Roulette: Expected payoff

Question

Assuming you are playing roulette.

The probabilities to win or to lose are:

\begin{align} P(X=\mathrm{win})&=\frac{18}{37}\\ P(X=\mathrm{lose})&=\frac{19}{37} \end{align}

Initially 1$ is used. Everytime you lose, you double up the stake. If you win once, you stop playing. If required you play forever.

We can calculate two expectations:

Win somewhen:

$E[X_{win}]=\lim_{n\to\infty}1-(\frac{19}{37})^n=1$

The expected payoff:

$E[X_{payoff}]=\lim_{n\to\infty}\left(p^n(-(2^n-1))+(1-p^n)\right)=1-(\frac{38}{37})^n=-\infty$

Terms:

• $p^n$ is the probability that the player loses all $n$ games. The invested (=lost) money is then $2^n-1$. $\Rightarrow p^n(-(2^n-1))$
• $(1-p^n)$ is the probablity that the player wins one of the $n$ games and stops playing. The player wins then $1$. $\Rightarrow (1-p^n)*1$

This result confuses me: We have the probability of 1 to win eventually, but the expected payoff is $-\infty$. Whats wrong here? A teacher said me that the expected payout should be 1, because somewhen you will win. I'm a bit confused, maybe i just calculated something wrong?

Thank you

Answer 1

Hmm, I am not sure I follow your formula for expected payoff, but here is how I would calculate it:

There is a $\frac{18}{37}$ chance of winning on the first turn.

There is a $\frac{18}{37}*\frac{19}{37}$ chance of winning on the second turn.

... There is a $\frac{18}{37}*\frac{19}{37}^{i-1}$ chance of winning on the $i$-th turn.

When you win on turn $i$, you have put in $2^i-1$, and you get a payout of $2^i$, for a net winnings of 1 (of course!)

So:

$$ E = \sum_{i=0}^\infty \frac{18}{37}*\frac{19}{37}^i = \frac{18}{37}*\sum_{i=0}^\infty \frac{19}{37}^i = \frac{18}{37}*\frac{1}{1-\frac{19}{37}} = \frac{18}{37}*\frac{1}{\frac{18}{37}} = 1$$ (of course!)

Answer 2

This is the celebrated St Petersburg paradox. The resolution is simple: you start with a finite amount of money, and you risk losing it all. Once your money runs out, you have to stop playing.

"If required you play forever": not in my casino you don't! After your money runs out, you will be politely escorted to the exit.

So if you start with a huge bankroll, then the probability of losing it all is very small; but the amount you lose is commensurately large. For instance, you might be risking a million roubles for the sake of winning $1$ rouble, which is foolhardy enough; but your odds of losing it all are even greater than one in a million, because of the house advantage.