If a man has 37 matches in the box in his left pocket and 21 in the box in his right pocket and repeatedly grabs either box at random to take a match out, what is the probability that he gets to a point where he has one match in each pocket?
Is this a Binomial Dist and ans is $\frac{56C_2}{2^{56}}$?
look at the first $56$ matches he reaches for. We have success if and only if he reaches for the right pocket $20$ times and the left $36$ times. So we have a string of $56$ letters each L or R with equal probability and we want the probability that exactly $20$ of them are R.
The answer is $$2^{-56}{56\choose20}$$