The casino offers a certain win-lose game, where you have $p$ chance of winning. You can bet any amount of money, and if you win you get twice your bet; otherwise, you lose your bet. If you use the optimal strategy, what is your chance of doubling your money, as a function of $p$?
I came up with the following incorrect solution:
You have a 100% chance of winning if $p>\frac{1}{2}$ and $p$ chance of winning if $p<\frac{1}{2}$. Suppose that $p>\frac{1}{2}$. Then each time you bet exactly half your money. If you have $x$ dollars, you end up with $\frac{3}{2}x$ if you win and $\frac{1}{2}x$ if you lose, hence your expected outcome is $\frac{3}{2} xp + \frac{1}{2}x(1-p)$ which equals $xp + \frac{1}{2}x$. So if $p>\frac{1}{2}$, then the total is greater than $x$.
Given that each game on average gains you money and you can play an arbitrary number of games, of course you should have 100% chance of doubling your money.
Similarly, if $p<\frac{1}{2}$, it can be shown that no matter how much we bet, we lose money on average. Then on average our money will tend to go toward zero, so we're better off just going all-in at the start, with $p$ chance of doubling our money.
I do not understand the proper solution, but I think this solution is incorrect; however, I'm having trouble pinpointing where my proof falls apart. Thanks.
Edit: for reference I've included a screenshot of the given solution.
Best Answer
This answer only examines the betting strategy the OP advocates (always bet half of what you have) and some of its variants. With this strategy:
Note that there is no theorem saying that every positive submartingale reaches every level almost surely, hence arguments based on averages of increments only, cannot suffice to reach a conclusion.
In the strategy the OP advocates, the fortune performs a multiplicative random walk whose steps from $x$ are to go to $\frac32x$ and to $\frac12x$ with probability $p$ and $1-p$ respectively. Thus, the logarithm of the fortune performs a usual random walk with steps $\log\frac32$ and $\log\frac12$, whose constant drift is $m(p)=p\log\frac32+(1-p)\log\frac12$ hence $m(p)=p\log3-\log2$ has the sign of $p-p^*$ where $p^*$ solves the equation $3^{p^*}=2$, hence $p^*=\frac{\log2}{\log3}=.6309...$
If $p\geqslant p^*$, $m(p)$ is nonnegative hence the logarithm of the fortune reaches the set $[C,+\infty)$ almost surely for every $C$. In particular, one doubles up almost surely.
If $p<p^*$, $m(p)$ is negative hence the logarithm of the fortune has a positive probability $q$ to never visit the set $[\log2,+\infty)$, for example. This means that with probability $q$, one will bet an infinite number of times without ever getting broke nor doubling up. In effect the fortune at time $k$ will go to zero like $a(p)^k$ when $k\to+\infty$, with $a(p)=\frac123^p<1$.
There is no easy formula for the probability $q$ to never double up but the probability to double up $n$ times decreases exponentially like $\exp(-b(p)n)$ when $n\to\infty$, where $b(p)$ is the unique positive solution of the equation $p(3^b-1)=2^b-1$.
If $p<p^*$, the strategy where one always bets half of what one has fails. What happens with the strategy where one always bets a proportion $r$ of what one has? The same analysis applies and shows that one doubles up almost surely, for every $p\geqslant\wp(r)$, where $p=\wp(r)$ solves the equation $(1+r)^{p}(1-r)^{1-p}=1$ (note that $\wp(\frac12)=p^*$).
The other way round, for every given $p>\frac12$, a strategy which wins almost surely is to bet a proportion $r$ of what one has, provided $p\geqslant\wp(r)$. Since $\wp(r)\searrow\frac12$ when $r\searrow0$, one gets: