Not able to understand the solution of this puzzle

[musicgold]

Hi,

I am not able to comprehend the solution of a problem from a puzzle book.

Your friend shows you an urn and tells you that there is one marble in the urn. The marble is either black or white. (Assume that the marbles are identical in shape and any two marbles with the same color are indistinguishable.)

She then drops one black marble in the urn. Then sticks her hand in the urn and takes out one marble randomly, which turns out to be black. Now she asks you what are the chances of drawing a black marble from the urn.

According to the book the probability of drawing a black marble is 66.6%.


My answer was 50%. Here is my logic.
1. Assume that initially there was a black marble (b1) in the urn. So when the friend drops another black marble (b2), the urn has b1 + b2. Now when she takes out one black marble , the probability of drawing another black marble is 100%.

2. Now assume that initially there was a white marble (w1) in the urn. So when the friend drops another black marble (b2), the urn has w1 + b2. As she takes out a black marble, it has to be b2. So the probability of drawing a black marble is 0%.

Therefore, my answer is the average of the 100% and 0% chances.

 

[tamtam402]

nevermind :D

 

[Whovian]

I think the problem is you're assuming there's a 50% chance of having a black marble and a 50% chance of having a white marble at the start of the experiment. You're not given any probabilities, you're just given that your friend draws a black marble.

This is related to the concept of in Deal or no Deal (at least, I think that was the game show), it's a better idea to pick the other door.

 

[DonAntonio]

I think the problem is you're assuming there's a 50% chance of having a black marble and a 50% chance of having a white marble at the start of the experiment. You're not given any probabilities, you're just given that your friend draws a black marble.

This is related to the concept of in Deal or no Deal (at least, I think that was the game show), it's a better idea to pick the other door.

Well, in fact the solution is 0.75 under the assumption that the probability of the first marble in the urn being black is 0.5...!

This problem's solved pretty easily with a probability tree: first there's a prob. of 0.5 to have a black marble in the urn and,

of course a prob. of 0.5 the marble is white. After that, we ll know the second marble added to the urn is black, so now

we have the branches:

B --> N , with probability $\,\displaystyle{\frac{1}{2}\frac{1}{2}=\frac{1}{4}}\,$ , and the only other relevant branch

which is $\,N --> N\,$ , with probability $\,\displaystyle{\frac{1}{2}\cdot 1=\frac{1}{2}}$ , so the final prob. to draw a black marble is $\,\displaystyle{\frac{1}{4}+\frac{1}{2}=\frac{3}{4}}$ , as

the book correctly states.

DonAntonio

 

[Whovian]

Actually, the book states it's 2/3, not 3/4. Maybe the book's rubbish?

 

[DonAntonio]

Actually, the book states it's 2/3, not 3/4. Maybe the book's rubbish?

Perhaps, or perhaps it contains an error, or perhaps I made a mistake. It'd be interesting to know what book is that...?

DonAntonio

 

[DonAntonio]

Perhaps, or perhaps it contains an error, or perhaps I made a mistake. It'd be interesting to know what book is that...?

DonAntonio

Rats! Re-reading the OP I realize now what part I misunderstood: the question is "what is the probability that the second marble in

the urn (the one that's left after the first one is drawn) is back"..! They don't mean the first marble already drawn!

Well, then: we havet thus 3 possible cases:

1) The urn contained a white marble, a black one wass added which was then first drawn: probability to draw again a black marble: 0

2) The urn contained a black marble (say, N1) and another black one (say, N2) was added. Probability to draw again a black marble after N1 was first drawn: 1

3) As in (2), but now we want the prob to draw a black marble after N2 (care here!) was first dranw: 1

As we can see, in two out of the three cases above we'll have a black marble left in the urn after the first one was drawn and, thus

the prob. is 2/3 = 0.6666, as the book says.

And btw, this indeed reminds the game Whovian mentions, which I recall as being Monty Python's contest game (one can even google this).

DonAntonio

 

[musicgold]

3) As in (2), but now we want the prob to draw a black marble after N2 (care here!) was first dranw: 1

This is not clear to me.

 

[viraltux]

The book is right,

One way to see this is as follow, imaging she does not draw any marble and ask you about the chances to draw a black one, the you could say you have two scenarios;

1 - W B
2 - B B

Then you would say 3 B out of 4 possibilities, so the chances to get a black marble will be 3/4.

Now, you friend takes one B away and you have:

1 - W
2 - B B

or

1 - W B
2 - B

So your chances to get now a black marble are 2/3. If you don't see it this way just repeat the experiment over and over, then you have

she takes B1 from B1W
she takes B1 from B1B2
she takes B2 from B1B2

So in the long run you have W,B2,B1 so, 2B and one W, and therefore the probability is 2/3.

