the possibility would be 2 in 6, or 1 in 3.

    • thequiet1 says:

      Your second scenario assumes the picker knows which stone he chose first. Nothing in the question suggests this to be the case. I believe the question is worded such that the stones can be considered to be picked up simultaneously, and I believe Richards answer only including 5 possibilities instead of 10 supports that.

    • Hannes Naude says:

      “If you completely open your fist and remove one blue stone, you know that there was at least one blue stone, so there are 2 possible combinations that can be dismissed (a and d). The chance of the other stone being blue is 2 in 10, or 1 in 5.”

      Careful here. Yes, 2 combinations can be dismissed, but the remaining 10 are no longer equally likely. The 8 combinations that include only a single blue stone should be weighted only half as heavily as the 2 that contain 2 stones.

      The answer is therefore 4 in 12 = 1 in 3.

      But this is a hard way of gettting to the answer. If the choice is random (which appears to be the only reasonable thing to assume) then the answer will be the same whether the second stone is picked up at the same time as the first or only after the first has been revealed.

      Clearly if the second stone is only selected after the first has been revealed as blue, then we are simply selecting one option at random from b y w. and the answer is still 1 in 3.

    • jh says:

      Sure, I think 1 in 5 is the correct answer.

      But if you make exercise to read “look inside my fist and remove a blue stone” to mean “peek inside my fist and see that the stone I looked at was blue, and then remove it”, of course that means that the picker has to decide on which stone he is going to look at before doing it.

      I worked out all the possibilities taking into consideration the order of the stones to make sure all the possibilities were equiprobable. It turns out the result is the same anyway. 2 in 10, 1 in 5, same thing.

    • jh says:

      I’m sorry, Hannes, I have tried but I cannot see why the 10 left possibilities are not equiprobable. They were before looking at the selection.

      Imagine we set up an experiment with the four stones, the box, a pencil and a sheet of paper:

      1. Take two stones from the box.
      2. Look at the stones. If none of the stones is blue, return them to the box and go back to 1.
      3. If there is only one blue stone write down “1”, if there are two, write “2”. Return the stones to the box and go back to step 1.

      After thousands of repetitions, you will have a list with ones and twos. 20% of the numbers will be “2”. One fifth.

      But if the experiment is set up like this:

      1. Take one stone from the box.
      2. If it is not blue, return it to the box and go back to step 1.
      3. Take a second stone from the box. If it is not blue, write down “1”, it is blue, write down “2”. Return the stones to the box and go back to step 1.

      It is clear that it is not the same. In this case, you would end with 33,3% twos.

    • Hannes Naude says:

      Hi jh

      I agree 100% with your calculation. I interpreted your sentence

      “If you completely open your fist and remove one blue stone”

      as

      “If you completely open your fist and remove one stone at random and it happens to be blue”

      while you apparently meant

      “If you completely open your fist looking for a blue stone and remove one”.

      Under my interpretation the 10 possible cases are not equiprobable, while under the second they are as you point out with your two experiments.

      My point in the larger context is that there is nothing in the original problem that makes
      “If you completely open your fist looking for a blue stone and remove one”
      a more likely interpretation than
      “If you completely open your fist looking for a white stone and remove a blue one”
      or
      “If you completely open your fist looking for a yellow stone and remove a blue one”

      Therefore the only reasonable interpretation of the original problem is
      “If you completely open your fist and remove one stone at random and it happens to be blue”
      BUT under these conditions the given answer is wrong. The correct answer is 1/3.

  18. jh says:

    Ups, my last post was meant to be an answer to thequiet1’s post.

  19. Yat says:

    Just an other problem for everybody who thinks that the fact that the guy choses the blue marble randomly or not is not important.

    I have two bags with marbles. The first bag contains one blue marble and 999 red ones. The second bag contains one red marble and 999 blue ones. They both look exactly the same, you don’t know which is which. I pick a marble from a bag, it is blue. What is the probability that the bag I picked the marble from was the first one?

    It should be obvious for everybody that if I specifically wanted to pick a blue marble, then I can have found it in either bag, so you have no information, the bag is either the first one or the second one, with 50% probability.

    It should be obvious too that if I have picked the marble randomly in the bag, then picking a blue marble from the second bag is not so surprising, whereas picking it from the first one is much less probable. Therefore, knowing that I randomly picked a blue marble from this bag indicates that it is much more probably the second bag.

    What happens in the original problem is exactly the same, but at a different scale. There are 5 “bags” (ie possible handfuls) with two stones, one contains 2 blue stones, the others contain only one blue. If I pick a blue one, you need to know if I choses randomly or not to define the probability that I picked from the “bag” with two blue marbles.

    • AMWhy says:

      Too many people assuming unnecessarily.
      Simplify the question. What is the probability that two picked stones are both blue? 1 in 6. You are now told that one stone is blue. What is the probability thatboth stones you picked are blue? 1/5 as the red/ White option has just been elliminated.

      Worded again. You take two stones. Chance of B1 and B2 is 1 in 6.
      You then find out (drop it, someone tells you, whatever) that 1 of the stones you picked is blue. That means of the 6 possible choices you could have picked, red and White isn’t one of them. This leaves you with B1 R, B1 W, B2 R, B2 W and B1 B2. Five choices of which one has both blues. Hence 1 in 5 chance.

    • Yat says:

      How is that a reply to my comment ???

      Your first wording adds an assumption to the original problem. You assume that the fact that you are being told that one stone is blue is an answer to the question “is there a blue stone in your hand”, whereas it could also be “give me the color of one of the stones in your hand”. It’s the same difference as in my problem, between “pick a random marble from the bag” and “pull a blue marble from the bag if there is at least one”. Probabilities are not the same.

      Your second wording actually makes the opposing assumption, you just solved it wrong. I am specifically refering to the “drop it” part in the parenthesis. If you drop one stone, it is chosen randomly between the two stones you have in your hand, and it is a blue stone. If you have only one blue stone in your hands, the one you dropped could also have been the other one. So there are situations where you have a blue stone in your hand, but the experience does not match the problem statement.

      Nothing better than the facts, here. Just try it yourself, pick two “stones” (or pieces of paper or whatever), then randomly remove one of these two stones. If it is blue, then you are in the same situation as the problem, so count it as a valid experience. Then make your stats on these valid experiences. I can guarantee that on the long run you will see that the second stone is blue with probability 1/3. Just try. Seriously. You will obviously be surprised.

    • Mike Torr says:

      Yat,

      I agree that it is possible, given sight of the contents of the bags, to choose a particular colour.

      I even agree that there is an equivalence between this and the problem of the four stones.

      I disagree that we need to know how the choice was made to calculate the probability.

      I think your definition of probability must be different from mine. Probability has to include a point of view, and this puzzle is assuming the point of view of an external observer: the knowledge in the head of the person selecting the stones is not relevant. Note: I didn’t say it had no meaning, I said it wasn’t relevant.

      When calculating probabilities, one has to ignore unknown factors – doing so is the very essence of probability theory. I posted about this somewhere else on this page, when I spoke of breaking down every possible event in the universe until you had a huge number of 0 and 1 probabilities. The information about how the stone was chosen is part of the “space of unknowns”, and by definition cannot be part of the calculation.

    • Yat says:

      There is still the same hole in your reasonning : you tell that you ignore unknown factors, and yet you make a huge assumption on one of these (you apparently think you don’t, but the method you use do). Seriously, don’t you want to have a calm, linear discussion by mail ? Am I allowed to post an email address here ?

    • Mike Torr says:

      I would advise against posting your email address on any web page, ever. Not a good idea!

      I would love to have a discussion about this offline and I’m sure I would enjoy it even more in person, but I’m sorry to say I don’t have the time it would probably take. I actually think we would never agree, in the same way that Creationists can never agree with people supporting evolution. I think we are starting from different assumptions about what probability actually is.

      Sadly, I’m really going to have to stop soon and go back to work. It’s way past lunch time :(

    • Yat says:

      The point of using email is that we don’t need to have this discussion right now. We can take a day to answer each message, it will be much more efficient.

      So here is another thing on which we disagree : if extracted from this comment page context, we take our time to examine eachother’s argument, I think we will probably come to an agreement. From my experience, this type of discussion very rarely fails, whereas the stats are the other way around when the discussion is in public. I assume that you are intelligent, so I don’t see any reason why we would never agree.

      Don’t worry about my adress, this is a spam address anyway, I was just wondering if it was allowed or not. So here it is : gyro.gearloose(at)free.fr

      Please feel free to start the conversation. If you want to do it but not today, just tell me here that you are ok with this.

    • Carl says:

      As an observer to this fascinating discussion, I think you should keep it here out in the open. I’ve been following most of the comments and I’m thoroughly enjoying thinking about it.

      I must admit that my first response to the problem was that it depends on what the selector intended/did, but after reading this discussion I’ve almost completely come around to Mike Torr’s (and the official) explanation.

      I’m not sure that I can add anything by way of explanation that hasn’t already been said, but if I think of any novel insights I’ll be sure to post them here.

  20. Rob j says:

    I originally had 1/3 but then changed it to 1/5.
    One thing that seems to be confusing people is that the puzzle doesn’t start until a blue stone is drawn. Any probability involved in getting us to that point is irrelevant. The conditions at the point of being asked the probability are that one blue stone has come out of the fist.
    There are 5 different scenarios which allow this condition, and only 1 which leads to a 2nd blue stone. 1/5

    • I think we’re all trying to be too clever here. (Me included). I was in the 1/5 school but not any more. If you read back through what you’re written you state “Any probability involved in getting us to that point is irrelevant.” I agree. However – the “5 different scenarios which allow this condition” must also be irrelevant! (This is what we all saw and is why we thought the answer must be 1 in 5) – BUT as you’ve just pointed out the 1/5 IS the probability that got us to that point. From this point onwards (ie the point at which one blue stone has been revealed) and the point at which the question is asked; there are only 3 stones left. One of those three is Blue. What is the probability of that blue stone being in the fist? It has to be 1/3

      I think (and I stand ready to be shot down here) that the 1/5 answer is only true if the question is “What is the probability of drawing two blue stones if you get another free go when you draw no blue stones?”

  21. Bletherskite says:

    I like many others got 1/3 but having read the question again and the comments I’m now understanding it like this:-

    If I had picked one stone out of the box, and it was blue, then had to pick another stone I would have a 1/3 chance of picking a 2nd blue stone.

    As I have picked up two stones to begin with, the chance of them being both blue (after checking that one of them is blue) is 1/5.

    Therefore with the question phrasing 1/5 is the correct answer.

    • Yat says:

      You made an assumption here : “after checking that one of them is blue”. Nothing in the problem statement tells that the guy checks that one of the stones is blue to specifically remove this one if there is one. What is said is that the guy checks the colors of the stones, then removes a blue one. Nothing more.

      We have no reason to assume that he actually was looking for a blue stone. Without knowing more, it would be safer to assume that he randomly removed one of the stones in his hand.

  22. Mike Torr says:

    OK, I’m going to have one more try at clarifying this for those of you who still think 1/5 is wrong.

    When we calculate the probability of an event, we can only assume what is presented to us as evidence. It is not valid to start speculating about other elements in the process, such as the internal thoughts and motivations of a human agent in the scenario. Think about it for a moment: assuming the universe is deterministic (OK, that’s perhaps debatable, but bear with me) – in theory EVERY event is predetermined, and therefore its probability is either 0 or 1. It only makes sense to assign a probability >0 and <1 to an event if we are taking into account the fact that we can't know everything that happened "behind the scenes". As our knowledge of the situation changes, the value will change accordingly.

    In this puzzle, part of the "behind the scenes" stuff, assuming the observer cannot read minds, is the thought process that led to the selector choosing a blue stone. Therefore we must discount this, and the answer is based only on the facts known to the observer; i.e. that a blue stone was removed. The decisions leading to this blue stone being removed are part of the "unknown" process, and therefore must be ignored when calculating the probability.

    If you still disagree, I'm afraid I have nothing else to say: we'll just have to differ on it :)

    • Yat says:

      No, we can’t differ without one of us being wrong. If you actually try to reproduce the experience, and actually count the number of times the other stone is blue and the other stone is not, you will be able to approximate an actual probability. And you will be surprised to see that it converges to 1/3 when you consider the removed stone was randomly chosed, to 1/5 when you consider the guy was looking for a blue stone in the first place.

      It is not a question of opinion, it is measurable fact.

    • Mike Torr says:

      How can one reproduce this experience and get a quantifiable result, when part of the unknown information space involves a person’s thought processes? See my other post on this.

      You’re right: we can’t differ without one of us being wrong. Nevertheless, we differ.

    • Yat says:

      This is exactly the point. In the problem statement, we don’t know if the guy removes a blue stone every time there is one, or just randomly removes one of the two stones.

      But both situations can be reproduced, and give different results, so indeed, one of us is wrong : you are. And I bet you did not even consider actually trying the experiment for yourself.

    • Mike Torr says:

      I give up, Yat – you’re totally missing the point, and you ARE wrong.

      It is NOT possible to “try this yourself”, because the problem as stated doesn’t give you all the information you’d need to set it up.

      This lack of information does NOT, as you seem to think, prove you correct. It just means that one has to calculate the probability without using the missing information – that’s what probability MEANS.

      I can’t keep up this debate much longer because I’m at work and I’ll lose my job, but I’m afraid you are wrong about this.

    • Yat says:

      Yes, the lack of information is the reason why you can’t give a definite answer. Saying I am wrong won’t make you less wrong. the problem does not allow you to conclude that the probability is 1/5 any more than it allows you to conclude that the probability is 1/3.

      I hope it is confortable for you to stay wrong as long as you dont know it.

    • Mike Torr says:

      OK, there’s clearly no point arguing any more, is there? Let’s drop it.

    • Yat says:

      Your choice. Being wrong, there will clearly be a time when you are out of arguments. Has this point been reached yet ?

    • Mike Torr says:

      No.

    • Yat says:

      Cool. Then give me something new. Or just try to answer my arguments. Or the bags with 1000 marbles things… anyway, but just stop repeating the same things over and over whereas I have answered to every wording of it.

  23. tommathias says:

    1/6.

    “I bring my hand out in a fist, look inside my fist and remove a blue stone. What are the chances of the other stone in my hand also being blue?”

    That isn’t different from saying: “What are the chances that I pick 2 blue stones?”

    12 possible picks:
    a) W-Y
    b) W-B1
    c) W-B2
    d) Y-W
    e) Y-B1
    f) Y-B2
    g) B1-W
    h) B1-Y
    i) B1-B2
    j) B2-W
    k) B2-Y
    l) B2-B1

    Options i and l are those that have 2 blue stones. 2/12 = 1/6.

    What have I missed?

    • Mike Torr says:

      @tommathias – you’ve missed the fact that the information available to the observer changes when the blue stone is revealed. At that moment, the set of possibilities reduces by two because W-Y and Y-W are eliminated.

  24. tommathias says:

    @Mike Torr: Eureka, so the question is about the second of 2 distinct events: first stone observed and second stone observed, not “What are the chances that I pick 2 blue stones?”

    So, with event 1 being observed as a blue, the only options available for event 2 of the original 12 are:

    b) W-B1
    c) W-B2
    e) Y-B1
    f) Y-B2
    g) B1-W
    h) B1-Y
    i) B1-B2
    j) B2-W
    k) B2-Y
    l) B2-B1

    Of those, options i and l are those that have 2 blue stones. 2/10 = 1/5.

    Correct?

    • Mike Torr says:

      That’s right – although many here are going to disagree :)

      It’s literally true that the probability changes at the instant the blue stone is shown to you – the same way that the probability of winning a game of poker would change if you knew exactly the order of the cards in the pack (or even just the next 20 of them…) Probability is information-dependent.

    • Yat says:

      If we consider the removed stone was picked randomly (which we can’t deny based on the problem statement only), then you forget one thing : those 10 events are incomplete, because once the two stones are picked, there is a second random draw between the two stones.

      For exemple, in the b case, the guy could either pull the W stone or the B1 stone. So this makes 2 options :
      b1) W-B1 => W is removed
      b2) W-B1 => B1 is removed.
      You can duplicate every one of your 10 options this way, and now you wil see that there are additionnal options to eliminate. For example, here, b1 is not a valid option.

      You will be left with 12 valid options, and in 4 of these 12 options, the remaining stone is blue.

    • Mike Torr says:

      Yat – what makes you think that the phrasing of the question implies the blue stone was picked randomly? All it says is that the selector pulls out a blue stone. It says nothing about how or why. And THAT is why we can’t take this information into account.

    • Yat says:

      Mike Torr, I agree with you 100% on this, and that is exactly why it is not possible to chose between the 1/3 and the 1/5 solution without making an assumption. You are the one with the affirmation that the answer can only be 1/5. I am the one saying that the problem is incomplete and that we need additionnal information to know the answer.

    • Mike Torr says:

      I am not affirming that there is only one answer to this – PROBABILITY THEORY is affirming it!

      You think there are two answers, but that is because you are peeking inside the space of unknown information and worrying about what it contains. Unknown information is not used in the calculation of probabilities – that’s what probability MEANS.

    • Yat says:

      I’m sorry, is there any point in this ?

      You are making assumption about unnown information, I just accept that there is unknown information.

      You litterally say “we don’t know why the host removes a blue stone, so he _was_ looking for a blue stone, and would have done so in any situation with at least one blue stone”. You do. And it is a pretty big assumption about unknown information.

      You can’t know that the host was looking for a blue stone, so you can’t give 1/5 as an answer.

    • Mike Torr says:

      “You litterally say “we don’t know why the host removes a blue stone, so he _was_ looking for a blue stone, and would have done so in any situation with at least one blue stone”. You do.”

      Er… no, I don’t. I say nothing about what the host’s motivations were. I cannot say anything about them because they are unknown in the puzzle as stated.

    • Yat says:

      So why do you make your calculations based on the fact that this affirmation is true ?

      Hey, we have discussions spread all over the comments page here, it is painful to handle and it pollutes the said page. What about an exchange of addresses, so we can try and take our time to continue this discussion, with more detailed explanations ?

  25. song sparrow says:

    I assumed that the box is opaque and the rocks are drawn randomly, although that was not stated in the problem.

    Originally, I got Richard’s answer with the same reasoning. I thought that the probability that mattered was the chance of randomly drawing two blue rocks from the container, which is 1/5.

    Upon further reflection, I see what others are suggesting that the actions of the chooser matter when he reveals that the first rock drawn is blue.

    If the chooser randomly draws two stones and randomly shows that the first stone is blue, there is a 1/2 chance of that happening. (There are twelve possible color pairs (WY, WB1, WB2,YW, YB1, YB2, B1W, B1Y, B1B2, B2W, B2Y, B2B1) , and six of them have blue as the 1st color.) After that, there is a 1/3 chance that the second stone will also be blue. Thus the overall chance of this happening is 1/6.

    If the chooser randomly draws two stones and makes a conscious decision to reveal a blue stone if he has one, then we can also consider the original color pairs where the blue stone is the second stone. That raises the original odds of drawing a pair with a blue stone to 10 out of 12 or 5/6 (only WY or YW do not include a blue stone). Out of those 10, 2 can be two blue stones, or 1/5. Thus the overall chance of this happening is still 1/6.

    • song sparrow says:

      I amend my answer to say that once we know that one rock is blue, the 1/2 or 5/6 chance from the original draw is now irrelevant because we’ve excluded the possibilities where WY is drawn.

      So while the overall chance of 2 blue stones is 1/6, once we know that one is blue, the odds of the other being blue is 1/3 or 1/5.

      Thus, now I agree with the other commenters. I had to write it out to understand it.

    • song sparrow says:

      So in sum,

      The chances of randomly drawing two blue stones from the 4 is 1/5.

      The chances of randomly drawing two stones, being able to reveal the first as blue, and having the second also be blue is 1/6.

      But after the two stones are randomly drawn, once the first stone is revealed to be blue, the chances of the second stone also being blue is 1/3 or 1/5, depending on whether the chooser reveals the first stone as blue randomly or on purpose.

    • song sparrow says:

      Wish we could edit! Correction to first option in my last post: The chances of randomly drawing two blue stones from the four is 1/6 not 1/5. It changes to 1/5 or 1/3 once we know that one stone is blue, and the final probability depends on whether the chooser displayed the first blue tone randomly (1/3) or on purpose (1/5).

  26. Kristian says:

    I didn’t eliminate the option with no blue, so I got 1 in 6.

  27. Confmat says:

    First answer I got was 1/5, using Bayes method.

    Then I read the comments and realised you could also get 1/3, if you evenly choose between the stones. Or 100% if you try to avoid choosing stone unless there is no choice.

    I came up with a general solution.
    If a is the probability of choosing the blue stone instead of the non-blue stone then the probability, p, that the other stone is also blue is:
    1 / (1 + 4a)

    Some examples:
    Always choose blue: a = 100% so p = 1 / (1 + 4(1)) = 1/5
    Avoid blue: a = 0% so p = 1 / (1 + 4(0)) = 1
    Choose evenly: a = 50% so p = 1/ (1 + 4(0.5)) = 1/3
    Because I’m a geek: a = 39% so p = 1 / (1 + 4(.39)) = ~39%

    • Yat says:

      I wish this simple, clear and complete explanation was enough for everybody to accept the fact that the way the removed stone is chosen actually is relevant ;-)

    • Confmat says:

      Thanks for the praise Yat!

      And if anyone makes runs the experiment then we can reverse my formula to figure out how they chose their stones :)

  28. Anonymous says:

    Love watching the fallout to this weekends puzzle. The observant will notice that some of the people who refuse to accept the official (correct) answer this week, did exactly the same thing last week. (he says – stirring the pot a little more….)

  29. Mike Torr says:

    This is beginning to feel WAY too much like the “0.9999999…. = 1″ flame wars…

  30. MaciejK says:

    The puzzle clearly states that we picked up 2 stones and have them both in our hand.
    It also clearly asks about the probability of the second stone in our fist being blue (NOT about the probability of drawing 2 blue stones) which makes all the speculation about the protocol meaningless.

    The puzzle is perfectly unambiguous and i seriously hope that those who disagree are not mathematicians.

    • Mike Torr says:

      So do I MaciejK. By the way, I’m not a mathematician by trade but I did graduate in maths.

      Here’s some light relief for anyone feeling stressed by this week’s “fallout” (click my name for the link) :)

  31. Eddie (@edzeteito) says:

    The one thing we can all agree on is that Richard never comes back to clarify or adjudicate. It feels a little as if he’s playing God to see what the humans make of his ‘creation’ (the puzzle). And in this respect I think the “1in5ers” are the theists and the “1in3ers” the atheists.

    • Camel Ali says:

      Eddie,

      Richard has already clarified and adjudicated. His word is at the top of the page, and lo it is 1/5 !

      I’m a believer. I was a little “agnostic” on Friday as I felt that there was some ambiguity in the question. But after reading some of the comments on this page, I’m a convert.

      For me Mike Torr’s comment “Unknown information is not used in the calculation of probabilities”, helped me to see the light.

    • Mike Torr says:

      Eddie I think you’re right about Richard – this whole Friday Puzzle thing is clearly a psychology experiment, and has been from the start. I expect Richard will write a paper on it eventually :)

  32. skeeto says:

    I encourage anyone who thinks Richard’s answer is wrong to write a simulation of the game (10-20 lines of code). It’s one of the best ways to understand these kinds of problems.

    • Yat says:

      Been there, done that. If you did and found 1/5, I bet you just assumed that the host would remove the blue stone every time there was one !

    • Mike Torr says:

      skeeto, I don’t think that’s possible with this puzzle because one of the contributing factors is a human mind, which is not possible to simulate… in fact, I think that may well be the ultimate source of all the disagreement here!

    • Camel Ali says:

      @ Yat

      You don’t need to assume that the host would remove the blue stone every time there was one. You just need to know that there was a blue stone, and he removed it.

    • Yat says:

      @ Camel Ali
      Ok, so in your simulation, you pick two stones randomly, then you randomly remove one of these two stones. If it is blue then the situation matches the problem, else we reject the experience. Are you ok with that ?

  33. This isn’t a puzzle, its a maths question. booo hissss

  34. bev says:

    Surely this is the wrong answer. You’ve established that the 1st stone pulled out of hsnd is blue. That leaves only a blue, yellow and white stone left. Therefore a 1:3 chance of having 2 blue stones.

    • MaciejK says:

      That would be true if he picked up one stone, revealed it and then went on to pick the second one.
      The puzzle clearly states that he picked up 2 stones and has them both in his fist.

  35. Camel Ali says:

    @ Yat

    No. Whether the stones were picked randomly or deliberately is irrelevant in this puzzle. We are told that a blue stone was picked from the hand. This implies that the hand did not hold both a yellow and white stone. It implies that the hand held a blue stone and one other (5 possibilites). Of these possibilities, only one has two blue stones in the hand.

    • Yat says:

      Did you actually read anything that was said here ? I already answered this misunderstanding several times, in several different ways. I even gave my address hoping for the possibility of a calm, linear discussion… feel free to use it if you still believe 1/5 is the good answer after reading the comments.

    • Camel Ali says:

      @ Yat. I have read everything that was said here. I do not misunderstand your arguements, I simply disagree with them.

    • Yat says:

      Yes, that is a good motive to not answering them. Thank you for your very constructive contribution.

  36. Mike Torr says:

    I have to go now, everyone. It’s been fun but I have too much to do. I won’t say anything else because I’ll only start something again ;)

  37. MaciejK says:

    Let me put you out of your misery Yat.
    The point you are raising would be relevant if we were looking from the perspective before picking up any stones and asking what are the chances that the guy will show us two blue stones.
    In this puzzle we are looking from the perspective where we know that one of the stones in his fist is blue and are asking what are the chances that the other one is also blue.

    • Yat says:

      My misery, yes… thank you…

      We not only know that one of the stones in his fist is blue, we also know that, for some reason, this is the stone he decided to show us / remove. That’s why it is a different question. I don’t remember asking the guy “hey, is there a blue stone in your hand ?”. He just happened to tell me there was a blue stone, and to remove it. For all I know, he could have removed the other as well.

      Your argument has already been responded to several times by several persons (including me). So are you just looking to see how long this can run in circles, or are you actually looking to try to resolve this issue ?

    • MaciejK says:

      “for some reason, this is the stone he decided to show us / remove.”
      He didn’t decide, it is simply the way this puzzle is set up. When you include the possibility where he shows us white or yellow stones first you are no longer talking about this puzzle.

      “I don’t remember asking the guy “hey, is there a blue stone in your hand ?””
      You don’t have to ask him anything.

      “For all I know, he could have removed the other as well.”
      No he couldn’t, that is not what this puzzle is about.

    • Yat says:

      thank you for these unargumented affirmations. Hey, my turn :

      You are wrong.

      Cool, huh ?

    • MaciejK says:

      My argument is that your logic doesn’t apply to this puzzle.

    • Yat says:

      Well, if this is an argument, then here’s mine :

      Yes, it does.

      Discussions are sure a lot simpler when we don’t need to justify our affirmations ! So, wanna discuss this, or just keep saying “I’m right your wrong I’m right you’re wrong…” ? I

  38. Charles says:

    W Y B1 B2 x x
    Y W B1 B2 x x
    Y B1 W B2 x
    Y B1 B2 W x
    W B1 Y B2 x
    B1 W Y B2 x
    B1 Y W B2 x
    B1 Y B2 W x
    W Y B2 B1 x x
    Y W B2 B1 x x
    Y B2 W B1 x
    Y B2 B1 W x
    W B2 B1 Y x
    B2 W B1 Y x
    B2 B1 W Y
    B2 B1 Y W
    W B2 Y B1 x
    B2 W Y B1 x
    B2 Y W B1 x
    B2 Y B1 W x
    W B1 B2 Y x
    B1 W B2 Y x
    B1 B2 W Y
    B1 B2 Y W Odds:
    20 4 0.2

    As long as the formatting keeps, here is a complete summary of the conditions. A set of 4 variables has 24 possible orders. If you imagine the first two columns as “in the box” and the second two as “in the hand”. Of the second pair of columns, those with at least one blue have an X in the following column, totaling 20. Those with two blue total 4. 4/20 is the same at 1/5.

