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Ramsey_KJ
Member since Sep2304

Oct3004, 06:02 PM (EST) 

"An Urn and 2 Balls"

Here is a probability problem that is ancient but a little demanding that I got off the MAA web sight. Suppose that there is an urn with two balls inside, both of which may be either white or black. On each draw, a ball is removed at random and then put back into the urn. If on the first two draws, a white ball is drawn, what is the probability that on the third draw a white ball is also drawn. Hint, it isn't 50/50. Have a Good Day KJ Ramsey 

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TheSmith
guest

Aug2905, 04:49 PM (EST) 

1. "RE: An Urn and 2 Balls"
In response to message #0

I get 2/3. There's four ways that the urn could have been set up: BB, BW, WB, WW. (This is a little sticky, but I imagined that was marble was slightly smaller than the other, thus making them distinct... much like making the two dice different colors to see that there's two 7's using a 3 and a 4.) Since a white was drawn at all, we know it's not BB, so that gives us 3 possible urns, each with equal probability of having happened. So there are essentially 6 marbles that could be drawn all with equal probability and 4 of them are white. 4/6 = 2/3. 

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Graham C
Member since Feb503

Sep1505, 10:17 AM (EST) 

2. "RE: An Urn and 2 Balls"
In response to message #1

>I get 2/3. > >There's four ways that the urn could have been set up: BB, >BW, WB, WW. (This is a little sticky, but I imagined that >was marble was slightly smaller than the other, thus making >them distinct... much like making the two dice different >colors to see that there's two 7's using a 3 and a 4.) > >Since a white was drawn at all, we know it's not BB, so that >gives us 3 possible urns, each with equal probability of >having happened. But that's not true, surely? If you drew two white balls in succession, the probability of the urn originally containing WW is higher than BW or WB. I get at first look* that the probability of the initial urn being WW is 2/3 and the probability of WB or BW (added) is 1/3. If it is WW then the probability of the next being white is 1; if it is BW or WB the probability is 1/2. So the total probability of the next ball being white is (1*2/3 plus 1/2*1/3) or (2/3 plus 1/6) or 5/6. The longer you go on drawing white balls the greater the probability the urn only contains white balls. (You've partially acknowledged this by allowing for the probability of BB to be zero, rather than the 1/4 that would have been assumed before no balls had been drawn.) > So there are essentially 6 marbles that could be drawn all with > equal probability and 4 of them are white. 4/6 = 2/3. * Using Bayes. Writing P(ww) for the a priori probability (1/4) that the initial distribution was both white, and P(WW) for the a priori probability (3/8) of drawing two white balls in a row you get P(wwWW) = P(WWww)*P(ww)/P(WW) = (1*1/4)/(3/8)


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iliaden
Member since Aug1405

Apr0206, 10:23 PM (EST) 

4. "RE: An Urn and 2 Balls"
In response to message #1

>I get 2/3. > >There's four ways that the urn could have been set up: BB, >BW, WB, WW. (This is a little sticky, but I imagined that >was marble was slightly smaller than the other, thus making >them distinct... much like making the two dice different >colors to see that there's two 7's using a 3 and a 4.) Up 'till here, I aggree, but it's the following part that has a problem. > >Since a white was drawn at all, we know it's not BB, so that >gives us 3 possible urns, each with equal probability of >having happened. This is wrong. Since you already know that ONE of the two balls is white, you must consider that there are two possibilities that are thrown away: the BB and the BW OR the BB and the WB, depending which ball you consider being drawn: the first one or the second one. You cannot keep both BW and WB because then we are unsure about the color of both balls, yet we are certain that one of them is white. My answer is 1/4 and it is explained in another posting. Ilia


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iliaden
Member since Aug1405

Apr0206, 10:23 PM (EST) 

3. "RE: An Urn and 2 Balls"
In response to message #0

I get 3/4 I simply consider the probability of getting a black ball. Since one of the two balls is white, there is a 50% probability of the second one being black. Meanwhile, there is also a 50% probability of the second ball being drawn, otherwise, it will be the first ball, which is white. thus the probability of a black ball being drawn is 50%*50%=25%, or 1/4 the probability of a white ball being drawn is thus 1(1/4)=3/4. Ilia


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mr_homm
Member since May2205

Apr0406, 10:26 AM (EST) 