But the interesting thing about this problem is why it is so unnatural for us humans to see the right the answer.

 

[musicgold]

I think my question is not clear.

Initially the urn has only one marble. There is a 50% chance that it is black and a 50% chance that it is white. The friend adds one black marble to the urn.
Then she randomly draws one marble for the urn, which happens to be a black marble. Now she asks : what is the probability of drawing a black marble from the urn now?

 

[viraltux]

I think my question is not clear.

Initially the urn has only one marble. There is a 50% chance that it is black and a 50% chance that it is white. The friend adds one black marble to the urn.
Then she randomly draws one marble for the urn, which happens to be a black marble. Now she asks : what is the probability of drawing a black marble from the urn now?

2/3

 

[jedishrfu]

there is a related problem of a princess facing marriage to a king that she hates. The king suggests a game to make it fair. If she can select the white stone from the pot then she can go free but if she selects the black stone then she must remain as his queen.

She notices that the king has cheated and placed two black stones in the pot. What should she do?

 

[viraltux]

there is a related problem of a princess facing marriage to a king that she hates. The king suggests a game to make it fair. If she can select the white stone from the pot then she can go free but if she selects the black stone then she must remain as his queen.

She notices that the king has cheated and placed two black stones in the pot. What should she do?

Kick his nuts and run.

 

[viraltux]

All right musicgold,

In fact I find more interesting why people don't see the solution that the problem itself.

In this case your mind is going around and around the fact that there is only one marble left, that marble is either white or black, and therefore the probability for it to be black is 1/2, right? Well... wrong!

Think about this experiment, imagine a room full of people in which 2/3 are women and 1/3 are men. Then I pick one randomly and I ask you about the probability of being female.

Would you say, "well, that person is either male or female, therefore 1/2"? No, you wouldn't because you can see clearly here that the experiment favors females with 2/3. Well, so does the girl in your problem, she favors black marbles when she places one in the urn, then the probability to draw a black marble is 3/4, and when she draws one black marble she still favors black but not so much, this time 2/3.

You need to think how the experiment favors one case or the other, not just paying attention at the final layout; just the same way you would not say "well that random person must be male or female therefore 1/2" you cannot say "well that random marble must be black or white therefore 1/2"

Clear now?

 

[Whovian]

there is a related problem of a princess facing marriage to a king that she hates. The king suggests a game to make it fair. If she can select the white stone from the pot then she can go free but if she selects the black stone then she must remain as his queen.

She notices that the king has cheated and placed two black stones in the pot. What should she do?

Take a black stone, and eat it (or put it out of view some other way.) Suggest to the audience that by looking at the remaining stone, they can see what colour the original one was. To note that the other one was also black is to admit cheating.

But this is getting off topic.

 

[cosmik debris]

And btw, this indeed reminds the game Whovian mentions, which I recall as being Monty Python's contest game (one can even google this).

DonAntonio

I think you mean Monty Hall, although Monty Python's would be a lot funnier.

 

[DonAntonio]

I think you mean Monty Hall, although Monty Python's would be a lot funnier.

Of course...lol. So I recalled not accurately, as well.

DonAntonio

 

[DonAntonio]

This is not clear to me.

The 3rd possibility is: we have two black marbles in the urn but this time the first one drawn wasn't N1 but N2...these are DIFFERENT, identifiable

black marbles for this problem's sake.

DonAntonio

 

[jedishrfu]

Kick his nuts and run.

Good answer except she gets brought up on assault charges and beheaded.

The best solution was to pick a stone and drop it quickly then say well you can what stone I picked by what's left and of course the king won't say he cheated cause it isn't kingly and so she gets to leave peacefully.

 

[moonman239]

Take a black stone, and eat it (or put it out of view some other way.) Suggest to the audience that by looking at the remaining stone, they can see what colour the original one was. To note that the other one was also black is to admit cheating.

But this is getting off topic.

This.

As I recall, there was a story about a king who had taken some people captive. One of their wise men faced the king, who gave him challenges and told him he and his people could go if he completed the challenges. In the last game, the king said he would write or have written "Stay" on one piece of paper and "Go" on the other. If the guy picked the one that said "Go," he and his people could leave.

The wise man, however, was not fooled. He knew there'd be no paper that said "Go." Both of them would say "Stay." So he took a piece of paper and ate it. He turned over the remaining one and said, "This one says 'Stay,' so the one I ate must have said 'Go.'"