    I think the question is unambiguous, two pieces are pulled, and the observer takes a blue one. This means it qualifies for the first row of X’s, and of those X’s how many are in the final column of X’s. Nowhere does it add any other limiters.

    • Charles says:

      Bummer. What do you know, the formatting did not keep.

    • Yat says:

      Still the same problem here : “at least”. Why do you assume that in every combination where the host has at least one blue stone in his hand, he will remove this one ?

      One of the other possibility here is to double every option, with the possibility that the first stone is removed, and with the possibility that the second stone is removed. Then you can eliminate the combinations where the removed stone is not blue and get 1/3.

    • Rob J says:

      This is a great show of all the possibilities and it shows how the two camps (1/3 or 1/5) get their answers.

      Let’s make an assumption that the person randomly chooses a stone from the fist.
      If we take the third column as “the one that the person randomly chooses” then there are 12 where a blue is the right answer. The others are invalid to the puzzle parameters.

      Then of those 12, 4 have a blue in the 4th column (the hidden one). This shows how, if drawing the blue stone from the fist was random then the probability of the second stone being blue is 1/3.

      However if the choice wasn’t random, and the person could choose either, there are, as you show, 20 choices that have at least one blue. There are still only 4 choices where both are blue, hence 1/5.

      This list illustrates both arguments quite well.

      However, I think the wording of the puzzle strongly suggests it is not random “I bring my hand out in a fist, look inside my fist and remove a blue stone. “. The person was acting on information. He could see inside his fist when pulling out the stone.

      My question is, though, to an observer, who doesn’t know if the choice was random or otherwise, what is the probability?

  39. Tomas Blomberg says:

    Hannes Naude got it right by showing how ill phrased the problem is. We need to know what strategy the person has if the two balls are different. Will he always name blue if there is one? If not, the five cases does not have the same probability.

    Read all Naude’s posts to understand that the puzzle needs to be phrased differently to have an answer.

    A better phrasing would be “I get asked ‘Do you have a blue ball?’ to which I answer Yes after peeking at the balls in my hand.”

    • Yat says:

      Yes, that is exactly it. Will you have a better luck explaining that to these guys ? ;-)

    • MaciejK says:

      We don’t need to know his strategy because the question isn’t about the chances of the guy showing us two blue stones given his intentions/strategy.

      It is about the probability that he randomly drew two blue balls once we know that at least on is blue.

    • Yat says:

      “It is about the probability that he randomly drew two blue balls once we know that at least on is blue.”

      No, once we know that he removed a blue stone. This is not the same information. To solve the problem correctly we need to know the protocol which lead to the removal of this blue stone.
      See the difference between those two events :
      “I randomly pick two stones, there is at least one blue stone”
      “I randomly pick two stones, there is at least one blue stone, I remove the blue stone”
      If you don’t see the difference, then here it is : the first is the union of the second and the following one :
      “I randomly pick two stones, there is at least one blue stone, I remove the other”. To pick the solution 1/5, you need to assume that the probability of this third event was zero. This is making an assumption about the protocol. You can’t avoid it.

    • MaciejK says:

      The puzzle states:
      “I bring my hand out in a fist, look inside my fist and remove a blue stone.”
      I think we can safely assume that the probability of:
      “I randomly pick two stones, there is at least one blue stone, I remove the other” is zero.

    • Tomas Blomberg says:

      MaciejK, are you saying that you will _always_ know that there is a blue ball present the times there is a blue ball in the hand?

      If so, the strategy of the person is to always name a blue ball when one is present.

      If his strategy is to name either of them at random when only one of them is blue, you will not always know that there is a blue ball present, even if there is one.

      The probability of both balls being blue is different in those two cases. I even made a computer simulation to make sure. Try it: Make sure the person only names blue in around 50% of the cases where only one of the balls is blue. Count the number of cases where he has said “blue” and the number of cases where he actually held two blue balls. It should be clear that this is not the same as if his strategy is to always name “blue” when a blue is present.

      If you don’t want to do a simulation, simply write all the possible cases twice. Remove half of the cases where only one ball is blue, since he would not have said “blue” in those cases. Remove all the cases where no blue ball is present. Count!

    • Yat says:

      Yes, flushing all knowledge of provbabilities is a method too. So, if no probabilities are conditional and you need only to base your reasonning on what you see without even trying to understand anything, there the following reasonnign gives an interresting answer too :

      One blue stone was removed, there are three stones left, one of them is blue. => 1/3.

      Of course, this reasonning implicitly assumes that the blue stone is randomly chosed, the exact same way yours assumes that the host always will remove the blue stone if there is one.

      Oversimplifying sometimes just gives you bad answers…

    • Yat says:

      (obviously, my last comment was answering MaciejK…)

  40. Camel Ali says:

    @ Yat

    Following on from your previous comments. I agree that if the first ball was picked at random from the hand instead of deliberately by the host, then the probability of the second ball being blue would be 1 in 3. However, this is there is no mention in the question that the ball would be picked randomly, so why would we assume it so?

    We can assume nothing more than we are told. Two balls are in the hand and one of them is blue. We can imply from this that the balls are not White and Yellow, but one is blue and the other is either blue, white or yellow. In this puzzle this is all the information given, and this is all we use to calculate the probability. If the question was worded differently, of course it could allow for a different answer.

    • Yat says:

      Ok, I am glad to read the first paragraph. This narrows down the disagreement.

      Why would we assume that the balls would be picked randomly ? Well… maybe because it seems to be the smallest assumption : if we don’t know how he chosed, then it seems more faire for me to assume that this choice is random than assuming he was looking specifically for a blue stone. But that’s not my point. Yes, I see more reasons to assume that the choice was random, but it is still an assumption. My point is that the question is incomplete because we need to make an assumption about this choice. Assuming that the choice is random is an assumption, assuming the host is looging for blue stones is one too, and assuming any other protocol is an assumption too. We can’t find an answer without adding assumptions to the problem statement.

      When you try to remove these assumptions and just ignore the question, you don’t avoid any assumption, you just make one implicitly. This is conditional probability, when you solve the problem the way you do, you attribute a probability of 100% to the event “knowing that there is only one blue stone, the blue stone is removed”. Even if you think you don’t, that’s what your method imply. Just roll your result the other way around using bayesian probability, you will land on 100% for this. The assumption is implicit, but you see, it is there.

    • MaciejK says:

      We assume 100% probability to the event “knowing that there is only one blue stone, the blue stone is removed” because the puzzle says: “I bring my hand out in a fist, look inside my fist and remove a blue stone.”
      We don’t need to question that …. it is simply a piece of information offered to us and it is the way the puzzle is set up.
      We would be worrying about his intentions if the questions was:
      “What are the chances that he will show you two blue stones?”

    • Yat says:

      Yes, we need to question that, because we are working with conditional probabilities, and we need to know the probability that he choses to remove the blue stone knowing that there is exactly one in order to calculate the probability that the other stone is blue knowing that he removed a blue stone.

      It would help you to try and use real probabilities instead of just hand-waving a solution where you eliminate bad options. This type of method can work, of course, but you must be very careful with what you are doing. Just look at Hannes Naude’s analysis (just below, I think).

    • Camel Ali says:

      @ Yat

      Hannes Naude’s analysis below, is a solution based upon eliminating bad options. It also assumes that the blue ball is randomly removed from the hand. It is misleading to assume anything that is not implicitly stated in the question.

    • Yat says:

      And yet that is exactly what you are doing when you use the other method : you assume that the blue ball is always the one removed when there is one, which is misleading. Same thing. Exactly. Just a different assumption. So no solution is more right than the other. That’s my whole point.

  41. Hannes Naude says:

    OK. I’ll try to show why the 1 in 5 solution is wrong using Richard’s own method of enumerating the possibilities and eliminating all of those that do not accord with the facts as given. To make formatting a little easier I refer to the stones as A,B,C and D. Lets say A and B are the blue ones. The first two columns of the makeshift table below are the stones in my hand. the third column is the stone that I withdrew. In the fourth column I mark with an X any possibilities that can be eliminated (because the drawn stone was not blue)

    A B A *
    A C A
    A D A
    B A B *
    B C B
    B D B
    C A C X
    C B C X
    C D C X
    D A D X
    D B D X
    D C D X
    A B B *
    A C C X
    A D D X
    B A A *
    B C C X
    B D D X
    C A A
    C B B
    C D D X
    D A A
    D B B
    D C C X

    Twelve possibilities remain, of which 4 correspond to hands with 2 blue stones.
    4/12 = 1/3.

    Clumsy way to get to the answer, but maybe convincing to those unwilling to consider alternative reasoning. The core difference with Richard’s reasoning is that unless there is reason to believe that the chooser actively seeks out blue stones, we must count the hands with 2 blue stones twice in our calculations, because there are twice as many ways in which they could satisfy the facts as given.

    • MaciejK says:

      If the question was about the probability that “the chooser” reveals two blue stones in a row then his intentions would be relevant.

      Here we only ask what are the chances that he randomly drew 2 blue stones. We also have one additional information which is that one of the stones in his fist is blue and that is the point where we start.

    • Yat says:

      @ Maciejk

      We also have another information : the blue stone is the one he showed you and removed.

      If the only information we have is that there is one blue stone in his hand, then the problem won’t be the same : for a start, you would have at least a probability of 50% that the remaining one is blue, simply because we don’t know that this is the one which have been removed. See how that kind of thing can be important ?

    • Tomas Blomberg says:

      An absolutely perfect explanation, Yat.

    • Yat says:

      @Tomas
      Well, thank you, but obviously it is not :6/

    • MaciejK says:

      Sorry Yat but I have failed to understand your last post.

    • Tomas Blomberg says:

      So how about some better phrasing of the problem? I suggested one earlier where someone asks “Is (at least) one ball blue?” to which the answer is “Yes!”.

      Another phrasing would be that the guy is allergic to blue and nothing else. He picks two balls and you see him break out a rash.

    • Yat says:

      What I was trying to say is that if the only informations you had were the ones you mentioned, then you wouldn’t know that the blue stone is the one that has been removed. This is an additionnal information that you can’t ignore, and that is not part of your set.

      If you only know that one of the stones is blue, then obviously the one that has been removed can be the other. Therefore, you have a 50% chances that the blue stone that you know was in the hand of the host is still there.

  42. MaciejK says:

    Listen Yat ….. I’m just a layman. But the most mathematical sounding explanation I can give is that you needlessly include certain outcomes in the sample space.

    • Yat says:

      Think so ? So what about my other solution ? You remove one blue stone, three stones are remaining, one of them is blue. Two are still in the bag, one is in my hand. The probability that the one is my hand is the blue one is obviously 1/3.

      Oh wait… you want to needlessly include something else ?

    • MaciejK says:

      Well….. you did it again .. I have no idea what you are talking about.
      Does anyone know any good mathematical forums we can post this problem on?

    • Yat says:

      wait… _I_ did it again ? What did you miss in my explanations ? You see one blue stone is removed, you are asked about an other one. If you consider that this blue stone has been removed, then you are left with three stones, including one blue stone. 1 blue stone among 3 stones. 1/3… is it not clear ?

      I would advise XKCD, of course.

    • MaciejK says:

      But he picks up two stones at the same time which he can do in 6 different ways (later one of them is dismissed).
      If he picked up one of them, then the remaining 3 stones would have 1/3 probability of being picked.

    • Yat says:

      See ? Now you are needlessly including elements irrelevant to the problem.

      If we know nothing about the protocol used, all we know is that the remaining stone is just one of the 3 remaining stones. Nothing more.

      How are these 6 different ways relevant if we chose to just ignore everything about the protocol ? I have 2 blue stones, one white and one yellow in a bag. All you know is that I removed a blue one from the set and that I have another one in my hand. What happened before is irrelevant, isn’t it ?

    • Tomas Blomberg says:

      MaciejK, below is Wiseman’s table twice. He and you assume that in 100% of the cases where only one ball is blue, the color “blue” is named. I’ve therefore marked the cases where “blue” would be named.

      W-Y
      W-B1*
      W-B2*
      Y-B1*
      Y-B2*
      B1-B2*
      W-Y
      W-B1*
      W-B2*
      Y-B1*
      Y-B2*
      B1-B2*

      In 2 out of those 10 cases both are blue. If instead, on an average, the person only names “blue” half the times when there is only one blue ball present, these are the cases when “blue” is named:

      W-Y
      W-B1
      W-B2
      Y-B1
      Y-B2
      B1-B2*
      W-Y
      W-B1*
      W-B2*
      Y-B1*
      Y-B2*
      B1-B2*

      Only 6 cases where “blue” is named and 2 cases where both balls are blue.

      How can it be clearer that the problem is badly phrased? I think the second case here is what should be assumed unless specifically stated that “blue” will _always_ be named when possible.

    • MaciejK says:

      Everything that happened before he showed us 1 blue stone is irrelevant. That is why the protocol is irrelevant.

    • Yat says:

      MaciejK
      Ok, so we agree on 1/3 ! Nice !

      Whait … no ?
      Hmm… why don’t you answer, then ?

    • Camel Ali says:

      @ Yat

      No. We also know that he picked up 2 balls in his hand, before removing one blue ball. This is relevant and can be done in 6 ways as MaciejK has said.

    • MaciejK says:

      We don’t assume that in case blue-nonblue blue is named. It is simply part of the setup of the puzzle. We need not worry about the cases where he names nonblue in blue-nonblue because he already named one blue and it is our starting point.

