I am reading this wikipedia article : https://en.wikipedia.org/wiki/Martingale_(betting_system)
I don't understand section mathematical analysis of a single round. Single round is defined to be series of loses ending with win or lose. So elements of sample space is {W, LW, LLW, LLLW, LLLLW, ... (n-1)LW, nL} where for example LLW means lose lose win and (n-1)LW means n-1 loses and win in that order. Let $q=\frac{1}{2}$ for simplicity. Why is probability of losing all money $\frac{1}{2^n}$ and not $\frac{1}{n+1}$. $\frac{1}{2^n}$ suggests that there are possibilites such as WLW(n-3)L but after a win game is over.
I understand that rules could be that after a win we bet again and it is still the same round but then when calculating expected value should be different because if we win twice we earn $2$ coins if we win three times we earn $3$ coins and if we win $k$ out of $n$ times we earn $k$ coins.
What is wrong?
While the sample space is, as you wrote, $\{W,LW,...,(n-1)LW,nL\}$, the probability measure is not the uniform measure. Not all elements of the sample space are equally likely. Here, $P(W) > P(LW)$, for example. Even though there are only $n+1$ elements, the probability of $W$ is still $\frac 12$, the probability of $LW$ is still $\frac 14$, etc. Indeed, the probability of $n$ losses is still $\frac 1{2^n}$. This doesn't imply that there are $2^n$ elements in the sample space.