Suppose there are N people standing in a circle and each one of them has a ball in their hand. When the referee says throw, they are supposed to throw the ball at random towards another person in the circle and at the same time supposed to catch the first incoming ball. If a person receives >1 ball, those fall on the ground. Question is what is the expected number of balls that will fall on the ground?
This question has appeared in one of my past interviews. Currently my take on this is (although it doesn't seem correct):
So can I assume $X$ to follow $\operatorname{binomial}(N-1, p)$ where $p = \frac{1}{N-1}$? In that case, $$\mathbb{E}(X) = \frac{N-1}{N-1} = 1.$$ So the expected number of balls on the ground for person $i$ $\mathbb{E}(X-1) = 0$? That makes the overall expectation $0N = 0$?
Your $\mathbb{E}[X]=1$ is the expected number of balls thrown at an individual.
But this does not tell you how many that individual is expected to drop, which is $\mathbb E[\max(X-1,0)]=\mathbb{E}[X-1]+\mathbb P(X=0)= 0+\left(\frac{n-2}{n-1}\right)^{n-1}$,
telling you that the expected total number of balls dropped is $n\left(\frac{n-2}{n-1}\right)^{n-1}$.