    • Yat says:

      Why is this relevant exactly ? That happened before he showed us 1 blue stone. I am taking you literally, here, so help me a little !

    • Yat says:

      (last comment was for Camel)

      MaciejK, ok, so why can’t our starting point be that we saw him remove one blue stone ?

    • Camel Ali says:

      @ Yat

      It is relevant because putting two balls in his hand can be done 6 ways as Richard stated in his answer.

    • MaciejK says:

      Well of course it is relevant. It would be meaningless to say:
      “What are the chances that he picks up two blue stones? … ooohh and also when he tried for the first time he picked up a blue one.

      Showing that one of them is blue is there to narrow the outcomes, and the question is about the chances of the other one also being blue to make that relevant.

    • Yat says:

      @camel
      And removing one ball from his hand after the pick can be done 2 ways.

      So… obviously this wasn’t an answer for me, but that’s not important. Please explain why the 6 ways thing is relevant, and not the 2 ways thing.

    • MaciejK says:

      It is our starting point that we see one blue stone.

    • Yat says:

      @ MaciejK
      I am sorry, I dod not found any argument in your message… so let me just follow your 2nd sentence :
      You know what narrows the outcome even more ? The fact that the blue stone actually is the one that has been removed. Are you even starting to understand that there is a big difference between the information you think you have and the information you actually have ?

    • MaciejK says:

      “And removing one ball from his hand after the pick can be done 2 ways.”
      And the puzzle states which way it has been done: He removed the blue one.

      “Please explain why the 6 ways thing is relevant, and not the 2 ways thing.”
      We have an answer to the 2 ways thing.
      6 ways is relevant because it is the number of ways you cans select 2 stones from a group of 4.

    • Yat says:

      @ Maciejk
      Ok, then go back to my comment from 5:04 pm. You told the same thing, I answered.

      So, now, are you even going to try and explain me why my solution is wrong ? I am starting to think you run in circles on purpose…

    • MaciejK says:

      Your solution is not wrong … it simply is an answer to a completely different question.

    • Yat says:

      @Maciejk :
      I am sorry, something went wrong in my last post… it was not answering your last message… so here is an update :
      No, the puzzle does not state which way it has been done. It juste states the result.
      So… no, You don’t have an answer to the 2 ways thing.
      And your explanations about the 6 ways thing does not explain why it is relevant, it only explains why there are 6 ways…

    • MaciejK says:

      “No, the puzzle does not state which way it has been done. It juste states the result.”

      We don’t need to know which way it has been done because the puzzle doesn’t ask: “What is the probability that he will show you 2 blue stones in a row”.

      All we need is the result because we only care about the probability of the other stone being blue.

    • Yat says:

      My super simple solution ? The answer to a different question ? Are you kidding me ?
      The problem clearly states that there are two blue stones, one white and one yellow in the bag at the beginning. It also clearly states that he removes a blue one. It then asks what is the color of another stone. I am just mimicking your way of solving the problem : ignoring everything that is not super obvious, and cutting to the simplest set of informations to manipulate. The fact that two stones are picked first is exactly as relevant as the fact that the host picked the blue stone from the two that were in his hands. So of course, my answer is wrong, for the exact same reason yours is. This is normal, there is no right answer because the problem is incomple. So explain to me exactly why my answer is wrong, and I wil mirror your own arguments to explain why yours is !

    • Yat says:

      MaciejK, this is getting messy, if you really want to discuss this and not only trying to prove me wrong without considering that you may be, then please send your next answer to gyro.gearloose(at)free.fr. We will be able to take our time to answer one thing at a time and focus on the important things.

    • MaciejK says:

      I have posted this puzzle on mathhelpforum.

  43. Lazy T says:

    The question did not stipulate that the box was opaque so I think it was glass,two blues could then be picked 100% of the time.
    Sloppy question, sloppy answer.

  44. Hannes Naude says:

    OK. Here’s another way to see that 1/5 is wrong. Consider the following simpler but analogous problem.

    I toss two fair coins.

    Enumerate the possibilities and calculate the probability of two heads coming up.

    HH
    HT
    TH
    TH

    probability is 1/4.

    I now reveal one coin to you and it shows heads. Enumerate the possibilities again and state the probability that the other coin also shows heads?

    By RW’s reasoning we can eliminate only the TT possibility, making the new P(HH)=1/3. Alarm bells should be ringing.

    But I can change the question and say that the coin I revealed was tails and ask for the probability that the other coin is also tails.

    By the exact same logic this answer is also 1/3. But now we have shown that irrespective of what coin I reveal the probability that the other coin is the same is 1/3. Those alarm bells should be ringing even louder now. This is an absurd result.

    Going back to our original enumeration we see that this probability should be 1/2. as HH and TT are 2 possibilities out of 4.

    Since the result is wrong the method has to be wrong. The same method is equally wrong for the problem as originally stated.

    • Hannes Naude says:

      Sorry. Typo. The last TH in my enumeration should obviously be TT.

    • This is an excellent explanation of why 1/5 would be wrong in a certain situation: If our hero chooses two stones at random without looking, then reveals one of the two stones at random without looking first, and it happens to be blue, the chance of the other matching is, indeed, 1/3.
      BUT… the coins are not a perfect analogy because they do not reflect this possible scenario: Our hero picks two stones at random, then looks at both stones. Our hero has a policy that if there is at least one blue stone in his hand he will show us this one, so he does so. The chance of the other being blue is 1/5.

      So Richard’s answer is only correct if the hero is deciding which of his two to show us. Because we have no reason to believe he’s CHOOSING blue stones as often as possible then the natural answer to the given problem is 1/3.

      If we were playing a game where we bet on the chance of it being a pair, and he hides both stones behind a screen before revealing one and giving us a chance to change our bet then perhaps we’d be right to assume that the chooser has a preference for minimizing the odds of a match. This is where the Monty Hall problem comes into this – please search for it if you don’t see why the chooser’s mental state matters. Richard’s answer is ONLY correct if the children’s looking at both stones in his hand and always reveals a blue one.

      So, in the fight of the decade, Yat was right (and also rather inelegant and rude, in my opinion).

  45. jh says:

    Imagine Mr. Wiseman at the stage in front of the audience. He produces an opaque box and tells the audience there are four stones, two of them are blue.

    He grabs two stones from the box, opens his fist so he can see the stones while the audience cannot.

    Now, he picks one of the stones and shows it to the audience. It is blue.

    Did Mr. Wiseman pick the blue stone at random? Remember that he only picked it up after looking. If Mr. Wiseman is going to pick one of the stones at random, why does he look at the stones first?

    Well, it is at least subject to discussion if it is possible to pick one stone from two at random while you are seeing them.

    If we define P as the probability that Mr. Wiseman has a plan and will pick a blue stone if available, the answer to the problem is:

    P/5+(1-P)/3

    If we are confident that Mr. Wiseman will pick the blue when available, P=1 and the answer is 1/5. On the other hand, if we think Mr. Wiseman does not have a plot and simply takes one of the stones randomly, P=0 and the answer is 1/3.

    From the wording of the problem I find most likely that Mr. Wiseman will pick the blue stone when available, but I’ll concede I am not 100% sure.

    So, my guess is P=0.85, and the answer is 0.22.

    • Yat says:

      clever ! :-)

      I think you should not omit the other way to pick non randomly : remove a blue stone only when you don’t have a choice.

    • Hannes Naude says:

      Very nice.

      But I fail to see any subtle hint in the problem phrasing that Mr. Wiseman is partial to blue. We have no more reason to assume that than to assume that he is partial to white or yellow.

      Therefore I maintain that 1/5 is not a valid answer to the problem as originally stated.

    • jh says:

      You’re quite right, didn’t occur to me the option that Mr. Wiseman would always try to avoid showing a blue stone and would do it only if he had no option. In this case the chance of the other stone being blue would be 1. If:

      P=possibility that Mr. Wiseman plans to show the blue stone if available

      Q=possibility that he hates blue stones and would show a blue one only if has no other chance

      1-P-Q=possibility that he picks the stone randomly

      The answer is P/5+Q+(1-P-Q)/3

      As Hannes says of course there are other possibilities as well.

  46. wraakh says:

    I thought the hand-thing was completely redundant. We know he removes a blue stone before we’re asked to make the guess. Removing one blue stone leaves one blue stone out of three possibilities. That makes it a third. We’re not asked to consider what the odds would be in any other scenario than the one described.

  47. lehcyfer says:

    It’s 1/3 because the story declares that first stone taken out of closed fist (so that the second stone was invisible) was blue. This leaves one of the other three stones in which one is blue. Hence one in three.

  48. Chris V. says:

    The answer should be 1 in 3. If you are counting the fact that you can take B1 or B2 with each pairing, then you have to consider that you could take B1 out first or B2 out first, giving 6 possible answers including at least 1 blue, 2 of which include 2.

  49. Matt says:

    I did 3 types of simulations in python, since that is what all of the cool kids are doing these days. The code (and output)can be found here for people to inspect by others for correctness: http://pastebin.com/GK9Fyvrq

    I think the people have the probability of each of the assumptions worked out pretty correctly — but I didn’t check every comment. Below are the three assumptions I made for each simulation, and the results (approximate). I did 10,000 runs each.

    1. Pick combination of two stones (order in pair doesn’t matter), check both stones, proceed only if at least one of the stones is blue. Probability of two blues: 20% (1 in 5).

    2. Pick combination of two stones (order in pair doesn’t matter), randomly pick a stone from in your hand, proceed only if that stone is blue. Probability of two blues: 33% (1 in 3).

    3. Pick permutation of two stones (order here matters), check first stone, proceed only if first stone is blue. Probability of two blues: 33% (1 in 3).

    So, #2 turns out to be the same model as #3, which makes sense and I did just to confirm that the two models are equivalent. But as #1 differs from #2, it does matter if both stones are inspected, or if one of the two stones in our hand is chosen randomly and happens to be blue. That too makes sense per Bayes formula, as getting a blue on a random choice gives us more certainty about the other stone than does someone looking at both stones and telling us that at least one of the stones is blue. We are more likely to randomly pick a blue stone if both of the stones are blue than if only one is, hence weighting the probability heavier that we chose from the two-blue scenario than the only-one-blue scenario.

    Which scenario Mr. Wiseman intended, I don’t know… when I originally solved it, I assumed scenario #1.

  50. Neily says:

    I just went with what I saw in the explanation. Two stones are pulled out, fist opened, one blue taken out. That leaves three possible stone colours that are in there. 1 Yellow, 1 Red, and 1 Blue. 1 in 3 chance that the other stone in the hand is blue. I didn’t differentiate between the two blues which is why I discounted two of the 5 combinations in the stated 1 in 5 answer above.

  51. Camel Ali says:

    Been thinking about this off and on all day, have enjoyed the heated debate, and weighed up all the arguements. I’m still with Richard and his 1 in 5.

    Forget for a moment, all the assumptions about whether the processes involved were random or deliberate, and what does it boil down to. To pick up 2 of the 4 stones in your hand there are 6 possible combinations as Richard has demonstrated. In showing us that 1 of the stones is blue, all that really tells us for definite is that the combination of stones in his hand is not White/Yellow. So that outcome no longer satisfies the conditions of the problem, leaving only 5 possible combinations. Only one of these gives us two blue stones in the hand, therefore the probability has to be 1 in 5.

    I think some people have tried to overthink this problem, and made it much more complex than it actually is. However, you have given me much entertanment this afternoon, so Cheers!

    • Tomas Blomberg says:

      You can also be sure that he doesn’t hold a White and Blue and decided to say “White”.

    • Camel Ali says:

      He doesn’t say anything, he shows us that he has a blue stone.

      PS, I’m surprised and a little impressed that everyone has refrained from double entendres regarding holding your stones in your fist. :-)

    • Rob J says:

      Nicely stated. To an observer, who is unaware of the motivations or knowledge of the perpetrator (for want of a better word), this seems the most logical outcome.

    • Tomas Blomberg says:

      Camel Ali, ok, you know that we are not in a case where he has W and B and show W. Right?

      In the long run, if the experiment was repeated, do you think it ever can happen that W and B are picked and W is shown? Of course it can happen! Do the simulation a couple of thousand times, if you can’t reason your way to the result in that case.

      This is the same as the classic case where someone has heard Martin Gardner’s thorough phrasing: I see someone with two dogs and ask “Is at least one male?” and get the answer “Yes”. What is the probability that both are male? Answer: 1/3.

      Most of the people here has probably heard that and then somewhere sees it posed as: “I have two dogs. At least one is male.” What is the probability that both are male? Answer: 1/2 since we can’t assume something other than a random strategy if this scenario played over and over. Not everyone would tell you that at least one is male if there is one of each kind.

      Do the simulation in both cases. Count the number of cases which describe each scenario. Count the number of cases where both are male. Who doesn’t agree that this is the way to estimate a conditional probability the Monte Carlo way?

  52. [...] der Murmelmaterie jedenfalls bekannt vorkommen. Aber bitte erst knobeln und nicht direkt in der Lösung nachgucken! Teile und herrsche:E-MailDruckenMehrFacebookTwitterDiggStumbleUponRedditGefällt [...]

  53. MaciejK says:

    As I said I posted the puzzle on mathhelpforum … there is one answer as of now … and it even uses mathematical jargon

  54. John Loony says:

    I don’t understand why there is so much argument and kerfuffling about this. The correct answer is 1/3, for the reasons I explained at the top of the thread.

    Anybody who agrees with me about this is normal, intelligent, logical, sensible, reasonable and normal.

    Anybody who disagrees with me is insane, evil, illogical, deranged, mad, crazy, bonkers, and is quite obviously an alien hippopotamus from the Andromeda Galaxy who has come to Earth to destroy civilization.

    • Matt says:

      I think this is a civilized way to subvert Godwin’s Law (http://en.wikipedia.org/wiki/Godwin's_law ).

      I really liked this puzzle for its ambiguity. It made me think of some of the alternate assumptions about the stone-handler, and how that would change the outcome of the probabilities. Really drove home how to handle Bayes’ Formula and sequential probabilities.

