I have just launched a new experiment on magic – if you have 2 minutes it would be great if you could take part. Details here.

On Friday I set this puzzle….

You meet a nasty dog in a dark alley. The dog has a bag and the bag has a pebble in it. The pebble is either black or white. The dog then adds a white pebble and shakes up the bag. He then asks you to take a pebble from the bag without looking inside it. What are your chances of taking out a white pebble?

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

UPDATE: Having thought about this further, and had over 50 emails about the topic, I am of the opinion that the original answer is wrong. Sorry about that and thanks to everyone for pointing out the correct solution!

Original answer:

There are 3 possible scenarios:

1) The initial pebble was white. Thus both of them in the bag are white and so you will be left with a white pebble.

2) The initial pebble was black. Thus there is one white and one black pebble, and you remove a white pebble.

3) The initial pebble was black. Thus there is one white and one black pebble, and you remove a black pebble.

Thus there is a 2/3 chance of getting a white pebble.

Actual answer: .75 (see below)!

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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No! Wrong!

There is a 50% chance that both are white and a 50% chance that one is black, and one is white.

Of the second scenario there is a 50% chance that you pick a black one and a 50% chance that you pick a white one.

So there is a 75% chance that you pick a white!

That was what I got too.

This puzzle is similar, but slightly different. Did this cause the confusion?

https://richardwiseman.wordpress.com/2010/02/05/its-the-friday-puzzle-45/

3/4

That’s what I thought.

I got this too.

Yep, it’s 75% chance for the same reason M gave.

I’m with you guys.

Yep, Richard, time to revise your GCSE level maths 😉

I agree with M’s assessment of the probabilities. There are four possible scenarios.

1) The initial pebble was white. You pick the initial pebble, which is white.

2) The initial pebble was white. You pick the second pebble, which is white.

3) The initial pebble was black. Thus there is one white and one black pebble, and you remove a white pebble.

4) The initial pebble was black. Thus there is one white and one black pebble, and you remove a black pebble.

The probability you will pick a white pebble is 75%.

But your first two scenarios are the same from the probability point of view (yes they are different actual scenarios but not in the terms we are looking at it from) . You end up with a white pebble, we don’t care if you get the “first” or “second” white pebble, just that you got white.

So that then leaves us with 3 scenarios hence the 2/3 rds

But Ben, the first scenario can happen 50% of the time, while the second two scenarios each can only happen 25% of the time. So it’s still 75% white.

Ben, if there are 99 white pebbles and 1 black pebble in a bag then would you say there are 2 scenarios “from the probability point of view” if we don’t care which white pebble is chosen?

That’s what I was going to write. Thanks for saving me the effort.

Mike,

Yep, that’s what I got.

Btw, Ben: actually from a probability point of view those two events are unique. We can’t count them as one possibility just because to the outside observer the outcome appears the same. If you try to calculate probability as you’re suggesting, you should quickly find that your calculations don’t match your actual results.

Option 1: a white is added ; 50% chance

Option 2: a black is added ; 50% chance

@ Option 1: pick a white one : 100% chance

@ Option 2: pick a white one : 50% chance

pick a black one: 50% chance

Black = 50% (option 2 by the dog) x 50% (black pebble)= 25%

On the other hand, the fact that this actually happens is rather close to 0% as dogs don’t talk….

2/3 was an answer I rejected early on… I believe it is 75% also!

To be honest, I’m more worried about why a nasty dog is walking around with a bag of stones.

2/3 is wrong, it should be 75% as other pointed out.

The answer is 75%.

Agree with those who say 75%. Yours aren’t equally likely outcomes.

I think 75 %.

Original = O New = N

either

O = white and N = white

or

O = black and N = white

The possibilities are O or N, so if N is picked, we know it is automatically gonna be white.

So,

O = wh but N picked = 100%

O = bl but N picked = 50/50

O = wh and O picked = 100%

O = bl and O picked = 50/50

So, adding up the variables, isn’t it 75% probable, as 100+50+100+50 = 300/4 = 75?

This answer is surely wrong. The reason is that it is counting only one sub-scenario when the initial pebble is white, whilst counting two different when the initial pebble is black.