Anyways, OP, maybe running a Monte Carlo simulation would help.

 

[D H]

I think my question is not clear.

Initially the urn has only one marble. There is a 50% chance that it is black and a 50% chance that it is white. The friend adds one black marble to the urn.
Then she randomly draws one marble for the urn, which happens to be a black marble. Now she asks : what is the probability of drawing a black marble from the urn now?

You do not know that there is 50% chance that it is black initially. That 50% probability your naive best guess, based on a complete lack of information. After performing the experiment you have evidence that let's you alter that initial naive guess. I'll look at two variations on the problem to illustrate this point.

Suppose instead of drawing a black marble, the marble pulled from the urn had been white. Now the answer to question what is the probability of drawing a black marble from the urn now? is 1. There is no doubt that the remaining marble is black. You have to alter your initial guess based on this overwhelming evidence. The evidence is not quite so overwhelming when the drawn marble is black, but it is still useful information, and it still does let one improve upon that initial naive guess.

Suppose you repeated the test many times: a black marble is added to the urn and a marble is randomly drawn from the urn. You do this ten times, and each time the marble drawn from the urn is black. This accumulation of evidence is quite overwhelming. The answer to the question what is the probability of drawing a black marble from the urn now? is 1024/1025.

 

[viraltux]

Hi DH,

You do not know that there is 50% chance that it is black initially.

Yes he knows, that is not part of his answer but part of the problem.

Suppose you repeated the test many times: a black marble is added to the urn and a marble is randomly drawn from the urn. You do this ten times, and each time the marble drawn from the urn is black. This accumulation of evidence is quite overwhelming. The answer to the question what is the probability of drawing a black marble from the urn now? is 1024/1025.

If you don't know the probability of the first marble being black then you can never know the probability to draw a black marble; if this is the case you can only give an estimation within a certain confidence.

 

[SW VandeCarr]

Assuming one ball is has a 0.5 probability of being black vs white and you've drawn a ball, there is only one ball left in the urn. The possibilities are the original black ball you put in, the other ball that happens to be black, the other ball that happens to be white. Therefore there are three possibilities, two of which are a back ball. So the probability you would have drawn a black ball is 2/3..

 

[D H]

Hi DH,
Yes he knows, that is not part of his answer but part of the problem.

No, he doesn't know this. Once the initial ball is placed in the urn its color is fixed. It is either white or black with a probability of one. The key problem here is that the initial state (initial color) is unknown. It is a hidden variable. That 50% is just a guess. This is an estimation problem, something that Bayesian statistics are quite good at addressing. Another problem is that drawing a black marble is imperfect information. (In contrast, drawing a white marble yields perfect information.)

Bayes law provides a way to deal with this imperfect information and with the unknown initial state. A simple treatment is to use Bayes law as is, assuming some reasonable prior probability regarding the hidden state. More advanced Bayesian techniques provide the ability to express the uncertainty in that initial guess. Some of these more advanced techniques even provide the ability to indicate that the initial state is completely unknown in the form of a singular a priori covariance matrix. That, however, is way beyond the scope of this thread.

Suppose you repeated the test many times: a black marble is added to the urn and a marble is randomly drawn from the urn. You do this ten times, and each time the marble drawn from the urn is black. This accumulation of evidence is quite overwhelming. The answer to the question what is the probability of drawing a black marble from the urn now? is 1024/1025.

If you don't know the probability of the first marble being black then you can never know the probability to draw a black marble; if this is the case you can only give an estimation within a certain confidence.

I should have qualified this part of my response in post #21 as once again assuming an a priori probability of 50%. A different prior will result in a different posterior probability for this repeated experiment.

For example, suppose the initial ball was randomly selected from an urn containing 99 white balls and 1 black ball. Now the prior probability is measly 1/100 but the posterior probability after drawing ten black marbles via the experiment is 1024/1123 per Bayes law. It's only when the prior probability is very small that this accumulation of evidence no longer strongly indicates that the remaining marble is black.

 

[viraltux]

I love the smell of a flame in the morning...

No, he doesn't know this. Once the initial ball is placed in the urn its color is fixed. It is either white or black with a probability of one.

Well, I'd say the initial ball's color is fixed even before is placed in the urn, unless of course it is a Michael Jackson kind of ball which are known to be able to change its color spontaneously.

The key problem here is that the initial state (initial color) is unknown. It is a hidden variable. That 50% is just a guess. This is an estimation problem, something that Bayesian statistics are quite good at addressing.

