There is a coin with a probability $p$ of heads, and $1-p$ of tails. Tosses are independent of each other. When you bet an amount of money $x$, you receive $2x$ if it lands heads, and you lose what you bet if it lands tails.
We are going to use the following strategy:
You start betting an initial amount $x_0$. Every time you lose, you bet twice what you bet in the previous throw. You keep doing this, until you win, at which point you retire, or start again, betting $x_0$.
The interesting case of course is when $p < 0.5$.
The question is: How profitable is this strategy? Intuitively I feel that eventually I can always win an arbitrarily large amount of money. But that can't be entirely right ("the house always wins!"). So what am I missing here? What's the expected win in each run?
Also, does this sort of strategy have a name in the literature?
P.S.: Found this later, https://en.wikipedia.org/wiki/Martingale_(betting_system)
Best Answer
So you are betting $1, 2, 4, 8,...,2^n$ dollars in each succesive round you do not win; that is, the prize of getting to the nth round is $\sum_{i=1}^{n}{2}^{n-1}={2}^n-1$.
If you win the nth bet, you get $2, 4, 8,..., 2^{n}$ dollars. To that we have to discount what you had to bet to get there, so your net win given that you win at round $n$ is $2^{n}-2^{n}+1=1$ dollar.
Now, we better calculate how much money we are expected to need in the betting process: it would not do to have to leave the table if we empty our pockets.
Let $X$ denote the money spent until we win. Then $X={2}^Y-1$, where $Y$ is the number of rounds until we win.
$Y\sim Geom(p)$, and so $P(Y=n)=p(1-p)^{n-1}\quad n\in\mathbb{N}$.
Thus: $$ E[X]=E[{2}^Y-1]= E[{2}^Y]-1=[\sum_{i=1}^{\infty}2^iP(Y=i)] -1= [\sum_{i=1}^{\infty}2^ip(1-p)^{i-1}] -1=p[\sum_{i=1}^{\infty}2^i(1-p)^{i-1}] -1 $$ The series only converges if $2(1-p)<1$; that is, if $p>1/2$. That means that you should be willing to bet infinite money in order to make the bet profitable, and all of that for a measly dollar.
In case you are curious, the expectancy when $p>1/2$ is: $$ E[X]=\frac{2p}{1-2(1-p)} -1= \frac{2p}{2p-1}-1=\frac{1}{2p-1} $$ But then again, is fairly obvious that when $p>1/2$ the game has positive expected value, so just playing with any bet is profitable.