If we assume the initial scenario has an equal likeliness of the pebble being black or white (we are not told explicitly) then there are only two possible configurations when the (white) pebble is added with equal probability (50%).

1) WW (both pebbles are white)

2) BW (the first pebble is black)

So if we take a pebble from in scenario (1) we might choose either the first or second white pebble which means that each of these two scenarios have a 25% possibility (50% in total).

If we take the second scenario, we might take either the black or white pebble with equal probability. Each has a chance of 25%.

So, adding this up there are 3 possibilities of choosing white (at 75%) and 1 possibility of black (at 25%).

So the chances are three-quarters that the pebble chosen is white.

I’m now going to run off to write a quick simulation, but I’m pretty confident this analysis is right.

The probability is surely 75% as others have

i got 75% too.

Basically there are 2 possibilities:

1. the bag has one white stone and one black stone.

2. the bag has 2 white stones.

for option 1 the chance to extract a white stone is 50%

for option 2 the chance to extract a white stone is 100%

so the chance to extract a white stone is 75%.

I realise you we don’t care if you get the first or second pebble, but the point is that there are two of them and Richard’s mistake (and yours) is to count this as just one.

yes, i don’t care if i get the first stone or the second one, because in the case with 2 white stones, i can’t determine that.

Another explanation:

– for the bag the probability to extract a white or black stone is 50/50.

– when you add the white stone you modify the probability to:

50% to extract the white stone you just added

50% to extract the stone that was already in the bag

so, the probability for the white stone is:

50% + 25% = 75% ( 25% – because it is a 50/50 chance that the stone from the bag to be white )

I decided to go practical.

Using an excel sheet and random function, I generated 10000 cases. I generated first pebble black or white at random, added one white pebble and then selected one pebble at random. Out of 10000 cases, I selected 7462 times the white pebble.

So the answer is 3/4 as most of the viewers believe and not 2/3 as the author suggests.

yep. it’s 75%.

1. you have a 50:50 chance (0.5) to either pick the white stone or the unknown stone.

2. the unknow stone has – we assume – a 50:50 chance (0.5 again) to either be black or white.

0.5 * 0.5 = 0.25 for the black stone.

-> 1 – 0.25 = 0.75 = 75% for the white stone

2/3 rds is correct

No it isn’t

Thanks for the inciteful analysis, guys!

*insightful

But well said!

D’oh!

Or perhaps I meant http://www3.merriam-webster.com/opendictionary/newword_search.php?word=incite …?

🙂

I just ran a monte carlo simulation for 10,000 iterations and got 74.76% white, which is well within the confidence interval for an answer of 75%.

Richard Wiseman is right !

Without the information that there is at least one white pebble there would be four possibilities:

WW

WB

BW

BB

The chances of taking out a white pebble would be 1/2.

But since BB is not possible, we only have these three possibilities:

WW

WB

BW

The chance is therefore 2/3

WB isn’t possible either: the extra information tells us that the SECOND stone is white.

So the only possibilities are WW and BW.

Which gives 75% white again.

Yeah, what Ed said. I knew the answer was 75% the yoctosecond I read the puzzle, thought about it for 10 seconds and kept my 75%.

i wonder what schrodinger would have to say on this…

I agree with all of those who say a 3/4 chance. However, we are all relying on the assumption that the probability that the initial pebble is white is the same as the probability that it is black, however without of any other information this is a fairly reasonable and logical assumption to make.

Oh no. It’s Monty Hall all over again.

Talking dogs posing questions with bags of pebbles? I believe that the whole thing must be a dream, and therefore, you have a 38% chance of pulling out a poisonous snake and 17% chance of pulling your own father out of the bag.

Using Richards logic, the chances of surviving jumping from a plane with no parachute is 50% – either you make it, or you don’t. Come on Richard, admit it that this puzzle/answer is a wind up. It’s at least 50% likely it is after all!

…however, if the dog was a threat if I didn’t pick White and if the pebbles were non-identical and i’d seen the added White pebble, then I’d put the odds of me picking the White pebble just added at close to 100% since a selection from a bag is not random as it’s not ‘blind’ since you can feel the choices.

So the answer is “any number between 0 and 100 inclusive.