  55. Carl says:

    Okay, I’ve now read pretty much all the comments, and I’m almost completely convinced that 1/5 is the correct solution. I think MaciejK made the clearest explanations followed closely by Mike Torr. Well done guys.

    Here’s my take on it, after some consideration…

    The simplest interpretation of the question is that you are told “at least one of the stones is blue”, in which case that’s a standard conditional probability question and the official answer is completely correct. Right? I don’t think anybody will disagree with that.

    The problem comes, however, with the ambiguity surrounding what the selector’s action tells you. I admit, it’s not completely clear that his action (showing you a blue) is exactly equivalent to saying “at least one is blue”, and I’m pretty sure that this is where all the controversy is coming from.

    However, after considering this for a while, I think I’ve come to the conclusion that the selector’s strategy, or “intention” is only relevant to how often he would show you a blue stone in the first place. But since we are told he definitely does show you a blue stone, then we can ignore it.

    Let’s say he will always show you a blue stone if possible. So, based on the six combinations, he will show you a blue stone 5 out of the 6 times. HOWEVER, of those 5 times, 1 will have a second blue stone.

    Let’s say he picks randomly. In this case he’ll show you a blue stone 3 out of 6 times. HOWEVER, on those 3 occasions, it could still be one of the 5 possibilities (we can only eliminate WY for sure), and the problem still stands.

    And for the situation where he will try NOT to show you a blue if possible, it’s a little trickier, and this is the only situation, I think, where knowledge of his intentions might come in to play. ie. if he shows you a blue, he must have another blue. HOWEVER, based on the question as presented, with only one trial, you have no reason to think that this is the case, and without any further information you have to assume it could still be one of the five remaining options (with WY eliminated). Because when you think about it, why would considering this possibility be reasonable under the circumstances? He could just as easily be using astrological charts to determine how to pick. Are we going to consider that possibility too? No, of course not, we have to assign the probability using all the available information, and no more, so if presented with a blue stone, we have no option other than to conclude there is a 1/5 chance that the other one is blue too.

    I can see what some of you are going to say: “What if you were betting on it, and he was using the ‘never show blue if possible’ strategy? You’d lose money hand over fist!” Well, yes, but that would only apply to a multiple-trial situation, in which you could take into account how often he shows the blue stone first, and then adjust your strategy appropriately after observing several trials. As I said earlier, the strategy of the selector really only effects the probability that he will show you a blue initially, and has no effect on the probability once a blue is shown, which, in this puzzle, it unequivocally is.

    Without any other information, and in a single trial, then the selector showing you one blue has to be equivalent to the statement “there is at least one blue.”

    I’m welcome to counter-arguments to this if my reasoning is wrong, but I will not accept:

    1) Two different answers. You can’t have two different probabilities for a single set of known facts. Probability is all about assigning numerical weightings based on incomplete knowledge, so two different values implies two different sets of known facts. If something is unclear, then it’s not a known fact and can’t be considered in your probability calculation, which is what Mike Torr was saying many times earlier on.
    2) Undefined/No answer. Clearly a probability exists. Just because we might be having trouble agreeing on what it is doesn’t mean no true answer exists.

    Anyway, that’s my summary of what I’ve learned/contemplated while wading through this massive discussion. I hope it might shed a little light and not just make the argument worse. Hmmmm….

    • jh says:

      Of course the answer depends on whether the stone is picked randomly or not. Let me explain. Let’s suppose Mr. Wiseman grabs two stones from the box and holds them in his fist. There are twelve equally likely possibilities:

      a) W-Y
      b) W-B1
      c) W-B2
      d) Y-W
      e) Y-B1
      f) Y-B2
      g) B1-W
      h) B1-Y
      i) B1-B2
      j) B2-W
      k) B2-Y
      l) B2-B1

      The stone closer to his thumb is the first (first column) and the one closer to the pinkie is the second one. For instance, e would signify that Mr. Wiseman is holding the yellow stone and one of the blue ones, and the one closer to his thumb is the yellow one.

      Now Mr. Wiseman has to decide on which stone he is going to look at. We are assuming this choice is random, so let’s imagine he tosses a coin: heads, he looks at the stone closer to his thumb, tails, he looks at the other one.

      In that case those two options would have a probability of P=0,5 each.

      Mr. Wiseman peeks into his fist at the previously randomly chosen side and reveals the stone. He produces it and it is blue.

      If the coin had been heads, and he had chosen the thumb side, we can dismiss a, b, c, d, e and f from the list, and we know it has to be g, h, i, j, k or l. From these, two of them (i and l) include a second blue stone, so in this case the chance for the other stone to be blue is 2/6, or 1/3.

      On the other hand, if the tossing had resulted in tails, he would have looked into the other stone. We know it is blue, so it has to be b, c, e, f, i or l, and we can dismiss the rest. Again, i and l include a second blue stone, so the possibility in this case to find a second blue stone in the fist would be 2/6 or 1/3.

      There are two possible scenarios, depending on the result of the coin tossing, each scenario would have a chance of occurring of 0.5, and in each case the chance for the remaining stone to be blue is 1/3, so the overall possibility of the remaining stone being blue would be:

      (1/3)*0.5+(1/3)*0.5=1/3

    • jh says:

      It is interesting to note that the result is independent on the method used to decide which side is going to be looked at. If P is the chance of Mr. Wiseman deciding on looking at his thumb side, and 1-P is the chance of him looking at the other side, the answer is:

      P(1/3)+(1-P)(1/3)=1/3

      If he was to toss a six sided dice and looked at his thumb’s side only when the dice resulted in a one, P=1/6, the answer is 1/3.

      Even if he decided to always look at the stone nearest to his thumb, P=1, the outcome is the same, 1/3.

    • Carl says:

      Okay, the answer to the statement “AT LEAST one is blue” is definitely 1/5.

      If he’d picked the way you described, then yes, 1/3 would be the correct answer. HOWEVER, we don’t know that that’s how he picked. He could have picked randomly, or by some preference. We just don’t know.

      Therefore, in the absence of any other information regarding the selection method, the only piece of meaningful information we have is: “at least one is blue”. Ergo, 1/5.

    • jh says:

      Agreed!

      The puzzle clearly says that the fist is looked into first, and then the blue stone is produced. Deducing from this that there is a previous choice (random or not) on which stone is going to be produced before looking is quite hard to believe.

    • Tomas Blomberg says:

      So from him lookin ginto his hand you deduce that he wants to show you a Blue ball, instead of that he wants to show you a Yellow ball but can’t find one. Why do you deduce that?

      You obviously can’t know his preferences if any from him looking into his hand before producing the Blue ball.

      If this problem gets posed to you by different people, do you agree that some might show a Blue ball because they want to, and some might want to try to show a Yellow ball, but can’t, because there is none in the hand?

      If you average all possible strategies, you end up with the same solution as if the choice was random. But if you want to verify that, do the simulation where you have lots of strategies and each time two balls are pulled you randomly select a strategy. Understand the difference? You don’t randomly select a ball, but you randomly select wich preference or aversion the ball picker has.

      Again, why would you assume that the ball picker desires to show you a Blue ball, when in fact his highest desire could be to show you a Yellow, but he can’t because there is no Yellow in his hand?

    • jh says:

      You do not have to suppose that the performer’s intentions was to show the blue stone if available.

      What we know is that he picked two stones from the box and after that, that he showed us a blue stone from his pick.

      It is like he’s saying “See? This is one of the stones I picked, and it is blue.” We do not know if the stone he is showing is the only one in his pick or the motivations he has in showing us this particular stone. What we do know is that his pick included at least one blue stone.

      The chance for the other stone to be blue is 1/5.

    • Tomas Blomberg says:

      That means that you are saying that

      P(Blue shown|W/B1) = 1

      Why would you assume that everyone that picks W/B1 will show you a Blue ball 100% of the time?

    • jh says:

      Let’s suppose the performer picks one stone with each hand, and then shows us the stone in one of his hands, without looking at the other one.

      There is a big difference. Now the information we have is about one particular stone in his original pick, and in this case the answer would be 1/3.

      But if he picks two stones, looks at them, and shows us a blue one, which is the way the puzzle is stated, the information we have is that at least one of the stones in his pick was blue, so the answer is 1/5.

    • Tomas Blomberg says:

      So it is clear that you, for no other reason than that he peeks and shows a Blue ball, believe that P(Blue shown|W/B1) = 1.

      If you don’t believe that. What is P(Blue shown|W/B1)? In words that means: the probability of everyone that would pick two balls showing Blue when there is a White and a specific of the Blue balls in his hand.

      You still believe that is 1?

    • jh says:

      I’m sorry, Tomas, the last post was only a follow up for the previous one.

      I’m not saying that the blue stone would be shown each time there is one. We just don’t know.

      But he showing us a blue stone is like knowing there is at least one blue stone in his original pick.

    • Tomas Blomberg says:

      Exactly! “We just don’t know.”
      Also, you seem to believe that us knowing that he holds at least one Blue means that we would always know when he holds at least one Blue. That is not true. Hence 1/5 is the worst possible answer since it assumes that we in _all_ cases would know if he holds at least one Blue. We wouldn’t.

    • jh says:

      No, we would not always know when the pick includes a blue stone. But he looked at both stones, and showed us one of them. In the puzzle a blue stone is shown, so we know his pick included one.

      Maybe in another occasion he has a white and a blue and he decides to show us the white one. At this point we are not speculating about his intentions, we just know that his pick includes a white stone in this case.

      If we had to guess the probability that the other stone is blue, from the list of possible picks:

      a) W-Y
      b) W-B1
      c) W-B2
      d) Y-W
      e) Y-B1
      f) Y-B2
      g) B1-W
      h) B1-Y
      i) B1-B2
      j) B2-W
      k) B2-Y
      l) B2-B1

      In this case we could dismiss the ones that do not include a white stone: e, f, h, i, k and l, and we’d know it would have to be one of the rest: a, b, c, d, g, and j. From these, b, c, g and j include a blue stone so the probability of the other stone being blue with the information available is 4/6 or 2/3.

      In my opinion we have to calculate the probability with the information we have available. From what the puzzle tells us, I believe the answer is 1/5.

    • Tomas Blomberg says:

      Now you are getting close. When you write
      “Maybe in another occasion he has a white and a blue and he decides to show us the white one.”

      What you write there definitely means that P(Blue shown|W/B1) is _not_ equal to 1. Do you agree that the most reasonable assumption is that P(Blue shown|W/B1) = 0.5? Not because we assume that his choice of balls is random, but that the popoulation of all possible ball pickers would cover all strategies. Or in your words “We just don’t know.” If we don’t know, the worst thing to assume is something as information dependant as P(Blue shown|W/B1) = 1.

      If so, the probabilities we have are

      P(Blue shown|W/B1) = 0.5
      P(Blue shown|W/B2) = 0.5
      P(Blue shown|Y/B1) = 0.5
      P(Blue shown|Y/B2) = 0.5
      P(Blue shown|B1/B2) = 1

      and I will simply assume that you know how to calculate P(B1/B2|Blue shown) from that.

    • jh says:

      This is not about the possible strategies that would follow “the population of all ball pickers”.

      If we assume there are a white and a blue stone in the fist, we do not know if the performer (I like to imagine this is Mr. Wiseman himself playing tricks on the stage in front of us, the audience) would always show the blue one, the white one, or randomly switch between the two.

      We just know that he looked at the stones, and whatever were his motivations, showed us a blue stone.

      Admitting that we do not know the probability of the blue stone being shown if one is present is one thing, but assuming that this probability equals 0.5 simply cannot be done.

      This probability would be 0.5 if the stone was chosen randomly between the two, or if Mr. Wiseman would always show the stone which is closer to his thumb, for instance.

  56. I think the 1 in 3 number is based on people thinking “I pulled out one, what’s the probability that if I pull another it’s (x).” But this is not the question being asked. The question is not “You have four marbles in a pot, you remove one, if it’s blue and you remove another, what’s the probability that’s blue too?” The question is, you remove a PAIR of marbles, what’s the probability that they are both blue, and then adding into that the fact that you KNOW at least one is blue. There are six possible pairs of marbles. Eliminate the one without a blue and there are five. One of those five is blue-blue. Therefore, it’s 1/5.

    • Edgar 2 says:

      Sorry, not quite right. The 1/3 answer is derived on the (in my view only tenable) basis that having chosen two at random from four you then choose one at random from 2. In that case the order in which you make the second choice matters. Of the ten possible sequences of outcomes in which there is at least one blue, 6 have a blue as the first stone. So as a result of us seeing the first stone as blue we now that there are precisely six possible outcomes overall. And of those two have blue as the second stone. Hence the probability is two sixths or one third.

      I would assert that to assume that the choice is made anything other than randomly leads us to an unanswerable question since we have no clue at all about what choosing strategy is in play; and since we assume that there is a definable answer (or set of answers) to a puzzle then we must dismiss the non-random choice alternative.

    • Steve Jones says:

      We can no more assume the choice is random than it is guided (indeed the wording of the question implies that a conscious choice was made, or why look in his fist). However, all that means is that there is no single unambiguous answer to the problem. Why people are seeking to show that there particular probability calculation is the single possible one, I don’t know.

      Simply, as worded, we cannot come up with a single probability.

    • Tomas Blomberg says:

      It’s not about assuming that the choice is random. He may very well have a strategy. The randomness comes in from the fact that we can’t assume that everyone that does the experiment has the same strategy. So on an average we can see the choice as random. It’s the best we can do with the info that is given.

      So the only assumption we make is: not everyone will have the same strategy. That’s where the randomness comes in that you percieve as that we have decided that his strategy is random.

    • Edgar 2 says:

      in response to Steve Jones’ comment above

      I think it depends on whether you believe that the act of setting a puzzle implies that an answer (or, for completeness, a definable set of answers) exists. I’d prefer (perhaps misguidedly, on the evidence!) to assume that RW also works on this basis. And with that assumption, choosing at random – ie no active chossing strategy – is the only way to proceed.

