A person puts some coins in 2 bins at random, and stops when one of the bins has n coins in it. What's the probability of the other having exactly k coins?
My work: $$P(k|n) = \frac{P(n|k)*P(k)}{P(n)}=\frac{1*P(k)}{1}$$
So probability of $P(k|n)=P(k)=0.5^{n-k}$
I am not sure I got the P(n) correct, as well as P(k), but I think P(n|k) is indeed 1, since there are always n coins in one of the bins.
Let $X$ be the number of coins in the other bin. The probability that you are looking for.
$$P[X=k] = \binom{n+k-1}{k} \frac1{2^{n+k-1}}$$
Explanation: you need to choose the $k$ times that you put a coin in the other bin from the $n+k-1$ possibilities.