That the initial state is unknown does not imply that the probability for a particular state is a guess. For instance, if I flip a perfect coin and I ask you "heads or tails?" you are not guessing that the probability for heads is 1/2, you know.

Bayes law provides a way to deal with this imperfect information and with the unknown initial state.

So does Frequentist / Fisherian aproaches.

A simple treatment is to use Bayes law as is, assuming some reasonable prior probability regarding the hidden state. More advanced Bayesian techniques provide the ability to express the uncertainty in that initial guess. Some of these more advanced techniques even provide the ability to indicate that the initial state is completely unknown in the form of a singular a priori covariance matrix. That, however, is way beyond the scope of this thread.

I would not call the Bayesian treatments to estimate parameters simple, oh... I feel so tempted to bite the bait, but for that we should start a new Thread "To Bayes, or not to Bayes"

For example, suppose the initial ball was randomly selected from an urn containing 99 white balls and 1 black ball. Now the prior probability is measly 1/100 but the posterior probability after drawing ten black marbles via the experiment is 1024/1123 per Bayes law. It's only when the prior probability is very small that this accumulation of evidence no longer strongly indicates that the remaining marble is black.

For Bayesians everything is a parameter which I don't think it has to be so, also Bayesian estimation methods do not allow probabilities 1 or 0, which, again, I don't think either it has to be so. But I'd say this is quite off topic and probably we're confusing the original poster musicgold.

So for musicgold's sake let's agree the book is right from a non Bayesian approach

 

[D H]

I love the smell of a flame in the morning...

The only one who is flaming is you. Stop now.

Well, I'd say the initial ball's color is fixed even before is placed in the urn, unless of course it is a Michael Jackson kind of ball which are known to be able to change its color spontaneously.

Of course. If we ran my repeated experiment and somehow drew a white ball at one point and then later drew another white ball, both a frequentist and a Bayesianist would say that there is something fishy going on here.

So does Frequentist / Fisherian aproaches.

You're right. Both approaches are useful. Neither approach is perfect unless the information is perfect (in which case, why use statistics at all?) In the words of Donald Rumsfeld, "Data is like a captured spy. Torture it enough and it will tell you anything."

also Bayesian estimation methods do not allow probabilities 1 or 0

That's nonsense. If the marble drawn from the urn was white rather than black a Bayesian approach would yield an answer of 1 in response to "what is the probability of drawing a black marble from the urn now."

So for musicgold's sake let's agree the book is right from a non Bayesian approach

The book is right from a Bayesian approach too, assuming a prior based on the principle of indifference.

Note that nowhere in the question (the blue text in the original post) is there any indication of the probability distribution for the initial ball. The principle of indifference is about all we have to go on.

 

[viraltux]

The principle of indifference is about all we have to go on.

I think I will apply this principle right now.

 

[DonAntonio]

The only one who is flaming is you. Stop now.


Of course. If we ran my repeated experiment and somehow drew a white ball at one point and then later drew another white ball, both a frequentist and a Bayesianist would say that there is something fishy going on here.


You're right. Both approaches are useful. Neither approach is perfect unless the information is perfect (in which case, why use statistics at all?) In the words of Donald Rumsfeld, "Data is like a captured spy. Torture it enough and it will tell you anything."


That's nonsense. If the marble drawn from the urn was white rather than black a Bayesian approach would yield an answer of 1 in response to "what is the probability of drawing a black marble from the urn now."


The book is right from a Bayesian approach too, assuming a prior based on the principle of indifference.

Note that nowhere in the question (the blue text in the original post) is there any indication of the probability distribution for the initial ball. The principle of indifference is about all we have to go on.

I think Viraltux has the correct approach in this case: for the contestant, the probability of the marble which is already in the

urn being black (white) is 0.5, unless there's some other piece of info available (like, say, knowing that the person "perpetrating" the game

has a huge bias for white marbles or whatever). So if a person participates for the first time in the game and heard nothing about it in

the past, then he can comfortably assume the prob. is 0.5 of the marble being black. Then, under this pretty natural (imo) assumption, the

evaluation of the prob. is, as already noted, 2/3.

DonAntonio

 

[D H]

That is exactly what Bayes' theorem says. With no additional knowledge, the best guess regarding the color of the hidden marble is that it is black with probability 1/2. Now we perform experiment of adding a black marble to the urn and randomly drawing a marble from the urn. If the drawn marble was white, we would know with 100% certainty that the remaining marble is black. That's not what happened. We drew a black marble. So what is the posterior probability that the remaining marble is black given that we drew a black marble? Bayes' theorem says that it's 2/3, the same answer as in the book.