Two men consequently flip a «fair» coin (possible outcomes are Head and Tail). Two men consequently flip a «fair» coin (possible outcomes are Head and Tail). A winner is defined to be a player who produces a Head first. What is the probability that the first player (he begins a game) will be a winner? Let X be a random variable which shows a number of flips in the series $TTT…H$ which describes one session of this game. What is the expectation E(X)? my solution for first : $$p=\frac{1}{2}+\frac{1}{2}(1-p).$$ $p=2/3$. what about second part ?

For the second part the reasoning is similar to that of the first part.

Let $E$ be the expected number of throws.

$$E=\frac12+\frac12(E+1),$$

since after the first throw either you stop (head, probability $1/2$) or you continue (tail, probability $1/2$) and in this second case your new expected number is $E+1$.

Thus $E=2$.