There are n marbles in a bag, player $1$ picks out $x$ marbles where $x \leq n$, marks them, and puts them back. Player $2$ picks out $y$ marbles where $y \leq n$. What is the expected amount of marbles that are marked after player $2$ goes?

I gathered that each marble player $2$ grabs has a $\frac{y}{n}$ chance of being marked and there are $\binom{n}{x}$ possible combinations of marking the marbles.

I'm at a loss on how to proceed and am just looking for a tip in the right direction (hence the self-study tag). Any help is appreciated!

  • $\begingroup$ what is the question of interest? does player $2$ mark what he picks as well? or does player $2$ just inspect ouf of those marbles that are picked, how many were previously marked by player $1$? $\endgroup$ Commented Sep 19, 2017 at 23:42

1 Answer 1


First, try some numerical examples to help. If you can code, definitely try simulations too.

If $x = n$, i.e. all marked, what's the expected number marked? How about $x = 1$?

Back to the general case. Look at any one marble player two picked out. What's the probability that was marked? I think you wrote $\frac{y}{n}$ by mistake.

Put this together with the greatest of tricks of expected value... linearity!


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