Everyone is assuming that the various alternative scenarios are equiprobable, i.e. that the probability of there originally being a white pebble in the bag was 50%. Yet, there is nothing in the description of the problem that indicates this is necessarily the case. As such, 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 %.

Indeed, but given that we weren’t supplied with this information, it’s a reasonable assumption to make that the probability of it being black or white are equal

It’s a fallacy to believe that if there are two outcomes from a process then they’re equally probable – so it’s not a reasonable assumption.

And if there are three outcomes, then you can’t assume equal (one third) probabilities either, as Richard assumes.

If information is missing you can’t just make it up. (I wonder if Richard has anything to do with setting A-level papers.)

I’d go with the 75% group too. The bag is either WW or WB, with a .5 chance of either. So to get the black is .5x.5.

I had 75% too, I wonder where Richard got this puzzle + answer from?

I have to disagree with you here Richard, and side mostly with M and the gang! While there are 3 scenarios here, and you defined them all perfectly, if you performed this experiment on a large group of people, scenario 1 would occur twice as often as either scenario 2, or scenario 3.

To be more specific, the probability of picking a white pebble is given by:

P(picking white pebble) = P(initial pebble white) * P(pick white pebble in this case) + P(initial pebble not white) * P(pick white pebble in this case).

In shorthand:

P(W) = P(A) * P(W/A) + P(A’) * P(W/A’)

P(W/A) is always 1 (picking a white pebble from a bag of two white pebbles),

P(W/A’) is 0.5 (picking a white pebble from a bag of one white and one black),

P(A’) = 1 – P(A)

So we get:

P(W) = P(A) * 1 + (1 – P(A)) * 0.5;

P(W) = 1/2 (1 + P(A))

This way, we can even figure out the probability based on different chances of the initial pebble being black or white, or reverse engineer your 2/3 answer to find what probability was needed:

If P(W) = 2/3;

1/2 (1 + P(A)) = 2/3;

1 + P(A) = 4/3;

P(A) = 1/3

For the probability of picking a white pebble to be 1/3, the initial pebble would have to have been white 1/3 times.

No no, the dog really likes white pebbles, and only rarely (in 5% of the cases) does he carry around a black one. So the probability to get a white pebble is 97,5 %.

75% is what I got too. The three possible outcomes listed aren’t of equal probability.

Sorry Richard, you’re wrong this time. As well as analysing it as above, you could also think about the prob of getting a black pebble. To get black you need to pick the uncertain pebble (50%) and it needs to be black (50%). So the prob of picking that pebble AND it being black is 25%. All other outcomes are white, so prob of white is 75%.

Assuming the initial pebble is equally likely to be black or white, the answer is 3/4.

The problem with Richard’s analysis is that it assumes the three scenarios he presents are equally likely. However, assuming the initial pebble is equally likely to be black or white, the likelihood of scenario 1 is equal to that of scenarios 2 and 3 combined.

An easy way to see that the answer must be 3/4 is to double all quantities involved. Thus the bag initially contains one black and one white pebble. Two white pebbles are then added. Clearly the probability that a white pebble will be drawn is 3/4.

From a frequentist perspective, it seems to me the answer is either 100% (one scenario) or 50% (another scenario). To a frequentist, probability is the long-term frequency of future events. The identity of the initial pebble, is not a matter of probability because it is not a future event. It is what it is (even if we don’t know what it is). If there is a white pebble initially in the bag and a white pebble is added, then the probability of drawing a white pebble is 100%. If there is a black pebble initially in the bag and a white pebble is added, then the probability of drawing a white pebble is 50%.

Here’s a similar example from a frequentist perspective: If I flip a coin and cover it up, what’s the probability that the flip yielded heads? Well to a frequentist, the question doesn’t make sense since the event has already occurred, it isn’t a matter of probability, and the coin is oriented whatever way it is oriented. It’s not a 50-50 scenario. But change the question to: What is the probability of GUESSING correctly heads or tails? Then the answer is 50-50 because the guess has not yet been made.