5. "RE: An Urn and 2 Balls"
In response to message #3

Hi Ilia, Your reasoning is correct, but you are finding the probability that the SECOND ball you draw is white. The problem asks for the probability that the THIRD ball you draw is white. More explanation below: >I get 3/4 > >I simply consider the probability of getting a black ball. >Since one of the two balls is white, there is a 50% >probability of the second one being black. >Meanwhile, there is also a 50% probability of the second >ball being drawn, otherwise, it will be the first ball, >which is white. > >thus the probability of a black ball being drawn is >50%*50%=25%, or 1/4 > >the probability of a white ball being drawn is thus >1(1/4)=3/4. > So when you draw that second ball, 3/4 of the time it is white, which is correct. Now, which of these white balls come from WW and which come from WB? Thinking about the case of drawing a black ball, you get the black ball half the time when you have BW and none of the time when you have WW. Therefore, when the second ball drawn is white, that ball comes from all the WW cases and half the BW cases. Therefore, after seeing that the second ball is white, it is twice as likely that you are in a WW case than in a BW case. Therefore, 2/3 of the time you will have WW and 1/3 of the time you will have BW. Now when you draw the third ball, you can only get black in half the BW cases, which is 1/2·1/3 = 1/6 of the time. therefore, the third ball will be white 5/6 of the time, which is the same result you would get from using Bayes' Theorem. Here's another way to see it, which is really Bayes' Theorem in disguise: Suppose you have a very large number N of urns, of which N/4 are WW, N/4 are BB, and N/2 are BW. This is your initial set of cases before you have drawn any balls. Now you draw the first ball and it is white. What cases are now left? All the WW, half of the BW and none of the BB. Therefore, you now have N/4 WW and N/4 BW urns. Now you draw the second ball, and it is white, too. What cases are now left? Again, all of the WW and half the BW, so you now have N/4 WW and N/8 BW urns. Now draw the third ball. The number of white balls you get will be all of the WW and half of the BW cases, which is N/4 + N/16 = 5N/16. The number of black balls you get will be half the BW cases, which is N/16. Therefore, white balls outnumber black balls by 5 to 1, so there is a 5/6 chance that the third ball will be white. By the way, to everyone who uses Bayes' Theorem: The prior distribution is important here. You must assume that initially the cases BB, BW, WB, WW are equally probable. This is just an assumption, and there's nothing in the problem that tells you this is true. The only truly correct answer to this problem is "You can't tell what the probability of the third ball being white is without knowing the prior distribution of the colors in the urn." For example, suppose I had 1 huge urn with 10^9 white balls and 10^9 black balls, and I chose two of them to fill my small urn. Then the probabilit's are equal (almost, they differ by 1 part in 10^9) of getting BB, BW, WB, or WW. But if my huge urn has 2·10^9 white and 1·10^9 black balls, then when I fill my small urn from it, I will find a different probability distribution among the cases BB, BW, WB, and WW. When I read about Bayes' Theorem, the prior distribution is always discussed, but whenever I see the theorem used (not just her, I mean in textbooks, especially physics textbooks) the prior distribution is always assumed to be uniform, and no justification whatsoever is given for that assumption. Even Dirac does it in his famous book on quantum mechanics. It must be the right assumption, because his physics comes out right, but WHY is it right? Is there some deep reason why it logically has to be that way? Then again, as you repeat the experiment, the prior distribution gradually ceases to matter, because you have more actual data known. If you pull a ball from the urn 100 times and it comes out white every time, then the probability that the 101st ball will be white is very close to 1, independent of the prior distribution. The ASSUMPTION about the distribution is gradually replaced by experimental KNOWLEDGE of the distribution after repeated experiments. So perhaps the prior distribution actually does not matter in physics, if the experiments are all repeated many times. This urn and balls question seems to lead into somewhat deeper ideas that you would expect for a middle school probability question! Stuart Anderson 

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iliaden
Member since Aug1405

Apr0606, 05:36 PM (EST) 

6. "RE: An Urn and 2 Balls"
In response to message #5

Hi Stuart, >Your reasoning is correct, but you are finding the >probability that the SECOND ball you draw is white. The >problem asks for the probability that the THIRD ball you >draw is white. This is my mistake, I haven't read the problem carefully. >So when you draw that second ball, 3/4 of the time it is >white, which is correct. Now, which of these white balls >come from WW and which come from WB? Thinking about the >case of drawing a black ball, you get the black ball half >the time when you have BW and none of the time when you have >WW. Therefore, when the second ball drawn is white, that >ball comes from all the WW cases and half the BW cases. >Therefore, after seeing that the second ball is white, it is >twice as likely that you are in a WW case than in a BW case. > Therefore, 2/3 of the time you will have WW and 1/3 of the >time you will have BW.
Thus here you agree that one of the BW and WB probabilities must be eliminated. I have encountered several people trying to proove me that I should keep both of them after the first draw. Right now I am confused. I can give you my explanation of why we SHOULD eliminate one of them, yet I am not certain about it. There goes:
When you draw the first ball, you assign a number to it. Let's say ball #1. we know that the ball #1 is White, while the color of the other one is unknown. In our question, the ball #2 is unknown, so it has a 50% probability of being white, and 50% prob. of being black. This means that there should be two variantes left: the WW and the WB. The BW is eliminated.
Am I the one who did the mistake, or were it the others? And another question: can you proove that if you keep drawing white balls, the odds of having a WW variation will go up? Thank you, Ilia Denotkine


