Talk:Banach's matchbox problem

Isn't the conditional probability is what is required? I believe the solution is wrong, given the statement of the problem. Piyush Sriva 23:43, 11 July 2007 (UTC)

The solution is correct. It's easier to understand if you consider the case where one pocket has infinite matches, so I've updated the wording with this. BPets (talk) 08:45, 13 June 2011 (UTC)

Why so complicated?

The probability that there are k matches left in the left box is equal to the probability that within 2N-k steps, we pick N times the right box. There are $$\binom{2N-k}{N}$$ ways to do this. So if L counts the remaining matches in the left pocket we get


 * $$P[L=k] = \binom{2N-k}{N}\left(\frac{1}{2}\right)^{2N-k}$$.

And if K counts the remaining matches in one of the boxes symmetry yields


 * $$P[K=k] = \binom{2N-k}{N}\left(\frac{1}{2}\right)^{2N-k-1}$$. — Preceding unsigned comment added by 77.12.198.255 (talk) 11:33, 11 May 2013 (UTC)