      Isnt’ it? Or am I missing something?

  57. Hannes Naude says:

    Nobody is arguing that the conditional probability of having two blue stones given that at least one is blue is anything other than 1/5. HOWEVER that is not what the question translates to.

    There is a difference between knowing that
    A) at least one of the two stone is blue
    and knowing that
    B) a blue stone was shown.

    B is a stronger statement than A since there are cases where the first still holds and the second doesn’t.

    For example he holds white and blue and shows the white stone. Then A is still true but B is not.

    The problem as phrased implies B rather than A.

    The probability of both stones being blue given that a blue stone was shown is 1/3.
    The probability of both stones being blue given at least one stone is blue is 1/5.

    I really can’t explain this any clearer.

    • Camel Ali says:

      Sorry Hannes, but if you show me a blue stone, then I know emphatically that at least one of the stones is blue! Basic logic. You also cite an example where the guy has a white and blue stone in his hand, but decides to show us a blue stone. This has no relevance to the puzzle, since the guy shows us a blue stone and not a white.

    • Camel Ali says:

      Sorry, meant to say, “the guy decides to show us a white stone”, not blue.

    • Tomas Blomberg says:

      Camel Ali, it is relevant for the same reason that you discard all the cases where Blue can’t be named at all. You should also discard the cases where Blue is not named, even though a Blue is present.

      I and others have already done the simulations for you in case your intuition steers you wrong. Was there anything wrong with the simulations? If so, how should a simulation of the problem be done, according to you?

  58. Tomas Blomberg says:

    Here is a simple simple MATLAB program with which I simulated three cases one million times each.

    Strategy 0 assumes that for some strange reason all people who would present you with this question would always show a Blue ball when one is present.

    Strategy 0 also calculates the probability in the case: Two balls are picked and someone asks “Do you hold at least one Blue ball?” and the answer is “Yes.” This case was _not_ posed by Wiseman, but is the traditional way to phrase these puzzles.

    Strategy 1 assumes that for a similarly strange reason as above, all people who would present you with this question would avoid showing a Blue ball.

    Strategy 2 assumes that strategies will vary (or no preferences) between people that will present this question to you.

    Output:

    >> blueBalls(0, 1000000)

    ans =

    0.2002

    >> blueBalls(1, 1000000)

    ans =

    1

    >> blueBalls(2, 1000000)

    ans =

    0.3338

    >>

    Source:

    function out = blueBalls( strategy, m )
    rand(‘twister’,sum(100*clock));
    balls = [0,0,1,2]; % 0 means Blue
    hits = 0;
    targets = 0;
    for i = 1:m
    shuffle = balls(randperm(4));
    select = shuffle(1:2); % pick two balls

    if strategy == 0
    % For some reason we assume that Blue will always be shown
    % when possible
    if or(select(1) == 0, select(2) == 0)
    targets = targets + 1;
    end
    end

    if strategy == 1
    % For some reason we assume an aversion to Blue so it
    % will only be shown when he must
    if and(select(1) == 0, select(2) == 0)
    targets = targets + 1;
    end
    end

    if strategy == 2
    % We assume no strategy so he is equally likely to show a
    % White ball as a Blue ball if he holds those
    ballpick = round(rand) + 1;
    if select(ballpick) == 0
    targets = targets + 1;
    end
    end

    if and(select(1) == 0, select(2) == 0)
    % Both are Blue
    hits = hits + 1;
    end
    end
    out = hits / targets;

  59. Hannes Naude says:

    HA. I knew this had to be explained somewhere.
    See
    http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
    for a closely analogous problem.

    • Matt says:

      The link Hannes gives here should be read by everybody who takes one of the sides of 1/3 or 1/5 at the exclusion of the other as a possibility. As Hannes said earlier, when the conditions are clarified, I don’t think anybody disagrees on the probabilities (well, anymore — I did see people above who said that there was no difference between looking at both and choosing a blue if it was there vs. randomly picking one in the hand and it came up blue. There is a difference).

      But the wording is ambiguous. “Looking in the hand” can mean at least the following reasonable assumptions: 1.) The handler looked at both stones and announced a blue if there was at least one blue in hand, and 2.) The handler randomly looked at only one of the stones in hand, and it happened to be blue.

      The math is settled. It’s the semantics of which model the words translate into. We won’t settle that ambiguity by argument, though it looks as if people will not stop trying.

  60. mm says:

    I also don’t see why some people claim the conditional probability is 1/5. If we model the problem in the following way p(both_blue | one_blue) = P(both_blue) / P(one_blue) then we get (1/6) / (1/2) = 1/3, doesn’t it?

    Why 1/6? So, we have P(both_blue) = 1/6 since 1 pair of blue out of newton(4, 2) = 3*4/2 = 6.

    On the other hand if we assume the ‘setting’ and don’t allow one_blue to be a random variable, then I would write P(both_blue; one_blue) = 1/5.

    To conclude, It seems the conditional probability gives you 1/3, and the ‘absolute’ probability gives you 1/5.

    • Tomas Blomberg says:

      By P(one_blue) I assume you mean P(at_least_one_blue). If so, the probability is the compliment of picking none of the blue. The probability of avoiding a blue as the first pick is 2/4 and, if that has happened, avoiding the blues again happens with a probability of 1/3. So the probability of picking at least one blue is 1 – 2/4 * 1/3 = 5/6.

      If you for some reason is looking for the probability of picking exactly one blue, you can pick it first or second with a total probability of 2/3.

    • mm says:

      No, what I had in my mind is:
      P(L=blue | R=blue) = P(L = blue, R = blue) / P(R = blue)

      where L – left hand, R – right hand; but I see your point.

      If we don’t have any knowledge which hand was used, but don’t have any preferences too, we could also put:
      0.5*P(L=blue | R=blue) + 0.5 * P(R=blue | L=blue) = 0.5 * 2 P(L=blue | R=blue) = P(L=blue | R=blue).

      I also think the following http://research.microsoft.com/en-us/um/people/minka/papers/nuances.html can shed some light on the problem (especially section ‘randomness is subjective’).

  61. Bleach says:

    What are the chances of the other stone being white? 1/3
    What are the chances of the other stone being yellow? 1/3
    What are the chances of the other stone also being blue?

    In my opinion the three should add up to 1

    the official solution assumes that Richard had the intention to pick the blue and ask about the other blue stone. Without this intention the chances are like above. We only know that the first stone shown was blue, not whether this picked was intentionally.

  62. Camel Ali says:

    I see that this debate is going to rumble on and on, and while I would like to continue, I do have other things going on in my life that require attention. One parting shot though. For those devising computer programs to solve the problem, the computer is only going to give you the result that you program it to produce. I’m not sure what merit that has in the debate.

    • Tomas Blomberg says:

      Camel Ali, your last comment regarding Monte Carlo strategies to probability problems shows that you clearly do not understand it and it obviously is not part of your profession.

      Monte Carlo estimations are not to confirm what you think the result is by altering the algorithm until it fits your desired result. They are to blindly estimate a relative frequency from a random population. It only fails in very specific probabilistic problems where the random number generator is not “random” enough. This is not one of those problems.

  63. Steve Jones says:

    Proposition

    All probability puzzle wordings are ambiguous with a probability approaching 1…

  64. Hannes Naude says:

    OK. Here’s one last attempt to clarify. Could anyone that still believes the answer to be 1/5, please post their answers to the following simple puzzle. Relevance will become apparent soon.

    I have three cards. One is white on both sides, the second is black on both sides and the third is white on one side and black on the other. I draw a card at random and reveal the one side to be white. What is the chance that the other side is also white?

    • Aci Virtanen says:

      33% chance that the other side is also white.

    • Steve Jones says:

      This depends on whether you were always going to show use the white side of the random card (if possible), or only showed a white face if forced (i.e. if the random card is double-white), or showed a face at randomly.

      To run through the numbers.

      Scenario (1) – you would always show a white side if available.

      In this case, there are two cards with at least one white face, and they form the sample set. Of these, one has white on both sides so the probability of the other side being white is 1 in 2 or 0.5. I suspect this is the scenario you are describing, but there’s sufficient ambiguity in the meaning I can’t say for sure.

      Scenario (2) – you will only show us a white side if there’s no choice.

      In this case, there’s only one of the three randomly chosen cards which will result in us seeing a white face and that card will have both sides white so the probability here is 1.

      Scenario (3) – you will show a random side of the card

      Taking the initial random pick, 1/3rd of those will be the double-black card and don’t form part of the sample space. Of the 1/3rd of picks that are the white/black card, only half will be revealed to us as a white face (net prob 1/6th). Of the remaining 1/3rd, double-white cards, then we will always see a white face (et prob 1/3rd). Therefore it’s twice as likely the random card will be the double-white one. So the probability with this scenario is 2/3 that the other side will be white.

      There are other scenarios that could be chosen where we might express the preference for choosing to display a white face (if we have a choice) as probability P, and a generalised solution could be produced for those using Bayes’ theorem.

    • Hannes Naude says:

      Very nice exposition Steve. I am baffled how people can read an explanation like that and still claim that the intentions of the chooser is irrelevant and all that matters is what happened rather than how it came to pass.

      The only quibble I have is with “I suspect this is the scenario you are describing, but there’s sufficient ambiguity in the meaning I can’t say for sure.” with reference to scenario 1. Why on earth would you suspect that? Is there anything in the original phrasing that leads you to suspect I favour white? The problem is perfectly symmetric.

      If I could prefer white, I could just as well prefer black (your scenario 2). If you accept that you don’t know what my preference is and that guessing is not a legitimate strategy, then the principle of maximum ignorance demands you assume a uniform distribution over my possible preferences (i.e. 50% chance I prefer white, 50% chance I prefer black). This is exactly the same as a random draw. So assuming that I have an unknown preference is the same as assuming that I draw at random.

      Therefore, even if you suppose that I know the colour on the other side of the card and have some unknown agenda in what I am showing you, the answer can only ever be 1/2.

      In the same way, in the absence of spurious assumptions about the specific preferences of the chooser, the answer to the original problem can only be 1/3.

    • Steve Jones says:

      I only say that was the scenario I thought was most likely because we are rather pre-conditioned to think there will be a “right” answer. The “I reveal one side to be white” could be interpreted as expressing a deliberate choice (indeed for many that would be the normal inference from such a sentence in English, albeit that’s probably a cultural artefact). If it said “I pick one side of the card at random, and it is white” then it would remove that slight hint of intent.

    • Hannes Naude says:

      Damn it. I meant:

      “Therefore, even if you suppose that I know the colour on the other side of the card and have some unknown agenda in what I am showing you, the answer can only ever be 2/3. (Your scenario 3)”

    • Hannes Naude says:

      Steve : “The “I reveal one side to be white” could be interpreted as expressing a deliberate choice.”

      I tried to stay as close to Richard’s phrasing as possible to keep the analogy between the two problems as close as possible. My larger point was that even if we assume with 100% certainty that I have an agenda (that is I definitely prefer either black or white), but we do not know what that agenda is, then the answer is still 2/3. Random selection of sides and random selection of preferences are equivalent, as Tomas has also pointed out.

      So you have to argue not only that my phrasing implies a deliberate choice, but that it implies a specific strategy with 100% certainty.

    • Steve Jones says:

      No – it’s no more reasonable to assume there’s no agenda than there is one. The best I can say is the odds in favour of the other side being white cannot be worse than 0.5.

      I should add that this is really more of a game theory question than a straight probability one. However, for it to be game theory then we’d need to have a definition of what counts as a “win”. For example, if the responder “wins” if he/she guesses the colour of the card’s reverse side. Then you can come up with optimal strategies for both the card picker and the guesser. As it is, the problem is ambiguous.

      Note that it is actually quite difficult to pose non-ambiguous real-world type probability puzzles unless they are trivial. Even the well-known ones such as Monty Hall (as originally defined), “Tuesday boy” and the prisoner problem are usually presented in ways where the protocol is not fully defined.

    • Aci Virtanen says:

      The only correct answer to the question is that there’s a 33% chance that both sides are white. The fact that you show me one white side is irrelevant. We’re are looking for the PROBABILITY.
      If you choose 1 card out of 3 at RANDOM the probability of any 1 of these 3 cards is chosen is 33% that doesn’t change under any circumstance. Why would you suddenly RANDOMLY choose 1 card 50% or 66% of the time?
      The knowledge what color 1 side of the card is doesn’t change the probability of a random event. The probability is always 33%.

    • Hannes Naude says:

      “…it’s no more reasonable to assume there’s no agenda than there is one…”

      Strange. The original puzzle read.
      “I put my hand into the box and pick up two stones”
      It was not explicitly stated that this draw is at random. Yet you have no problem with this assumption. But two sentences later when he reveals a stone it is suddenly reasonable to assume he prefers blue. Then the answer should be 100% since he will always pick up both blue stones. ;-)

      In all seriousness though. I don’t care if you assume an agenda. You just can’t assume that you know what the agenda is. Choosing an agenda at random and revealing a stone/card according to it is equivalent to choosing a stone/card at random.

    • Hannes Naude says:

      Aci: “The knowledge what color 1 side of the card is doesn’t change the probability of a random event. The probability is always 33%.”

      Sorry Aci, but you are way off base here.

      To intuitively see why you statement is false. Consider that I asked what the probability is that both sides are white. 33%. We agree at this point. Now I reveal that one side is black. What is the probability now? According to “The knowledge what color 1 side of the card is doesn’t change the probability of a random event.” it should still be 33%. If you still believe there’s a 1/3 chance that both sides are white after one side has been revealed as black, you are invited to my house for Poker over the weekend. Bring lots of cash ;-)

    • Aci Virtanen says:

      Did you read my message? “The knowledge what color 1 side of the card is doesn’t change the probability of a random event.” I’m referring to the original event of you picking 1 out 3 randomly. Think of it like this, I’m thinking of number between 0-100 Can you guess what it is?
      Whit this knowledge you have a 1% chance of getting it right if I tell you that it’s over 50 you now have a 2% chance of getting it right. OK I tell you my number is 67 now you have a 100% chance of getting it right. The PROBABILITY of me choosing the number 67 is still 1%. See my point? Any knowledge we gain after the original event (choosing the card) doesn’t change the ORIGINAL probability.
      You’re thinking of it as what’s the probability of me knowing what the color on the other side of the card is, that’s is different.

