# Profitable strategy in coin tossing?

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)

• This (well known) strategy is called "Gambler's Ruin" (google it). The point is $2^n$ grows so fast that before long you will have to pay a debt of more money than you could possibly have. – Gregory Grant Feb 8 '16 at 9:48
• I think it is profitable - no matter how smal $p>0$ - under the extra condition that the initial credit at your disposal is infinite. However, if you are in that situation then it is somehow impossible to increase your credit: $\infty+c\ngeq\infty$. – drhab Feb 8 '16 at 9:54
• Potentially interesting context for the other side: Kelly betting. – πr8 Feb 8 '16 at 11:52

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.

For a strategy to be profitable you'll need $p > 1/2$, but still your approach is not good anyway.

The flaw in your approach is that it assumes that you have infinite funds which you in reality doesn't. What happens when you have finite funds is that eventually you will get ruined (it's called the "Gambler's Ruin" by the way).

With that strategy you will on each suite of tosses either make a win before you go broke or you will go broke before you win. Even if you have a slight advantage you will put all your capital on stake at every suite and you will eventually loose.

If you for example have $2^n-1$ cents and start with one cent you will be able to bet at most $n$ times before you will have to win without going broke. The probability that you'll loose $j$ times in a row is $(1-p)^n$. On the other hand if you win after $j$ tosses you'll see that you've betted $2^j-1$ cents to win $2^j$, netting at $1$ cent profit. So you'll loose $2^n-1$ with $(1-p)^n$ probability and win $1$ with $1-(1-p)^n$ probability. This adds up to:

$$1(1-(1-p)^n) - (2^n-1)(1-p)^n = 1 - (1-p)^n - 2^n(1-p)^n + (1-p)^n = 1 - (2(1-p))^n$$

As you see the only way this becomes positive is if $2(1-p) < 1$ that is $1-p<1/2$ or $p>1/2$. But even then your approach is bad since you're risking your entire capital in doing so.

Actually the bet to place optimally when taking the possibility to get ruined into account is determined by the Kelly Criterion. The fraction to bet is

$$f = {2p - 1\over 2}$$

which means that if $p<1/2$ you would be indicated to place a negative fraction bet (which means that you should play at the other side). If I recall correctly you should never accept a bet more than twice that size (in case you can't chose the bet size), this is actually what the house is supposed to do when they place limit's on maximum bet.