Getting back to the pebbles, Professor Wiseman’s answer is correct, but only if you look at the problem from a Bayesian perspective on probability, which is degree of confidence in a particular assertion being true and not as a matter of long-term frequency of future events (as viewed by frequentists). To a Bayesian, the answer to my first coin flip question is 50%. Most people answer that question like Bayesians.

So how does a Bayesian analysis of Richard’s problem yield an answer of 2/3?

The only way a Bayesian analysis would yield a 2/3rds probability of a white is if the probability of the first pebble being black was 2/3rds (and the probability of the white being 1/3rd). As we weren’t explicitly given the odds then the most people have made the reasonable assumption that these were equal. If these odds are not equal we can only say the chance of picking a white pebble is somewhere between 0.5 and 1.

I stand corrected, Nick. Yes, 3/4.

Another way to solve it is, consider you run this experiment N times, generating stones, not removing them.

You would then have in you sack (roughly if N is large enough):

N white stones + N/2 white stones + N/2 black stones = (3/2)*N white stones and N/2 black stones. Or more simply a ratio of white to black of 3:1 This then means the probability of picking the white stone is 3/(3+1) = 75%

50% is the only correct answer, as anyone who reads the puzzle thouroughly can tell.

My reasoning? The dog is a NASTY dog. It wouldn’t put a white pebble in as the first pebble, as that would give 100% chance of the player (you) winning. Therefore the initial pebble must be black and the chances of selecting the white pebble are 50%

*thoroughly (line 2)

You are making the assumption that the dog cares whether you “win” or “lose”, and the assumption that the game has a winning or losing outcome. It doesn’t say either, only asks what the chance of a white pebble is, without saying whether this is the desired outcome or not.

the dog doesn’t appear to care whether a white or black pebble as there appears to be no difference in outcome. Indeed it’s not even the dog asking the question…

However, it does seem strange that a nasty dog asks you to pick a pebble rather than ordering you to do so.

it is possible the dog is so nasty that he knew you’d assume that and thus drop a white stone in with another white stone to ensure that your guess of 50% 50% would be incorrect.

a really nasty dog would use white and black cats, not pebbles

Tell you what….unless Richard decides to post a rebuttal of the many, many “75%ers”, it won’t help the sales of his puzzle book.

Let’s see….if 50% of his solutions were correct, but he pulls out one incorrect solution for a promo, what is the probability that I will actually shell out money? 75% or 66%?

One interesting point is that maths is not a democracy. FWIW I agree with the 75% crowd.

No, you are all wrong and richard is right. You are all making a major error. You can’t use the calculations or simulations u are all suggesting because you dont know the other color of the marble. So it doesn’t mean that 50% of the time it’s white and the other half its black, u don’t know! Because those odds don’t count u can’t count 2 of the same outcomes when ther are 2 white marbles in the bag. The odds are 2 out of 3.

No, you can’t just ignore a whole set of probabilities because you made an assumption on them!

you are quite right that the probability cannot be calculated unless you know the probability of the initial pebble being black or white. In that case we could only state that the chance of choosing a white pebble are between 0.5 and 1.0. However, it’s perfectly reasonably to make (and state) the assumption that the chance of the first pebble being white is 0.5. There is no such reasonable assumption to assume that the chance of the first pebble being white is 1/3, which is what you need to get an overall probability of picking a white being 2/3rds.

There are wordings of similar, but not identical problems which do yield 2/3rds as the answer but not this one.

I have also run a number of simulation runs of this problem and these all yield close to a 75% selection of a white pebble (to within normal statistical levels of significance).

Either you have to make an assumption about the probability of the original stone, or you can’t. If you do, the only reasonable assumption is 50/50, which gives an answer of 75%. If you don’t, all you can say is the answer is between 50% and 100%. In both cases Richard is wrong.

No, you are all wrong and richard is right. You are all making a major error. You can’t use the calculations or simulations u are all suggesting because you dont know the other color of the marble. So it doesn’t mean that 50% of the time it’s white and the other half its black, u don’t know! Because those odds don’t count u can’t count 2 of the same outcomes when ther are 2 white marbles in the bag. The odds are 2 out of 3.

