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We have k boxes, in each of them a white and b black balls. We select a random ball from the first, throw it into the second. Then we pick a random ball from it (i.e. from the second box), throw it into the third, etc. What is the probability that we will draw a white ball from the last box?

I am not really sure how to approach this question and what formula to use. I made a python program that showed me the answer is a/a+b. Now I am stuck with actually proving it.

I tried to make a decision tree and for smaller k I confirmed that the sum of white balls being picked on each row (row being the k) is a/a+b. Obviously the chance for black balls is b/a+b. Is the fact that if we put a=b and the chances are the same because the formulas for both colors are on each consecutive row multiplied by the same numbers somewhat useful?

My intuitive answer would be that if we pick really high numbers as a and b then the +1 in each box doesn't really matter. So as a and b approaches infinity the limit is just a/a+b.

RandomGuyOnMath
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    Why not work it out for small $k$ to get a sense of the pattern? Note: it's not clear what you mean be "I confirmed that the sum of white balls being picked on each row (row being the $k$) is $\frac a{a+b}$". – lulu Feb 26 '23 at 16:00
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    Could we solves this by mathematical induction. Let P(n) be the probabability that we pick the white ball from box n. Let's assume that P(n) = a/a+b. For P(1) it works. Now P(n+1) = P(pick a white ball from box n then pick a white ball from box n+1) + P(draw a black ball from box n then draw a white ball from box n+1). This equals P(n+1) = P(n) * 1 / (a+b+1) + a / (a+b+1). No we substitute a/a+b for P(n) and we get that P(n+1) is also a/a+b. – RandomGuyOnMath Feb 26 '23 at 16:31
  • Sure. If you think the answer is $\frac a{a+b}$ then induction lets you prove that. What you write in your comment looks good...if you like, you can post your own solution below. That way users here can check the details. – lulu Feb 26 '23 at 16:35
  • I made the decision tree and checked that the answer is a/a+b for smaller k. As you said I may not have been clear with that. – RandomGuyOnMath Feb 26 '23 at 16:35
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    Oh, no problem. And, yes. That's the correct answer and a sensible method for proving it. I do suggest posting your solution below. – lulu Feb 26 '23 at 16:36
  • I will post it soon. Thank you for your help. – RandomGuyOnMath Feb 26 '23 at 16:38

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You can simplify it greatly by treating it as a mixture problem.

Initially, each of the $k$ boxes is having fraction of white balls as $\Large{{a}\over{a+b}}$

Now, when a ball is transferred from one box to the next, the expected fraction of whites in the next box remains $\Large{{a}\over{a+b}}$, because each box is having the same initial fraction of white balls.

Ergo, however many boxes there are, when you draw a ball from the last box, $P(white \; ball\; drawn) = \Large{{a}\over{a+b}}$

true blue anil
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  • You can also see how the concept can still be used where each box does not have the same proportion of the "desired" type. https://math.stackexchange.com/questions/4615490/there-are-two-multisets-of-circles-and-squares-probabilty-of-pulling-a-circle-f/4615533#4615533 – true blue anil Feb 26 '23 at 16:54