    • Hannes Naude says:

      Aci, I suspect you are trolling.

      But just in case you are actually serious, you might want to go edit this wikipedia page:
      http://en.wikipedia.org/wiki/Bertrand's_box_paradox
      where they erroneously state that the answer is 2/3. Fools.

      Although I suspect that you’ll have to edit quite a few more pages to bring the internet into line with your novel definition of probability.

    • jh says:

      As the problem is stated it is quite clear that the answer is 2/3. If you blindly choose a card and show one of its sides randomly, it is like choosing from the six possible sides of the three cards. Three of them are black and the other three white. It is white, and each white card side has an equal possibility to show up, so 1/3 each. Adding up the possibilities of both sides of the all white card, we find the answer to be 2/3.

      Mr. Wiseman’s problem as I see it is not quite the same. The equivalent one with the cards would be like this:

      Mr. Wiseman has a set of three cards, one completely black, another one completely white and another one with one black side and the other one white.

      Mr. Wiseman draws a card at random and does not let us see it. He glances at the card, and finally he exposes the card showing its white side.

      This is not equivalent to the original problem. In this case one of the three cards was taken randomly, but its side was not shown randomly (at least that we know). The fact that Mr. Wiseman shows the card exposing a white face only tells us that the card he drew was either the all white one or the black and white one. In this case, the answer would be 0.5.

    • Hannes Naude says:

      1) “The fact that Mr. Wiseman shows the card exposing a white face only tells us that the card he drew was either the all white one or the black and white one. In this case, the answer would be 0.5.” I disagree with this. Allow me to introduce one more problem to clarify. There are three urns
      A contains 999999 red balls and 1 black ball
      B contains 1 red ball and 999999 black balls
      C contains 1000000 black balls.
      I pick an urn at random, look inside it and withdraw a red ball.
      What is the probability that it was drawn from A.

      According to you logic, the draw only tells us that the ball came from C. It could have come from A or B so the chance of it having come from A is 50%. You should be able to see that this is ONLY true if you know with 100% certainty that I always seek out a red ball whenever I can. Any doubt on this point lets the probability that the ball came from A shoot up.

      2) It is general practice in probability problems to assume a uniformly random drawing when it is not contraindicated. Case in point :”I put my hand into the box and pick up two stones.” is not specifically stated to be random, but you assume it is. On the other hand you feel that “look inside my fist and remove a blue stone” is not random. I don’t see the distinction. If you want to argue it has to do with the subtly hinted at “fact” that he could not see the stones inside the box, but he could see them in his hand then I would like to know what the answer would be if he could see inside the box?

    • jh says:

      My point of view is that we have to stick to the information we have available. I have to admit that I am not comfortable with the fact that taking the puzzle words as “the pick includes at least one blue stone” yields the same result as “every time at least a blue stone is in the pick, a blue stone will be shown”.

      But the fact is that it is irrelevant to speculate about the chances of a blue stone being produced if the pick had a white and a blue. Is there any chance that the white stone were produced? Maybe yes, but I do not care. I have to admit I do not precisely master Probability or Mathematics (by no means! I wish I did), I think that perhaps you are seeing the problem from a frequentist point of view, while mine is more Bayesian.

      The fact is that a blue stone was produced. That is all we know. If we could ask Mr. Wiseman to repeat the trick a few hundreds of times then we could take a frequentist approach and try to find the pattern in which stones of different colours are produced. But this is a puzzle and we are not allowed to set up an experiment of which we could draw any further information.

      And what we also know is that the stone was shown after Mr. Wiseman looked at his selection. We do not know why did he show us that particular stone, but I do not think that the best thing is to assume a uniformly random drawing.

      In your example, one of your urns has one red ball and 999999 black ones. If you asked someone to look at the content of the urn and take out whichever ball he desired, do you really think it is safe to assume that the red ball would be drawn only once each 1000000 times? Of course not, if the red ball is visible, it will draw the picker’s attention and there is a greater probability that he will pick the sparkling red little ball than the other dull black balls.

      Your puzzle is taken to the extremes, but it serves as a good example. I still stand that from a purely mathematical point of view, the fact that the ball was drawn from the urn and the drawing cannot be considered a random act, but a conscious one, equals as saying that the only piece of information we have is that the ball must come from one of the urns, so the possibility it came from A is 0.5.

      Of course this is way off from the results we would get from a real test in the terms the problem is stated. But the thing is: assuming the ball drawing is random (when it clearly is not) would result in a probability of 999999/1000000 it came from the urn A. But it would also be way off. As I said above, a red ball in a crowd of black ones will have a greater probability of being picked unless the pick is blind. To know this probability is a question of experiment taking into account things like, size of balls, size and shape of the urns, available lighting, gender and age of the picker, maybe cultural background and many other things.

      And about the puzzle not saying that the original two stone pick was random, you are right! It does not say so but I assumed it was (everybody did!). The wording could have been better, but there is a great difference between the two stone picking from the box and the stone being shown: the stone being shown was only shown after looking at the fist and seeing the outcome of the box picking.

      Anyways, you want to know what the answer would be if Mr. Wiseman did look into the box before grabbing the two stones. Well, I have to admit that from my point of view the problem makes no sense at all anymore. If there is no random action at all, the pick is just a matter of will from Mr. Wiseman and it is not possible to calculate any probability at all unless we make assumptions. And these assumptions would have to be that his picking from the box was a random one, even knowing that it wasn’t.

    • Hannes Naude says:

      Your distinction between random and non-random events is entirely artificial. If you add enough information, nothing in our everyday experiences is random. The throw of a dice, flip of a coin or spin of a wheel of fortune. All outcomes of human-scale events are pre-determined by physics (I specify human scale because at the quantum level there are truly random events.) If you had the info you could calculate what face is going to come up on the die or which side will show on the coin. The fact that you don’t have the info makes the outcome random, from your perspective even though it may be non random to someone who was given more info than you.

      What are the chances of me being involved in an accident tomorrow? Is it a random event? No, it is determined by my actions. Yet, my insurance company can attach a probability to that outcome and the fact that they are still in business tends to indicate that they are pretty accurate. The more information they have about me and my habits, the more accurate their estimate becomes, but even with no info (on me specifically) they can make an unbiased estimate.

      Similarly, I don’t care whether the draw was random or an act of will. The fact is that I do not know what goes on inside RW’s mind forces me to treat it as a random event (which, from my perspective, it is). I don’t care whether the shown ball is picked by the flip of a coin, throw of a dice or by some mysterious process inside RW’s cranium. They are all just random processes from my perspective. Given more information, I can refine my estimate of the distribution from which the picks are drawn and my answer may change (in the same way that the answer changed once the colour of the first ball was revealed). But in the absence of any such information, the only sensible assumption is that the distribution is uniform with no discrimination between colours whatsoever.

    • jh says:

      Well, my opinion is that in the absence of any information on why precisely the blue stone is shown the only sensible thing to do is making no assumption at all.

      Which stone would be shown in each case may be random from the audience’s point of view, but assuming that the probability of showing a blue stone in case there is another one of a different colour equals exactly 0.5 is going too far. It may be random from our side, but we have no means to estimate its probability.

      The fact that in this particular case Mr. Wiseman showed a blue stone means only that we have new information that reduces the sample space from 6 to 5 possible outcomes.

      I understand your point of view, but this is how I see it.

    • Hannes Naude says:

      “Well, my opinion is that in the absence of any information on why precisely the blue stone is shown the only sensible thing to do is making no assumption at all.”

      Agreed. What you don’t seem to realise, is that by taking your 5 possible outcomes to be equiprobable you are implicitly assuming that the chooser prefers blue. In your terms :
      “… but assuming that the probability (…) equals exactly 0.5 is going too far.”
      In your case you are assuming this probability to be exactly 1. This is certainly going too far.

      You are not faced with a choice between making an assumption and not making an assumption, but between assuming
      p(blue shown| one blue,one non-blue)=0.5 and
      p(blue shown| one blue,one non-blue)=1.

      The latter is entirely unreasonable in the absence explicit indication that it holds. The former is reasonable and obeys the principle of maximum ignorance.

    • jh says:

      I am sorry to come back so late, I’ve been really busy at work!

      Yes, I realize that considering the 5 possible outcomes as equiprobable, the answer is exactly the same as considering that in the case a blue stone is present in the pick it is going to be shown. I am not comfortable with this.

      But I believe that we are not to ask if there is any chance that in the case one blue stone were present with one of a different colour, Mr. Wiseman would show the non-blue one. We do not know and we do not care.

      The problem is a one-time show, and as soon as we learn that a blue stone was produced from the pick, we know that this scenario did not happen. We are positive on this: at least one of the stones is blue!

      The problem is more a question on which type of assumptions each person is ready to make. I guess our points of view are quite apart and are quite irreconcilable.

      I want to say that I really understand your point of view and I really enjoyed the debate.

    • Tomas Blomberg says:

      “The problem is a one-time show…”

      My juggler friend has balls with those same colors, picked two at random, peeked into his hand then showed me a blue. What is the probability that the other one is blue?

      There, you have now come across the problem once again.

      Eventually you’ll realize that to calculate probabilities you can assume that the scenario is repeated an infinite number of times even if you for some reason think the scenario will never happen again. It has proven utility, since it produces the correct answer in every case.

  65. SteveG says:

    There are a lot of spins on this one and it reminds me of the Monty Hall problem (which still throws me until I write down all the possibilities.

    But this seems less tricky.

    There are a lot of comments and I have to admit I haven’t read them all so someone else may well have come up with the same explanation:

    The answer is 1 in 5 technically, but 1 in 3 practically.

    Because, based on the question, there is no distinction between outcomes w-b1 and w- b2 and no distinction between y-b1 and y-b2. Therefore these four outcomes can be combined into two outcomes: w-bx and y-bx. The only other possibility is b1-b2 leaving a total of only three outcomes which include (any) blue ball. Thus the probability of there being (any) blue ball remaining in the hand is 1 out of those 3 possibilities.

  66. Tomas Blomberg says:

    After thinking a bit about where some people go wrong, the basic problem seems to be that they assume that the three cases

    W/B1 and Blue is shown
    Y/B1 and Blue is shown
    W/B2 and Blue is shown
    Y/B2 and Blue is shown
    B1/B2 and Blue is shown

    have equal probability.
    Since P(Blue shown|B1/B2) = 1 it means that these people also assume for example P(Blue shown|W/B1) = 1.

    Then of course the conditional probability P(B1/B2|Blue shown) = 1 / (1 + 1 + 1 + 1 + 1) = 1/5

    But why do these people assume all these probabilities to be equal? They assume that in 100% of the cases, someone would hold W/B1, the Blue will be shown. Why? I have no idea why P(Blue shown|W/B1) = 1 is more reasonable to assume for these people than P(Blue shown|W/B1) = 0. Can someone that desperately wants P(Blue shown|W/B1) = 1 tell me why P(Blue shown|W/B1) = 0 is out of the question?

    The resonable thing to assume is that on an average all possible strategies ballpickers can have, will even out. That means that P(Blue shown|W/B1) = 0.5, P(Blue shown|Y/B1) = 0.5, etc.

    You then see that P(B1/B2|Blue shown) = 1 / (1 + 0.5 + 0.5 + 0.5 + 0.5) = 1/3

    No one has been able to show that P(Blue shown|W/B1) = 1 is given in the problem description. One reason is that it isn’t given at all.

    • Tomas Blomberg says:

      [Correction] “…five cases…”

    • mm says:

      Nice!

      I am not native English, but perhaps there is some linguistic issue that somehow bias people to think about the problem in this way. For instance is the mental picture of the phrase ‘look inside my fist and remove a blue stone’ is equivalent to the following one ‘open one hand randomly, and the stone in this hand has happened to be blue’, or not?

      I also believe, another problem can be some people tend to think that since every statement in math is either true or false (or undecidable), you always get one ‘correct’ answer about the reality. Therefore one reasonable good explanation seems to be the ‘correct’ one.

      I am also interested in what people think about this problem, since for me P(blue shown | not_blue/blue) = 0.5, and P(blue shown | blue/blue) = 1 seems to be the most natural.

      Perhaps it is also better to show more explicitly how you got those numbers. So:

      P(A|B) = P(B|A) P(A) / P(B) by Bayes rule, and P(A) = sum_C P(A,C) = P(A|C) P(C) by sum-product rule. Therefore
      P(B1/B2 | blue shown) = P(blue shown | B1/B2) P(B1/B2) / P(blue shown) = P(blue shown | B1/B2) P(B1/B2) / [ sum_color1/color2 P(blue shown | color1/color2) P(color1/color2) ]

      Now if we assume P(color1/color2) = c (it is equally likely to get any pair of stones, see next assumption) then we can cancel out P(B1/B2), and P(color1/color2) to finally get
      P(B1/B2 | blue shown) = P(blue shown | B1/B2) /
      sum_color1/color2 P(blue shown | color1/color2)

      which, in turn, yields either 1/5, or 1/3 (or even more) depending on how you treat P(blue shown | color1/color2) as you said.

    • Tomas Blomberg says:

      “For instance is the mental picture of the phrase ‘look inside my fist and remove a blue stone’ is equivalent to the following one ‘open one hand randomly, and the stone in this hand has happened to be blue’, or not?”

