First, this week I started a new channel containing life changing ideas in less than a minute. It is called In59seconds, and I hope that you will come over and take a look.

Second, here is the puzzle. Please do **NOT** post your answer, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

A box contains two coins. One is a double-headed coin, and the other is an ordinary coin ( heads on one side, and tails on the other). You take one of the coins out of the box and look at one of the sides. It’s a head. What’s the probability that the other side also contains a head?

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.

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5 sec

4 sec

Please do NOT post your answer, but do say if you think you have solved the puzzle and how long it took.

There are two coins. You are looking at the head on one of them. It is either the double head or the tail. If the question was “what are the chances of pulling out a coin tails up?” the answer would be one in four (there are 3 heads and one tail), but if you are already looking at a head, then the other side is either the head or the tail.

3 seconds

Once you are looking at a head, the other side is either a head (if it’s the double header) or the tail (if it is the normal coin). There are only 2 coins so it is 50/50 which one you choose. If you choose the normal coin then the other side will be tails, if you have chosen the double header then the other side will be a head.

As I see it, when the coin is drawn, there are three possibilities regarding this first coin, each of which is equally likely:

Double-Headed Coin Heads

Double-Headed Coin Heads (the other head)

Ordinary Coin Heads

“Of the three, two of the possibilities will show heads on the other side; only one will show tails on the other side. So the probability that the other side of the coin is heads is two thirds.”

Hope this helps.

10 seconds

its aproblem of conditional probability similar to monty hall problem

hope this helps

I think I got it 😕 a quick guess, took a few seconds haha 😀

I hate probability questions but I have an answer in about 15 seconds although it’s probably (!) a toss-up as to whether it’s right or not (no spoilers intended)

Quick guess – quick think – about 10 -15 seconds for second answer

Got the wrong answer instantaneously and then managed to get the right one after about a minute. Either that or it is the other way around. This type of problem often causes discussion.

Same here, Steve. I got the obvious answer as soon as I’d read the question. The answer that is probably correct, however, came to me a bit later… when I realised how dim I’d been.

We’ll find out in a few minutes because it’s Monday and I’m late to the puzzle this week.

This Friday Puzzle is a “Red Herring”. The real Friday Puzzle is why this was posted at 09:56 instead of 08:30!

I think I got it – I probably haven’t because I was too quick for my probability capabilities, but if I have I will celebrate with cake! xx

********* Helpful WARNING for those new to the Friday puzzle *******

It’s possible that in the next few hours someone will post below the answer to the puzzle – yes, that’s in spite of Prof Wiseman’s polite request not to.

So, if you really want to solve the puzzle on your own, don’t read the comments. On the other hand, if you want a hint so heavy it will influence the tides, or you even want the outright answer, I’m afraid some self-important twit may oblige you soon.

Miles, you spelled ‘twat’ incorrectly

2/3

This one is not as straight forward as you might think! My initial guess was wrong, working it through took about 30 seconds or so. It’s a bit like the Monty Hall problem. Having a two headed coin in my pocket might have helped 🙂

The Monty Hall Problem is fundamentally different, in that it includes a third party who has more infromation than you. Unless I have missed the part that reads “You take a coin and Monty Hall shows you that it has a head on one side”, this is a simple probability problem.

It’s a bit like the MH problem because it’s not what your instinct initially thinks, there is more too it, and that alters the probability. No it’s not a TV gameshow.

I know Which Part of the puzzle did you not read: “Please do NOT post your answer”

The Monty Hall Problem is fundamentally different, in that it includes a third party who has more infromation than you. Unless I have missed the part that reads “You take a coin and Monty Hall shows you that it has a head on one side”, this is a simple probability problem.

Fool, it’s similar in that you have to balance partial knowledge of the situation (a goat behind a door, or that one side of the coin you took has a head) with prior probabilities

It is like the Monty Hall Problem, the difference here is instead of the 3rd party providing the information the puzzle itself gives you the information that the first side you see is a head.

Unless I have missed the part that reads “You take a coin and Monty Hall shows you that it has a head on one side”

That is exactly what the puzzle does, when it says the side you see is a head.

