## Friday, March 3, 2017

### probability: of balls and urns

This is an unsolved problem from "Probability, a concise course, by Y A Rozanov". Very instructive in how to perform probability calculations.

There are N urns numbered 1, 2, ..., N in a row, each containing r red and b blue balls. A ball is drawn at random from the first urn and transferred to the second, then a ball is sampled from the second and transferred to the third, and so on, until you finally sample a ball from the Nth urn. The question is, what is the probability of drawing a red ball from the Nth urn?

I've heard people give all kinds of answers, but here's the correct way to solve the problem.

Start from urn #1.
What is the probability of sampling a red ball? r/(r+b).
A black ball? b/(r+b).

So either a red ball or a black ball is sampled and put into urn #2.
So urn #2 might have an extra red ball (in r/(r+b) cases), or an extra blue ball (in b/(r+b) cases).
So while sampling from urn #2, in r/(r+b) cases, you are sampling from r+1 red balls and b blue balls, and in b/(r+b) cases, you are sampling from r red balls and (b+1) blue balls.
So the probability of sampling a red ball from urn #2 becomes:
r/(r+b) * (r+1)/(r+b+1) + b/(r+b) * r/(r+b+1)
Simplifying, we get, p(drawing red ball from urn #2) = r/((r+b)(r+b+1)) * (r+1+b) = r/(r+b)
Wow! So regardless of the complexity of the experimental setup, the probability of drawing a red ball from urn #2 remains r/(r+b) JUST AS IT WAS FOR THE DRAW FROM URN #1!

Now, we can simply re-run the experiment thinking of urn #2 as urn #1, and urn #3 as urn #2 as above, and just keep repeating it till we reach urn #N. The probability of drawing a red ball from urn #N is therefore also r/(r+b). Interesting question.