The game allows you to bet only \$1 at a time, and if you win, you end up \$1 richer, otherwise you end up \$1 poorer. Probability of winning is $p=0.45$. If you start with \$1 - what is expected number of games before bust (i.e., \$0 dollars)?


My initial approach was essentially just:
$\sum\_{n=1}^{\infty} (2n-1)(0.45^{n-1})(0.55^n) {\approx} 1.212$ where $2n-1$ is the number of games played (as it must be odd because you start with \$1 and therefore must lose one more game than you win.





| X = No. of Games | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| P(X) | 0.55 | 0.136125 | 0.0336909375 | 0.008338507031 | 0.00206378049 | 0.0005107856713 | 0.0001264194537 | 0.00003128881478 | 0.000007743981658 | 0.00000191663546 |



(i.e., a sum product of the table above)


This is apparently wrong and I'm unsure where my logic / thinking is wrong. Would love any insight!



Expected number of games until bust
Let $x$ be the expected time to get to \$0, starting from \$1.


What's the expected time to get to \$0, starting from \$2? It's $2x$. That's because to get from \$2 to \$0, we first need to get from \$2 to \$1, and then we need to get from \$1 to \$0. And getting from \$2 to \$1 behaves exactly the same as getting from \$1 to \$0.


Now we "condition on the first step". Starting from \$1, we make one step, and then we're either finished (with probability $0.55$), or we're at \$2 (with probability $0.45$). So we get
$$
x = 1 + 0.45 \times 2x.
$$
Solving this gives $\boxed{x=10}\,$.


(More generally, if $p<1/2$ is the probability of winning a dollar at each step, then $\boxed{x=1/(1-2p)}\,$.)


(As others observed, your original method doesn't work because your sum only counts one way of going bust after $2n-1$ games, for each $n$. But for any $n>2$, there are more. e.g. for $n=3$, you have two, WWLLL and WLWLL.)