You are right in the statement that there is no reason to assume that there is 1 in 3 chance of the pebble being white. But ther is also no reason to assume it’s 50% white, so all simulations are based on assumption. Without making any assumptions we go back to the 3 possible outcomes richard gave. Because u lack vital information to calculate it that way

Hence why most replies start with ‘Assuming a 50% chance of the pebble being black or white’ or similar words. You have to make an assumption to solve the problem at all, and 50/50 is the most reasonable one to make.

Yup, 75% according to this little Python program:

I’m afraid that proves nothing. That could conceivably generate 100% white results, but that wouldn’t mean the chances of drawing a white are 100%.

Dave S – Yes, while it is not proof per se, it is strong evidence, assuming the random numbers are good.

I posted the source so that everyone could see the mechanisms involved, which were as close as I could make them to the original within the Python language and 10 minutes. Repeated runs consistently give values around the 75/25 mark, none were anywhere near the claimed 67/33 split.

Congratulations, that is _the_ most inefficient piece of python code I have ever seen.

I got 75% but didn’t think it through too much. Followed the same logic as most people above did. I like the ones who have done a practical experiment to get the answer. Seems I got it right then.

I fear that, while most people’s statistical calculations are correct (that the probabilty of choosing a white pebble would be 75% in the narrow sense of the problem), people are forgetting an important fact subsumed within the context of the problem. The agent placing the pebbles in the bag is a nasty dog, and there is some evidence to suggest that dogs are colourblind, as such we have to assign values to the probabilities that one might choose a yellow pebble (with the dog thinking it to have been a white one), or perhaps a dark blue pebble (with the dog thyinking it to have been a black one). This involves a lot of number crunching like calculating the ratio of pebbles in the general vacinity of the dog that looked black to the dog but were various other colours. As such I think we can give Richard a bit of a break for missing the target by a little bit!

No mercy

The fallacy in Richard’s explanation is that he’s given all three of his scenarios the same probability: ‘there are three possibilities, two of them are what we’re looking for, therefore probability is 2/3.’

If you look at those cases closely you’ll realise that scenario 1 covers ALL cases where the initial pebble is white, while scenarios 2 & 3 must cover all cases where it’s black. Assuming (a priori) a 50/50 chance of white/black initially, the probability of scenario 1 must equal the probability of scenario 2 and 3 combined:

P(1) = P(W) = 0.5

P(2) + P(3) = P(B) = 0.5

Assuming P(2) = P(3) (fair, as we have an equal chance of pulling any pebble from the bag), that gives

P(2) = P(3) = 0.25

And so the probability of pulling a white pebble = P(1) + P(2) = 3/4

I have read somewhere in discussions of the Monty Hall problem and others that humans often fall for this fallacy as it’s tempting to just count the possibilities and assign them each an identical probability as Richard seems to have done.

Monday morning coffee needed? 😉

imagine a game of plinko:

drop your chip

/\

/ \

/ \

(1w 1b) ( 2w)

/\ /\

/ \ / \

(b)

looks to me that 3/4 times you plinko chip will fall on a and if you somehow see that 2/3 times that will happen,…Do share it with me.

BZZZT, thanks for playing Richard. The mathematically complete and correct answer is this: “The question as stated cannot be answered. The probability requested does not exist without knowing the initial probability of white/black for the first pebble. If we make the reasonable assumption of 50/50 for that value, then the final probability is 75/25. Other values will give different answers.”

aren’t there 4 possibilities? let’s say pebble A is in the bag and pebble B is the white one being added.

1. pebble A – white, pebble B – white

1a – you pick pebble A – white

1b – you pick pebble B – white

2. pebble A – black, pebble B – white

2a – you pick pebble A – black

2b – you pick pebble B – white

75% of the time you’ll pick white.

Thanks!! I will buy a lottery ticket tonight. I can only win or lose (2 different cases), and a 50% probability of winning a prize just looks great; I had never thought it this way, silly me.

You should buy two tickets just too make sure…!

Love how everyone gets hung up on the talking dog. I think this may be the more interesting aspect of the puzzle.

I make it 75% as well, and I think Veli is right, that Richard’s scenario 1 is in fact 2 scenarios because there is two ways of choosing a white pebble there. I suspect this is what he missed.

Susqueda: Are there only two tickets in that lottery? That’s the only way I can see of having a 50% probability of winning

it was irony

Yup, answer is wrong.

