Distribution and expectancy value ? _We let a fair coin $N$ time flipping with stake $m$. If it's 'head' we'll get ($2m$) twice the money back, if it's 'tail' we'll ante up twice ($4m$) till 'head' appears. After the $N-th$ time the game ends. Let $T $ be the time till the ending with $T(\omega)=min\\{n\ge 1:\omega_n='head'\\} \land N$ and $X(T)$ the whole deployed capital. Now I wanna calculate:_ $\bullet$ the expectancy value of the winning $\bullet$ the expectancy value of the time $\bullet$ the distribution of the winning $\bullet$ the distribution of of the time My attempt: I said it's a binomial distribution,so we know: $\Bbb P[X=k]=\binom{n}{k}p^k(1-p)^{n-k}$ here $p=0.5,n=N,k\in \Bbb N$ The expectancy value of the winning should be $2^N$ because the first $N-1$ times we have lost. The whole stake at the end would be just $m$. I don't know how to formlate the distribution....because the coin flipping is fair is it $\Bbb P[X=N]=(\frac{1}{2})^k$ ?

Assuming this is a conventional martingale (if you win, then stop; if you lose, then double your stake on the next round) with fair odds on a fair coin, then:

* the net amount you profit after stopping is your original stake $m$
* the probability you play the $n$th round is $\dfrac1{2^{n-1}}$ for positive integer $n$
* the probability you first win on the $n$th round and then stop is $\dfrac1{2^n}$ (a geometric distribution)
* the expected number of rounds is $\displaystyle\sum_1^\infty \dfrac n{2^{n}}=2$ (the reciprocal of the probability of winning a particular round, given that you play it)
* the amount you stake on the $n$th round, if you play it, is ${2^{n-1}}m$
* the expected amount you will have staked in total before winning is infinite: $\displaystyle \sum_1^\infty \dfrac{2^{n-1}m}{2^{n-1}} $



This final point is one of the reasons that it may be difficult to implement this strategy in practice. Typically a gambler has finite resources.