“It is like the Monty Hall Problem, the difference here is instead of the 3rd party providing the information the puzzle itself gives you the information that the first side you see is a head. ”

Not quite – the key point in the MH problem is that he knows some information you don’t and you can use that fact (ie *that* he knows, not *what* he knows) to change the probability. Here, there is no “hidden but known” information. Simply, after looking at one side of a coin, there a certain number of equally likely outcomes.

If you are saying that the probabilities change after looking at a coin, then that is of course true, as it is with most “drawing from a bag” problems!

I don’t understand. What does the TV broadcaster, explorer and marine biologist have to so with this?

@Hugh https://en.wikipedia.org/wiki/Monty_Hall_problem

There is an apparently counter-intuitive probability problem named after him, which says that you choose a door from three, behind one of which is hidden a car. MH then opens another door, showing that there is no car behind it and asks if you would like to change your choice to the remaining door. Probability says that you have a 2/3 chance of winning if you swap, caused by the fact that Monty always knows where the car is.

It’s very much like the Monte Hall problem. The difference is only that Monte Hall isn’t standing there asking if you would like to switch coins (or something). But figuring out the initial probabilities involves the same sort of thinking. It’s the difference between thinking that it’s 50:50 (which is what most people think who get the MH problem wrong), and 1/3:2/3.

Sorry – even though the answers happen to be the same, I still disagree. It is “the same” in that both are about probabilities which change in the light of new information, but the MH is more complex in a much more subtle way.

This problem is very simple. Before drawing, there are 4 possible outcomes (2 coins with 2 sides each): the observation of one side of the drawn coin happens to eliminate one of these outcomes. Two of the remaining happen to satisfy the question.

MH starts with only 3 equally likely outcomes. The actions of a third party (who you know has extra information, even though you never know the *actual* information) uses his knowlege to reduce the outcomes to 2. If MH’s choice of door was random, I would agree that the problem is the same (though the probabilities different – iff MH, by chance, doesn’t show the car, it is 50% for each remaining door). In the actual MH case however, we know he never chooses the car, so by swapping I only lose if I was “unlucky” enough to choose the car first time (1/3).

Think of it this way to see they are fundamentally different: in drawing the coin we *might* have found a tail. In MH, we will *never* see a car.

Actually, the only real difference between this problem and the Monty Hall Problem, is that the MHP includes the case where what you observed (Monty Hall opening a specific door) couldn’t happen. To make the problems truly the same, add a two-tailed coin to the bag and it is equivalent to MHP. We know you didn’t pick this coin, so the answer is the same with it added. Here’s how to see it is equivalent to MHP:

There are three possible games in the MHP. In one the car is behind the door Monty Hall opened. Just like the drawing the two-tailed coin, we know that didn’t happen so we eliminate that case entirely. In another, the car is behind the door you’d switch to, so Monty Hall had no choice about what door to open. Just like drawing the two-headed coin, in this case there is a 100% chance of observing what was actually observed, making the original probability for the two happening together 1/3. In the last case, where you picked the car, there is only 50% chance Monty Hall would open the door he actually did. Just like there is only a 50% chance to see a Heads if you picked the normal coin. So the combination of this case with what was observed has an original chance of 1/6, not 1/3. The probability of switching to the car is (1/3)/(1/3+1/6)=2/3. And note that I didn’t complete the solution to this problem.

For whatever reason, foolonthehill thinks that it is different because the 50/50 chance is the result of Montry Hall’s choice. That is irrelevant – it’s still a 50/59 chance.

Ahem – fat fingers hit the “9” key by mistake. It’s still a 50/50 chance.

@JeffJo I don’t deny that the mathematics happen to be (largely) congruous – your additional TT coin makes it match exactly, but simply having a probability of success which changes from 1/3 to 2/3 does not make a problem the same as MH.

I am trying to say that the subtlety of the MH problem is not shown in this problem: having chosen a coin, we make an observation about it, which obviously assists in identifying the coin. It is possible to observe a Tail, which would help considerably!

In MH, it does not matter which door we choose, it does not matter what we observe, the outcome is *always* the same:

– We should always change doors.

– P(seeing a goat) = 1

– P(success before seeing a goat) = 1/3 => P(success after seeing a goat) = 2/3

– We never know anything about the door we chose.

– In fact, we gain *no* new knowledge throughout the game.

– But it is important that MH knows where the car is.

