Here is the game: Suppose I have a n sided fair dice and a fair coin. Let k be the number of times I roll the dice before I have to flip the coin. Rolling a 1 on the dice "wins," the game is over if I flip the coin and it lands on tails. Landing on heads gives k more dice rolls, repeating until the coin lands on tails.
My question is, what should k be so on average I win exactly once before the game is over.
My intuition from thinking about some simple cases is leading me to believe k = .5/(1/n) but I don't have any convincing justification.
Best Answer
Let $M$ be the number of the first toss of tails.
Then $P(M=m)=\frac1{2^m}.$
The expected number of ones in $mk$ rolls is $\frac{mk}n.$
So the expected number of $1$ rolls is:
$$\sum_{m=1}^{\infty}\frac 1{2^m}\frac{mk}n=\frac kn\sum_{m=1}^{\infty}\frac m{2^m}=\frac{2k}n$$
So you need $k=\frac n2$ to expect exactly one. If $n$ is odd, there is no exact $k,$ but $k=(n+1)/2$ is the smallest that will have expected value $>1.$