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mr_homm
Member since May2205

Apr0706, 08:22 PM (EST) 

7. "RE: An Urn and 2 Balls"
In response to message #6

Hi Ilia, >When you draw the first ball, you assign a number to it. >Let's say ball #1. we know that the ball #1 is White, while >the color of the other one is unknown. In our question, the >ball #2 is unknown, so it has a 50% probability of being >white, and 50% prob. of being black. This means that there >should be two variantes left: the WW and the WB. The BW is >eliminated. Your idea is not really wrong, but it makes the problem harder when you try to think about the third ball. If you think of the first white ball as ball #1, you are correct in saying that BW is impossible. The trouble is that you can only do this trick ONCE. When you pull out the second white ball, you can't be sure whether it is the same one you saw before, so you can't eliminate any more possibilities. If you marked the first ball in some way, like writing a #1 on it, then you could tell the balls apart, but then you are collecting more information than the original question allows. When you try to find the probability that the third ball is white, you have to look at the cases where the second ball is white. There are two ways to get a white ball with WW and only one way with WB, so 2/3 of the time you must have WW and 1/3 of the time you must have WB when the second drawing gives a white ball. That means there is now a 2/3 chance of WW and a 1/3 chance of WB. So when you draw the third ball, the chance it is white is (2/3)(1) + (1/3)(1/2) = 5/6. This gives the same answer as I got, but you have to use different reasoning on the second and third time you draw the ball. The reasoning for the third ball would also work on the second ball, so you might as well just use that reasoning all the way through, so that the argument is cleaner. > >Am I the one who did the mistake, or were it the others? As I said, you are not wrong, but your way of thinking leads you to get stuck on the third ball. > >And another question: can you proove that if you keep >drawing white balls, the odds of having a WW variation will >go up? Yes. Let's say the probability WW is p and (BW or WB) is q, and you are just about to draw a ball. This is called the prior probability. Now you draw a ball and get a white one. What is the new probability p' of WW? Bayes' Theorem says P(WWW)P(W) = P(WWW)P(WW). P(WWW) = p', P(W) = p+(q/2) (because all of the WW and half of the WB or BW will give a W ball), P(WWW) = 1, and P(WW) = p. Therefore, (p')(p+q/2) = (1)(p), so p' = p/(p+q/2). But q = 1p, so this is p' = 2p/(p+1) = 2/(1+1/p). Now this is interesting, because it relates to a formula you mentioned in another thread. If you rewrite it like this: 1/p' = (1/1 + 1/p)/2 you can see that p' is the harmonic mean of p and 1. Therefore, p' is closer to 1 than p is, and this will always be true as long as p<1. Therefore, if you keep doing the experiment, in the limit, p' must go to 1. Now since P(W) = p+(q/2) = p+(1p)/2 = (1+p)/2, the probability that the next ball will be W is the arithmetic mean of 1 and p. But this also goes to 1 since p goes to 1. Hence it eventually becomes virtually certain that both balls in the urn are white and that the next ball drawn will be white. By the way, if you don't want to use Bayes' Theorem, you can look at it like this: Suppose you have 2^(N+2) urns, and 1/4 of them are WW, 1/4 are BB, 1/2 are BW or WB. You draw a ball from each, and look at all the cases where the ball is white. Then you will have 2^N WW urns, and 2^N total BW and WB urns. Every time you repeat the experiment, all the WW urns will pass the test (i.e., they will give a white ball) and half the BW and WB urns will pass the test. So after you draw the second ball, you have 2^N WW and 2^(N1) total BW and WB, and so on. So after drawing the nth ball, you have 2^N WW and 2(Nn+1) total BW and WB. Therefore the probability p = P(WW) that an urn is WW after the nth ball is drawn is 2^N/(2^N + 2^(Nn+1) = 1/(1+2^(1n) = 2^n/(2^n+2), which satisfies the harmonic mean equation for p' I gave above. Just put n+1 in the formula to get p' and n to get p, and it'satisfies the equation. This also shows that the probability p goes to 1 for large n, and gives an exact formula (which you could also get from the harmonic mean equation and the initial condition that p=1/2 for n=1), and therefore P(W) goes to the arithmetic mean (1+p)/2 = (2^n + 1)/(2^n + 2), which approaches 1 and also gives the values we know for the values of n that were checked by hand. For n=1, you get 3/4 and for n=2 you get 5/6 (remember, n is the number of balls that have ALREADY been drawn, so n=1 means we are about to draw the second ball, and it has a probability 3/4 of being white). This was a surprisingly interesting problem. Thank you for asking a very good, thoughtprovoking question! Stuart Anderson 

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