To highlight the importance of the last point, I’ll propose a slightly different version: Micky Howell is not such a good game show host, and always forgets where the car is. I choose a door, and Micky opens one of the other doors where we see a goat. Should I still switch doors?

Yes … you should still switch. It matters not whether the host knows where the car is and picks the other door, or picks the goat by chance. You should still switch, because there is still a 2/3 chance that the car is behind one of the two doors you did not originally select. Monte’s knowledge really doesn’t play that much a role.

Foolonthehill, I showed that more than just the mathematics was the same. The random elements are completely parallel; it just takes a twist of logic to see how. I didn’t want to get into it, because it takes a while to explain properly. Essentially, the importance isn’t THAT he knows where the cars is, it is in how he uses his knowledge to determine if there are one, or two, doors he could open. In the 3-coin example, it would be the same if I picked the coin, looked at both sides of it without showing either to you, and told your that there was a Heads. That I know both sides isn’t relevant; how I picked what to tell you is.

But you are wrong about the MHP- you do gain new information. The two obvious pieces are that one door’s probability went from 1/3 to 0, and another went from 1/3 to 2/3. But that 2/3 value requires an assumption – it’s one that you should make, but it is an assumption nonetheless.

You meant that you gained no new information about your own door, but that is wrong, too. It’s just that what you learned didn’t change the value of the probability, based on that assumption. The assumption is that Monty Hall is equally likely to choose either door when both have goats. If, instead, the probability was Q that he would choose the door that he did, the probability your door has the car is becomes Q/(1+Q). But you should assume Q=1/2, which means this value remains unchanged at 1/3. The information changed, but the value did not.

All of these problems, and the Two Child Problem as well, are examples of a generalized Bertrand Box Paradox. That’s where a random selection has at least one of two properties, but may have both. If you learn about just one property the selection has, the probabilities that it has just that one, or both, depends on how the property to reveal is chosen when the selection has both.

These properties are obvious in all of the examples I’ve mentioned, *except* the MHP. People try to make the properties out to be CAR and GOAT and, noting the same discrepancies you pointed out, they insist it isn’t an example. But it is if you choose the properties correctly: Once you have chosen a door, the door to its right may have a different prize, or the door to its left may, or both may. When Monty Hall reveals one, you learn that your prize is different than the remaining door, but not whether it is different than Monty’s. Monty’s knowledge about where the car is only helps him decide which of these properties your door has. It is unimportant to the solution because your door had to have one, and there had to be some mechanism (here, Monty’s knowledge) to determine them.

@foolonthehill

You have missed the part where I stated ‘a bit like’ as in similar but not the same, the addition of the ‘bit’ qualifier make the difference even greater. Saying its not the same is kinda pointless, as no one has made that claim.

Sorry, I wasn’t trying to start an argument – let’s agree that it is “a bit like” in that both are in some way counter-intuitive. Furthermore, I agree that the Bertrand Box problem (which this could be expanded to) and Monty Hall are “similar” in that they both relate to conditional probabilities on three possible outcomes.

I still think my point still stands about a key difference between the problems (which is what makes each of them individually interesting). In this case, you have a 1/2 (1/3 in BB) probability of picking HH (or TT in BB!), which is improved to 2/3 once you know *this* coin has (at random) a Head – it is counter intuitive because it *could* still be either coin, so first thoughts lead to 1/2.

MH has a probability of 1/3 of picking the car. The probability that this door is the car never changes. “Solving” MH means realising that being shown a goat and offered the swap is simply offering you *both* of the other doors, hence swapping gets you a probability of 2/3.

For what it’s worth, I never found BB counter-intuitive, as it’s simple to enumerate the scenarios – MH had me completely confused. I’ll leave it there.

(@idahogie – if Micky/Monty chooses at random, then swapping gives you no benefit. Simply, one in three doors that Micky opens will show the car!)

Foolonthehill, there are two “values” associated with the MHP; the prize (car/goat) and the door number (1,2,3). You are fixated on the prize when you say it is fundamentally different than BBP or this coin problem. My point is that it is identical when you look at the door number. Monty Hall’s knowledge is equivalent to being able to recognize the difference between a heads and a tails, which is just as required in this problem as knowing where the car is to the MHP.