Possibilities 1 2 and 3 are mutually exclusive events. 1 has probability 50%, 2 and 3 are equally likely and together have probability 50%. Thus, when you add up the probabilities of 1 and 2, you get 50%+25%=75%.

OH, I’m a different M to the first M!

I probably should name myself M (first) then… 🙂

It looks as if Richard confused this problem with the Monty Hall Problem. The main difference between those two problems is, that Monty Hall knew which door was right, and which wasn’t. In that case the chances change from 50:50 to 1/3:2/3. If Monty Hall had from time to time opened the door with the car behind (“Oops, bad luck for you!”) the chances would have remainded 50:50.

Assuming that the initial chance of the white pebble in the sack is 50% we will end up with a 75% chance – unless we assume that the dog retroactively changes the colour of the pebble according to the pebble you pick 😉

You never know what these wily dogs will do. Personally, I prefer cats.

The following C# program runs a simulation of what actually happens in the puzzle, and the answer definitely comes out as 25%:75%.

using System;

namespace DogPebbles

{

class Program

{

static void Main(string[] args)

{

char[] Pebbles = new char[2];

char ChosenPebble;

double WhiteChosen = 0, BlackChosen = 0;

Pebbles[0] = ‘W’;

Random R = new Random();

for (int i = 0; i 0.5)

Pebbles[1] = ‘W’;

else

Pebbles[1] = ‘B’;

if (R.NextDouble() > 0.5)

ChosenPebble = Pebbles[0];

else

ChosenPebble = Pebbles[1];

if (ChosenPebble == ‘B’)

BlackChosen++;

else

WhiteChosen++;

}

Console.WriteLine(“Black: ” + BlackChosen.ToString());

Console.WriteLine(“White: ” + WhiteChosen.ToString());

Console.ReadLine();

}

}

}

Arg – wordpress has mucked up my code.

Line 17 needs to be a loop over a large number of steps with the next line checking the value of the 50-50 random number…

The Python guy got there first. And still, this proves exactly nothing.

Not sure why you think that – this accurately simulates what actually happens in the puzzle, and if you run it over 100,000,000 simulations, you get extremely close to 75% every time.

Or is that a fluke? Every time?

Is there another way of simulating this that you could tell us about?

It “proves” nothing, but it does strongly “demonstrate” the likely solution.

Dumb question from me, but how does this differ from the classic “boy/girl” problem? In that case the man knows he has at least 1 boy + a boy or girl; we know we have at least 1 white (which we add) + a black or white (already in the bag). That problems gives it a 2/3 prob of a girl (‘black’) being his other child.

Just a bit confused on a Monday am.

If a father knows he has at least 1 boy + a boy or a girl, there are these possibilities

boy + boy

boy + girl

girl + boy

so in two of three cases a girl is involved. In this case you have two marbles in a bag. You know one of them is white, and you pick up a white. There are these possibilities

white + white -> pick up white 1

white + white -> pick up white 2

white + black -> pick up white

white + black -> pick up black

in 3 out of 4 cases you pick up a white. The difference is in the selection. In the boy/girl case you’re only asking “how often is there a girl?”, here you are asking “how often is there a white marble *and* how often will I get it when I pull out a marble at random?”. There is a black marble in the bag in 1 cases out of 2, but you get a white marble in 3 cases out of 4

To expand on what Anders said, the Boy/Girl problem is about conditional probabilities: if you hear that a family has two children and at least one son, then there’s a 1/3 probability that the other child is a son since you don’t know whether they’re talking about the elder or younger child; conversely, if you ring the doorbell of a family with two children and a boy answers the door, there’s a 1/2 probability that the other child is a son since in this case we have a way of distinguishing between the two children.

So, in this problem, knowing that the white marble was added at stage two distinguishes them and so doesn’t muck up the probabilities.

If we change the end of the question to: “You pull out a white marble. What is the probability that the original marble was white?”, then the answer would be 2/3.