      Even if the person’s choice is deliberate, showing a Blue ball after peeking could indicate:

      He always wants to show a blue ball.
      He never wants to show a blue ball.
      He always wants to keep a blue ball in his hand.
      He always wants to show a yellow ball.
      He never wants to show a yellow ball.
      He always wants to keep a yellow ball in his hand.
      He always wants to show a white ball.
      He never wants to show a white ball.
      He always wants to keep a white ball in his hand.
      He prefers to show blue over yellow, but white over blue.
      etc
      etc
      etc

      So if you sum over all possible strategies, you in effect get the same result as if he randomly selected which ball to show. The nice thing is that you get the same result. It doesn’t matter if he has a strategy or picks randomly. The only time we are allowed to make such an assumption is if that info is stated in the problem. It isn’t.

    • Anonymous says:

      The problem with this it that you cannot assume equal weight to all the possible strategies. Especially not after he shows you a blue stone.

    • Tomas Blomberg says:

      So, which of those strategies do you think is strengthened after seeing the blue ball? Definitely the one where he wants to hide the white ball, right?

      Feel free to do the exhaustive calculations of all possible strategies. I’d definitely agree on your answer.

    • Anonymous says:

      The problem with this is that you cannot assume equal weight to all the possible strategies. Especiallu not after he showes you a blue stone.

    • Anonymous says:

      “So, which of those strategies do you think is strengthened after seeing the blue ball? Definitely the one where he wants to hide the white ball, right?”

      Yes, all strategies where he wants to show the blue ball is strengthen.

      “Feel free to do the exhaustive calculations of all possible strategies. I’d definitely agree on your answer.”

      First we need to find a probability that the picker indeed has a fetish. Just because he can, does not mean he does. We would need to estimate this probability though some study. Then we could do some calcualtions.

    • Tomas Blomberg says:

      A study can not always be done, yet you, if this is your line of work, can be asked to calculate a probability. So, what is the probability, before the study is done?

      If you still want to complicate your calculations instead of realizing that covering all strategies is the same a pretending that the ball was randomly selected, assume a strategy is used with probability f and that there is no strategy (random selection of ball) with probability 1-f.

      If your f still exists after those calculations, I suggest integrating over f, pretending that it varies from universe to universe.

      Sometimes you have to do _something_ to produce a result before you try to, by experiments, estimate a distribution. Sometimes those experiments can’t even be done.

      Did I say that I don’t like the way the problem was posed? :) 1/5 is the most wrong answer I’ve heard so far.

    • mm says:

      Tomas Blomberg:
      ‘Even if the person’s choice is deliberate, showing a Blue ball after peeking could indicate:’

      In my post I was more interested in ‘why people believe in 1/5 so much: is it because this number was given in the solution, or maybe there is some specific interpretation of the problem, or sth. else’, not about the math behind the problem.

      So in other words why people assume so often that the person can see what is inside of his both fists, and always pick up blue if available.

    • Tomas Blomberg says:

      That is an interesting question, and I think for most 1/5 people it does not have to do with them assuming that a blue would always be shown when possible. I rather think that they see the five remaining cases and think “Of course these have an equal probability.” without realizing that that implies P(Blue shown|W/B1) = 1. So they simply stop thinking there and are satisfied with the answer without considering what P(Blue shown|W/B1) = 1 actually means.

    • Anonymous says:

      “A study can not always be done, yet you, if this is your line of work, can be asked to calculate a probability. So, what is the probability, before the study is done?”

      I am just way to lazy to figure that out. It would take way too much time to do the calculations properly. Besides, I doubt even five people would care about the result, I wouldn’t.

  67. Anonymous says:

    I have not read throgh all the responses, but I think we should view the selectors strategy is a random variable. If his strategy is “remove a blue stone if it is available” the probability is 1/5. If it is “remove a stone at random” it is 1/3. If it is “remove a blue stone if there are no other options” it is 1.
    Therefore all we can conclude with certanty is that the probability is between 1 and 1/5.
    We could also note that the probability that the selector’s strategy is “remove a blue stone if it is available” increases when he does remove a blue stone.

    • Tomas Blomberg says:

      “…but I think we should view the selectors strategy is a random variable.”

      That’s an excellent idea. If you do that, you’ll end up with 1/3 being the answer. Simply use a population of ball pickers with different strategies and take all those into consideration when calculating the final answer.

    • Anonymous says:

      “If you do that, you’ll end up with 1/3 being the answer.” I’m afraid that will not hold. We still have no idea of the distribution of their strategies. Thanks for the compliment though. :)

    • Tomas Blomberg says:

      So when you are presented with a probability problem, you stop calculating because you don’t have any apriori knowledge of the distribution? Of course not. Unless this is not in your line of work so you have the luxury of just ignoring some problems that are posed to you.

      Just don’t assume the most ridiculus case of all ball pickers always having a Blue ball showing fetisch.

  68. Anonymous says:

    I like the idea of taking a sample of ball pickers. That way we will get the best estimate of the true answer imo. But that does not imply that the answer will be 1/3, but it would probably closer to 1/3 than either 1/5 or 1. The answer could for example be 0,4 or 0,3.

    However, for Friday puzzles I think you should be able to assume that all information you will get to solve the problem is presented to you in the problem. I don’t think it is reansonable that you should having to conduct a study group is just to solve a Friday Puzzle.

    • Tomas Blomberg says:

      Agreed. It’s so simple to phrase it better:

      Someone randomly picks two balls. I ask if at least one is blue. He answers “Yes.” after peeking. What is the probability of both being blue?

      But phrasing it badly is always an eyeopener for some people who think 1/5 is the answer. In March 9, 2006, I finally understood that not all phrasings of such problems are equal. I’m forever in debt to the person who worked for several days to make me understand the differences between problems I thought were identical, when I quite stubbornly insisted on the “classic” answer. :)

  69. Bletherskite says:

    Wowser – this is a thread and a half!

    My original answer had been 1/3 but after reading Richard’s solution and then re-reading the question I am now firmly in the 1/5 camp. I don’t find the question ambiguous at all.

    My maths teacher in primary school used to tell me that I always managed to find really complicated solutions to quite simple problems – seems I’m not the only one ;-D

    • Tomas Blomberg says:

      When this is correctly phrased, the 1/5 answer is supposed to be the unintuitive and complicated yet correct answer.

      Laymen should fail the puzzle since they don’t complicate things and are supposed so think “The ball that is left in his hand is either Yellow, White or Blue so the answer must be 1/3.” to which the educated person laughs and shows why 1/5 is the correct answer.

      That’s what makes it such a good puzzle when correctly phrased.

      Ironically, 1/5 is the wrong answer when the problem poser has not understood how important the phrasing is. As is the case here.

    • Hannes Naude says:

      Noooo. Here we go again
      http://xkcd.com/386/
      ;-)

      OK so if you’re firmly in the 1/5 camp I would like to hear your answer to this question:

      Suppose there are three cards:

      A black card that is black on both sides,
      A white card that is white on both sides, and
      A mixed card that is black on one side and white on the other.
      All the cards are placed into a hat. I pull one at random and place it on the table. The side facing up is black. What are the odds that the other side is also black?

    • mm says:

      What is means to be more complicated? And in particular why 1/5 is less complicated?

      If we measure how complicated things are by the number of assumptions we made, then P(blue shown | not_blue/blue) = 0.5 seems to be less complicated than P(blue shown | not_blue/blue) = 1 since the later needs extra assumption about the person drawing a stone.

  70. Camel Ali says:

    LOL. Just checked back, and I see that some of you are still arguing over this one. Have fun :-)

  71. vesine says:

    This had me confused for about an hour because the 6 possibilities of drawing 2 stones is really the “results” not how the test can run. It’s a lot clearer (to me anyway) when I write out ALL the possibilities of how the test can run, including the draw order.

    There are 4 balls, 1 2 3 and 4.
    1 is w, 2 is y, 3 and 4 are b.

    1st pick, 2nd pick, result:

    1 2 NA
    1 3 fail
    1 4 fail

    2 1 NA
    2 3 fail
    2 4 fail

    3 1 fail
    3 2 fail
    3 4 pass

    4 1 fail
    4 2 fail
    4 3 pass

    So ignore any result that has only 1 and 2 as they don’t contain a blue ball (marked NA). There are 10 remaining possibilities, only 2 of which satisfy the condition “the other ball is also blue” (only balls 3 and 4). Hence the chance that both balls are blue after revealing one blue ball is 1 in 5.

    • Tomas Blomberg says:

      You are very close to a good way of solving it there, but you forgot the parameter of which ball is named.

      Write that list _twice_ and next to one list write that he names the first ball and next to the other list write that he names the second ball. Now do what you just did: write NA in all cases 1 or 2 is named.

    • Anonymous says:

      Not certain where 1 in 5 comes from
      Using your logic of the twelve options you list only the second six yield a blue ball first
      Of these six two are a pass
      That’s 1 in 3 to me

  72. Red_in_se8 says:

    Lets imagine that one of the blue balls has a number 1 printed on it and the other 2 printed on it.

    The possible combinations are

    Y + W

    Y + B1

    Y + B2

    W + B1

    W + B2

    B1 + B2

    Two balls are drawn out and you are told one is Blue.

    If it is B1 we now can eliminate any combinations that does not include B1 (Y + W, Y + B2 and W + B2). That leaves 3 so the answer is 1 in 3.

    If it is B2 we now can eliminate any combinations that does not include B2 (Y + W, Y + B1 and W + B1). That leaves 3 so the answer is 1 in 3.

  73. Red_in_se8 says:

    Although the way the puzzle is phrased there is no need to look at these combinations. Two balls are picked. You are told one is Blue. That leaves 3 balls and one is blue which means it is a 1 in 3 chance that the second is Blue.

  74. Red_in_se8 says:

    In Richard Wisemen’s answer he says

    ‘W-Y is eliminated, but all of the other possibilities involve at least 1 blue stone. However, only one of them involves 2 blue stones so the answer is 1 in 5.’

    But only 3 of the possible combinations involve the blue stone that has been revealed.

  75. Alain says:

    Surely there are actually 7 possibilities, if you take B1-B2 as a possibility there is also B2-B1 as a 7th possibility, therefore eliminating W-Y you have 6 possibilities with blue and 2 chances at getting Blue Blue (B1-B2 and B2-B1) that is 2 in 6 or a 1 in 3 chance of a blue blue combination.

  76. mikekoz68 says:

    “So Richard got it wrong” ya, the guy who writes books explaining the fallacies of the human mind has been corrected by a you-tube wannabe commentor lol The whole point of the puzzle is that people do not understand probability- it is not intuitive, Richard posts the correct answer of 1/5 and the more people argue over it the more he is right- probability is hard for us to grasp

    • Tomas Blomberg says:

      [sarcasm]That’s a _great_ argument.[/sarcasm] I tried a similar argument myself on a puzzle when I wrote “Oh yeah?!? What’s the probability that the answer to a classic problem in probability theory suddenly is wrong?!?”

      It turned out that the one phrasing the problem had misunderstood the classic problem, and I didn’t think it mattered at the time. So I was wrong in spite of the “clever” argument I made above. I’ve been on your side, so I know that it can take a while to understand how important the phrasing is in these puzzles.

      The correct conclusion you should draw is that Mr Wiseman doesn’t understand probability as well as you think he does.

    • You Tube Wanabee says:

      Here is an attempt to simply the whole debate. Elements in quotation marks are direct from the question.

      “I have a box and inside the box are four stones.”

      Let’s call the stones A, B, C and D

      “One stone is white, another is yellow, the third is blue and the fourth stone is also blue.”

      A and B happen to be blue.

      “I put my hand into the box and pick up two stones.”

      The combinations possible in choosing two stones are:-

      i. AB
      ii. AC
      iii. AD
      iv. BC
      v. BD
      vi. CD

      “I bring my hand out in a fist, look inside my fist and remove a blue stone.”

      Of the above combinations i, ii, iii, iv and v have a blue stone in them. Combination vi. does not. Hence there are 5 possible pairs that could produce the blue stone which Richard removes from his fist.

      “What are the chances of the other stone in my hand also being blue?”

      The following lists the possible ways of choosing a second stone, blue or otherwise, having first chosen a blue stone. The list shows the stones chosen in order.

      a. A B (from pair i.) – a blue/blue combination
      b. B A (from pair i.) – a blue/blue combination
      c. A C (from pair ii) – fist stone only blue
      d. A D (from pair iii) – first stone only blue
      e. B C (from pair iv) – first stone only blue
      f. B D (from pair v) – first stone only blue

      The above shows six scenarios – a. to f. Of these six two produce a blue as the second stone. This gives a probability of 1/3. This solution also involves the five pairs but does not end up with odds of 1/5. It also, I am afraid, counters Richard’s solution. The five pairs do not result in odds of 1/5 for the second stone blue. Richard’s solution ignores the fact that the first stone chosen could be either A or B. A can be followed by the other blue stone B. Conversely B can be followed by the other blue stone A.

  77. Another You Tube Wannabe says:

    Mikeoz68

    You might care to look at this link from earlier in the year:-

    http://richardwiseman.wordpress.com/2011/06/27/answer-to-the-friday-puzzle-110/#more-4357

    Richard made the same mistake in his first answer to this, somewhat simpler, puzzle, then later corrected it

    • Tomas Blomberg says:

      What’s interesting in that problem is that some people think it’s unsolvable, as we are not told the probability of the color of the initial ball:

      “…it is impossible to accurately indicate the probability of getting a white pebble. Imagine, for example, that the probability of a white pebble being in the bag originally was 0.001 %. This would be in accord with the description and yet would lead a totally different result than if that probability was 99.999 %.”

      Of course, it’s the same as in the problem in this thread; if we are not told the probability, assume that the strategy varies when the problem is repeated. In lack of other information, the only thing we are allowed to do is to assume that all strategies will even out in the long run, which is the same as using a probability of 0.5 in these cases. Any bias in any direction needs to be thoroughly justified, but there is no justification for any bias in the phrasing of any of the two problems.

  78. Virtualfog – Gaming & Coding Community…

    [...]Answer to the Friday Puzzle…. « Richard Wiseman[...]…

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    [...]Answer to the Friday Puzzle…. « Richard Wiseman[...]…

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