But if you find the BBP so simple, use it to answer this one: From an overheard conversation on the bus, you learn that a man has two children and that at least one of them is a boy. What are the chances both are boys?

@JeffJo (Having said I’d leave it at that….!) I need to say that I agree that that you are correct, in that they can be reduced to represent the same mathematical problem by choosing the suitable properties: in MH, you can say each door has a prize different to that on its Left (L), Right (R) or both (LR – the car). In the same way, you can choose a coin in BB which has a side matching the coin to its Left (L), Right (R), or both (LR)

The difference simply lies in interpreting the wording and to make it BB into MH we need adapt it by saying: you choose a coin (win by getting HH), and are then (intentionally) shown a Tail on one of the remaining coins. Should you swap to the remaining coin?

(Actually – this seems odd since, in MH, LR is the car, but in BB it is the HT coin. Why do I need to use HH (or TT) as the prize? I expect the properties I gave above need refining)

Anyway, let’s agree that once we have understood the problems, we can reduce them both to the same “probability question”:

– MH: What is the probability that the two doors you *don’t* choose have a car behind them?

– BB: What is the probability of finding a coin with both sides the same?

Thanks: I had never considered alternative properties for the doors.

@JeffJo Of course, it depends (and I know the Two Child problem too!).

If he tells me he has two children, there are 4 possibilities. If he answers yes to the question “Do you have a son?”, this tells me there are only three scenarios left, of which only one satisfies “both are boys”. If he simply tells me the sex of his (say) eldest child, then the sex of the remaining child is independent and remains 1/2.

The parallel with BB is with the former, with BG, GB, BB as the three scenarios against TT HH TH. Again, a problem I *intuitively* see as much simpler than MH.

I see what you are saying about flipping the “labels”. Rather than defining door A as the car, door B as Monty’s door & C as the remaining goat, We can define door A as my choice, B as Monty’s and C as the “swap”. This is a comparable set up to box A as HT, box B as TT and box C as HH – the observation of a H (ie that B is TT) is equivalent to knowing that B is a goat.

(I’m definitely going to leave this here – I have other things that need doing!)

I find it very interesting that Dave and Anonymous posted answers even though the rules of the game are explicitly not to do that. It would be good for them to let us know why they did it. One possibility is that they are new to this list and in a rush and don’t read instructions. That would be good both for the instruction maker and even more good for the people who may not be aware that they are careless.

They posted answers because they are intelligent enough to realise that people have the choice of whether they want to read the comments before trying to work out the answer (or not). I always try to work it out before reading the comments, because I am intelligent enough to understand that there might be spoilers in the comments.

Morris DeGroot, 1989

I just spent 5 minutes arguing with my brother over what the odds are.

In 48 hours, I will win the argument.

Well, I figured it out before scrolling down and seeing the inevitable bunch of muppets posting the answer

I’m not sure which is worse, the trolls who ‘inadvertently’ disclose the answers or the pseudo-intellectual, humourless, pedants who seem to a) take great pride in demonstrating their speed of thought (“Ha, solved in in 37 nano-seconds”) or b) showcase their ‘genius’ be referencing obscure sources or c) get sooooooooo stressed because the question was 0.00000000001% ambiguous.

To the latter, I say “go away” and leave these type of puzzles to us lesser dim witted mortals

Well, I’m not sure which is worse, to have people posting things they were asked not to post, to have people take such pride in their answers as to model themselves as geniuses, or to not have comment moderation turned on.

Dang, are you like ‘b’?

UC,

You have forgotten d) those people who complain about spoilers and say stuff like “what part of do ‘not post your answer’ do you not understand?”.

They are the worst….even worse than the Monty Hall obsessives – and that’s going some.

Good one

I think you’ve already spoiled this spoiler, eh Dave?

or did you and Anon cut ‘n’ paste the same source.

and after editing of the comments mine was left invalid,

then D’s neighbor spoils D’s secretary’s spoiler,

Does Dave’s dog have anything to say?

I might say “Arf”, but that would be incorrect as the proper answer is two thirds.

Classic puzzle that evokes a lot of controversy…like the Monty Hall problem. I’ll be interested to see the discussions after the reveal on Monday. Two different answers seem very plausible — One challenge is to explain why the incorrect answer is wrong.

or to explain why this puzzle is no different than just asking the probability of pulling the double-headed coin out of the box.

