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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.

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What makes you think this is incorrect? –  Ross Millikan Sep 29 '11 at 23:58
Looks good to me, too. (Of course the casino may not allow you to bet fractional dollars, so perhaps a more realistic strategy for the $p>1/2$ case would be to bet one chip at a time until you've either doubled your initial money or gone broke.) –  Henning Makholm Sep 30 '11 at 0:24
@gambel An interesting problem! What book is this from? –  Byron Schmuland Sep 30 '11 at 12:22
It's from the Wohascum County Problem Book, and it's supposedly one of the harder problems. –  gambel Sep 30 '11 at 14:03
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2 Answers

up vote 5 down vote accepted

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:

One never gets broke (one's fortune is always positive). One doubles up almost surely if $p$ is large enough but with a probability which is less than $1$ if $p$ is small enough.

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:

For each $p>\frac12$, the strategy where one always bets a proportion $r$ of what one has wins almost surely for every small enough positive value of $r$ (for example, $r\leqslant 2p-1$).

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Ah that seems to be right, it would indeed make sense to use the logarithmic scale if we are multiplying and dividing. –  gambel Sep 30 '11 at 22:54
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I will assume that you know about measure theory as the basis of probability theory. I'll first try to provide a precise formulation of the problem for $p \gt \frac{1}{2}$ and then try to explain what is missing in your explanation.

We have as probability space the space of bets $$ \Omega = \{-1, 1 \}^{\mathbb{N}} $$ where the event 1 denotes "win" and -1 denotes "loose", and the probability of $1$ is $p$. We have an initial budget of $z_0$ dollars. We can bet arbitrarily small amounts of money.

A betting strategy would be a sequence $(a_i)$ denoting the amout of money $\geq 0$ you bet, $$ (a_i) \in \mathbb{R_+^\mathbb{N}} $$ such that for all $n \in \mathbb{N}$ $$ \sum_{i=1}^n \omega_i a_i \ge -z_0 $$ This condition says that you cannot bet money that you don't have, i.e. once you have lost $z_0$ you have to stop (and bet $0$ dollars from that moment for all following bets).

When you write

you should have 100% chance of doubling your money.

your claim is that for $p \gt \frac{1}{2}$ there is a betting strategy $(a)_i$ so that $$ \text{winning condition: }P(\omega: \text{there is an index n such that } \sum_{i=1}^n \omega_i a_i \ge z_0) = 1 $$ Your betting strategy to back up your claim is

each time you bet exactly half your money.

Let's call this betting strategy $(b)_i$. You'd have to prove that the winning condition holds. This does not follow from the fact that the expected value of an individual bet is positive.

For your betting strategy, for example the event $$ \omega = -1, -1, ... $$ will trivially won't let you win. This is also true for the event $$ \omega = -1, -1, -1, 1, -1, -1, -1, 1,... $$ etc.

So there are non-winning events $\omega_i$ for your betting strategy: Has the set of all such events probability zero?

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