what is expected number of round?

Question

if we have a game that with the probability of $\gamma$ continue another round and with the probability of $1- \gamma$ finished I know the expected number of the round is calculated with help of geometric series and answer is $1/(1-\gamma$). but why? and how can we prove it?

Answer 1

It is not necessary to use geometric series.

Let $\mu$ denote the expectation of the number of rounds that the game will last.

Then:$$\mu=\gamma(1+\mu)+(1-\gamma)1=1+\gamma\mu\tag1$$leading to: $$\mu=\frac1{1-\gamma}$$

If we insist on using geometric series (not my idea) then let $X$ denote the number of rounds that the game will last and work out:$$\mathbb EX=\sum_{n=1}^{\infty}nP(X=n)=\sum_{n=1}^{\infty}n\gamma^{n-1}(1-\gamma)$$