I knew this would happen as soon as I read the puzzle on Friday. It’s 75% that you will draw a white pebble. However, this puzzle remarkably similar to Lewis Carroll’s Pillow Problem. The wording is the key. In the Lewis Carroll’s Pillow Problem, you draw a white pebble FIRST, and then you have to calculate the conditional probability that the other pebble inside the bag is white. As it is worded currently, the Friday puzzle doesn’t match the Lewis Carroll Pillow Problem because you are calculating the chance that the first drawn pebble is white.

Of course the nastiness of the dog is irrelevant. But there is a chance to confuse this problem with another closely related, but distinct, problem, that of the three doors. I think the answer is 3/4, but the answer to the three door problem is 2/3.

You have three doors. Behind one door is a car and behind the other two is a goat. You pick a door. The door isn’t opened. Then Monty picks another door behind which is a goat. Your first choice had a probability of 2/3 of being right, that is, the door with the car. Now, do you switch or stick with your first choice? You can’t do better than 2/3 but you could do worse.

The dog with the balls in the bag is not quite the same. The difference is what makes the answers different. Hence, 3/4 rather than 2/3.

erm, your first choice had a probability of 1/3, and if you switch you have a 2/3 probability of winning, but you left out one very crucial detail: Monty will always open a non-winning door among the remaining two doors, excluding the one you chose. Unless you include that piece of information, the probability calculation becomes drastically different

I wonder if this was actually a psychology experiment to examine what proportion of people seem to REALLY enjoy telling someone else that they made a mistake.

Amazing! This is becoming dangerous….

If it’s any consolation Richard, the very first question in Prof Ian Stewart’s “Cabinet of Mathematical Curiosities” has an error in it (at least in the hardback version I have) with the analysis in the answer flatly contradicting the wording in the question.

Great credit to Richard for correcting his original answer. To be fair it has generated some wonderful comments. Can’t wait till its friday again!

Well I did get the right answer but to be I’m really still more concerned about the talking dog!

Roll on Friday 🙂

2/3 chance

75. % probability.

Nobody is wrong.

That makes no sense! ‘chance’ and ‘probability’ in this context mean the same thing. The answer is 75%!

2/3 = ~66.7%, not 75%

There are only 3 pebbles to choose from.

If there were 100 a percentage would look better

Therefore you have a 2/3 chance.

75% works. Also.

I knew the answer was .75 and I knew that I would be told that was wrong. At least he changed it.

I had the right answer, but I would have gone along with the wrong solution if Richard hadn’t changed it. Such is the confidence I have in my math skills.

The answer is 75%. In simple puzzles like this, we only have to count the possible scenarios, and then see which of them fits the outcome we’re being asked to assess. The possibilities of what’s in the bag are:

W W

B W

If you withdraw a pebble at random, 3/4 of the time you’ll get white, and 1/4 of the time you’ll get black.

Actually, it’s still technically wrong, because we’re never told there’s a 50/50 chance of there being a white or black pebble in the bag originally, so cannot assume. All we know for sure is that there’s a greater than 50% chance of a white pebble.

Example: The dog, before walking up to you, used a random number generatior to generate a number between 1 and 50. On 1, he would have put a white pebble in his bag, otherwise, a black.

The probability is then 51%. By reversing that method (e.g. black on 1, white for 2-50), it’s 99%

We do not know how the dog selected the first pebble, so the correct statement is one of not knowing, not an exact number.

3/4 is right. Draw a tree digram if in doubt…

Spelling correction… 3/4 is right. Draw a tree diagram if in doubt…

Yeah 3/4 here too… assuming the nasty dog was honorable and picked the first stone by a fair method rather than rigging it

I think it’s 75%, but that is based on the assumption that the first pebble is equally likely to be black or white. All we’re told is that it’s either one or the other. It might not be 50/50. If it isn’t, then the probability of finally drawing a white pebble is in fact P(white) + 0.5P(black). If it’s 50/50, the answer will be 0.5 + 0.5(0.5) which is 0.75.

Obviously, if you are interacting with a talking dog, you are dreaming. Therefore your chances of pulling out a white pebble depends on how much your subconscious thinks you deserve it.

On the other hand, if you are actually a fictitious character embedded in a logic puzzle using probability, and we assume a 50% chance of the first pebble being white, then half the time you are guaranteed a white pebble and the other half the time you have a 50/50 shot at a white pebble. (50% * 100%) + (50% * 50%) = 75%

I can’t believe he got this wrong. It’s so simple!