It is very different. Just imagine it said the face you first see is a tail instead – does that change the probability of what’s on the other side? Is that the same as asking the probability of pulling out the two-headed coin?

the probability of both children being girls, if one is a girl, is different from the probability of the second child being a girl, if the first is a girl.

@foolonthehill

You are being provocative by asking what chance of a head if I see a tail first. Obviously, 100%. Therefore you must want us to conclude that (as the full probabilities must sun to 100%) if I see a head first, then the chance of a tail is 0%,

You are trying to be a trickster with us!

My other post outlines the true analysis needed to solve this puzzler.

Ken

Please let us know what your two comments are on Monday

Thanks

TMT

ctj

“the probability of both children being girls, if one is a girl” is axiomatically zero.

100%

Oops. Forgot that part. Might need some short term memory puzzles. (Sorry delete if I could)

Took about 20 secs or so.

Surely the answer’s not 100%. What if it’s the H/T coin you pick first?

In the UK, the tail sides of coins all have animals on them as they show coats of arms. You will always see a head if you turn a coin over, no matter if it’s a double headed coin or not.

As the question is about whether the side contains a head, not if it is the head side, Stu is correct.

You could be right, though that would a silly trick.

I skipped over the comments so I wouldn’t see any spoilers, so I don’t know if I’m right or not. I wrote out the four possible draws from the box, eliminated the one that is irrelevant, then figured the chances of the other side being heads for each of the remaining draws and added up. I am terrible at probabilities!

Couple of seconds but then I am notoriously bad at these. Think I have avoided the logic trap which may mean I’ve jumped right into it!

For those who are wondering what Dave posted earlier, here it is:-

As I see it, when the coin is drawn, there are three possibilities regarding this first coin, each of which is equally likely:

Double-Headed Coin Heads

Double-Headed Coin Heads (the other head)

Ordinary Coin Heads

“Of the three, two of the possibilities will show heads on the other side; only one will show tails on the other side. So the probability that the other side of the coin is heads is two thirds.”

Hope this helps.

I’ve got the right answer in a few seconds, but I foresee lots of arguments over it.

About 5 seconds

The probability is 1 in 4,000.

I do not see this is difficult. It is the same as saying there are four coins in the box. Three gold (ie heads) and one silver (ie tales).

I draw one gold (heads) therefore there are three coins undrawn (or unseen sides). These are two heads (ie gold coins) and one silver (ie the tales).

So for a silver to be next is a 1 in a third percentage.

(Note I have answered my own restatement of the puzzle not the original. This is permitted as I can see by earlier comments)

50% in 5 sec.

http://www.viralnova.com/box-of-money/

Interesting puzzle: it made me argue with myself over the likelihood of 2 possible solutions. I do hope the side that won got it correct! We’ll see on Monday. Have a great weekend!

very quick. 5 sec

Dave, Self Important Twit, Dave’s Neighbour, Anonymous, Dave’s Secretary,

May I say, on behalf of the whole Spoilerphile community, how touched we are by your contributions to the thread this week.

As an oppressed minority, we are truly thankful for the wonderful comments that you have made.

Fight the oppression of the forces of darkness (i.e. spoilerphobes)

The present day Spoilerphile refuses to die!

This little piece of Java code confirmed my suspected answer.

final int N = 1000;

final int NORMAL_COIN = 0;

final int DOUBLE_HEADS = 1;

final int HEADS = 0;

final int TAILS = 1;

int nrInvalid = 0;

int[] nrDraws = new int[2];

Random rand = new Random();

for (int i = 0; i < N; i++) {

int coin, sideUp;

do {

coin = rand.nextInt(2);

sideUp = rand.nextInt(2);

if (coin == NORMAL_COIN && sideUp == TAILS) {

++nrInvalid;

}

} while (coin == NORMAL_COIN && sideUp == TAILS);

++nrDraws[coin];

}

System.out.println("NORMAL_COIN: " + nrDraws[NORMAL_COIN]);

System.out.println("DOUBLE_HEADS: " + nrDraws[DOUBLE_HEADS]);

3/4 I think

My long run in laws!!! They generate a adorable few will not they?!?!?

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