Personally I was flipping between the two answers, so to resolve it I imagined the following:

The dog has 20 bags, 10 that have one black marble each and 10 that have one white marble each. He then adds one white marble to each bag.

10 bags have two white marbles, so I’ll pull a white marble from those no matter what.

Of the other 10, I have a 50-50 chance of getting a white marble, so statistically I should get 5 marbles.

10+5 = 15, and out of 20 that makes 75%.

An interesting sideline to this: can you reword the original question so that – with exactly the same pebble/bag scenario – the correct answer to the question *is* 2/3?

The dog, instead of offering you first pebble, pulls one out and it’s white, and invites you to calculate the probability that the remaining pebble is white as well

If I’m not totally mistaken, this removes one of the four possibilities, black/white => pulls out black, leaving only three scenarios, in two of which the remaining pebble is white, so 2/3

Not sure what you mean by “exactly the same scenario” though. If you mean that you still get first choice then the problem needs something else, I can’t immediately think what

How about the following fairly minimal change…?

You meet a nasty dog in a dark alley. The dog has a bag containing one white and two black pebbles. He asks you to take two pebbles from the bag without looking inside it. The dog then adds a white pebble and shakes up the bag. He then asks you to take a pebble from the bag without looking inside it. What are your chances of taking out a white pebble?

The correct answer to the question is now 2/3.

Everyone is wrong. you have fallen for the oldest trick in the book. its far too easy to say its 75% or there is a 3/4 chance of picking a white pebble, but you have all made the fatal mistake of forgetting the FULL question….

the first line clearly states “you meet a nasty dog…” by definition the dog is evil and is upto no good. leaving aside the fact the dogcan talk and is able to put a pebble into a bag and shake it up, (used his mouth or does this dog have thumbs?) he is a nasty dog and that means that the chances of you getting out of that dark alley alive let alone choosing a white pebble are slim to none – not 75%.

sorry folks but i win!

Yep. I got 0.75. My reasoning was similar to Roland’s. I pictured two boxes representing the two possibilities for what was in the bag. The first box contains WW and the second box contains BW. The probability of picking a W from the first box is 100% and the probability of picking a W from the second box is 50%. Since the probability of picking either of the two boxes is assumed also to be 50%, the overall probability of picking a W is:

0.5 x 1.00 + 0.5 x 0.5 = 0.75 or 75%

Still waiting for the catch……

Hummm, I got 100% because I will chose the wet ball and not the dry one!

I think that the “nasty dog” will probably bite your arm off before you even get to the bag, so does it matter what’s inside?

I agree with .75 but on a second thought I felt if the alley you have mentioned is dark how on earth can I see if the stone I picked up is black or white!

I think .75 is correct. But only in the sense that you can apply the indifference principle to the truth of any statement where you have no indicative knowledge.

This is a _very_ tricky use of the principle of indifference, where the way we express our knowledge affects the derived probabilities. The more possibilities we choose to enumerate in our formulation of a a problem, the less likelihood they each suddenly get. In this case: if the dog could have carried “a white or a non-white stone”, each would still have p=1/2, but if the options had been “a white, a black or a purple stone”, each would have p=1/3. Even though the first formulation implies the second. Ouch.

This sounds like a paradox, but not really. The reason p = 1/2 in the first case is that you have two stones. The reason that p =1/3 in the second case is that you have three stones. The probability of non-white in the first case is 1/2, but the probability of non-white in the second case is 2/3. Completely consistent.

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The case with M pebbles in the bag

0.5 chance for the M pebbles to be white -> 100% chance to extract one white pebble

0.5 chance for the M pebbles to be black -> 1/(M+1) chance to extract one white pebble

Summing it up the chance to extract a white pebble is (M+2)/(2M+2)

For M == 1 we have 0.75

Interesting. That means:

limit {(M+2)/(2M+2)] = 1/2 as M approaches infinity. In other words, for very large M (the number of white pebbles in the bag), a bag containing one black and M white is like a bag with all white pebbles. Not very surprising is it?

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