Roulette: Expected value for "double up" strategy and infinite many games

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:

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

This result confuses me: We have the probability of 1 to win eventually, but the expected payoff is $-\infty$. Whats wrong here?

Thank you

Answer 1

I wrote this assuming the payoff is double the bet, as in Roulette when betting on spaces with these odds. So a \$1 bet yields \$2, or a \$1 net. When repeated, one is starting with a \$1 loss, and doubles their bet to \$2. If they win, they receive \$4, but have invested \$3 in the series, meaning the net gain is \$1, the original bet.

The EV of any bet in this game with these parameters is:

$(bet * P(win)) + (-bet * P(lose))$

So $(1 * (18/37)) + (-1 * (19/37)) = -0.027$

You can expect to lose 2.7 cents per $1 bet, or 2.7% of whatever money you put down. This holds true for a bet of any amount. Not by coincidence, this is precisely the house edge in European Roulette. The house edge is simply the EV of whatever game you are playing.

So if you were to repeat this process an infinite number of times, betting double your previous bet, with an EV of -2.7%, surely you can see why you would lose in the infinite case.

I honestly have no idea what all your math stuff is doing (not because you're wrong, just because my own understanding isn't there), but I have a pretty good understanding of infinite. When you calculated your "win somewhen" equation, you could conversely do the same thing for "lose somewhen" and would come to the exact same answer. Weird stuff happens with infinite and I think what you've come across is a situation where you have a 100% chance of winning and losing, depending on what you're looking at.

In simple terms, if infinite is the length of time in which all that can happen, will happen, then there is a probability of 1 that any possible event will occur in that time frame. Since during infinite iterations of the game, everything that can happen, will happen, you have 100% chance of winning and losing during that time. This, however, does not change your expected value. It will always be -2.7% of your bet. You can always add to infinite, but the resulting "value" will also be infinite.

$\infty$ * -.027 = -$\infty$

P.S. this is my first post on StackExchange, so let me know if I broke any rules or whatever. Also, criticism of my reasoning is highly encouraged.

Answer 2

It is absolutely the case that you will win eventually with probability one. However this means your expected winnings is one, since when you inevitably win and walk away you win a single bet. I don't understand your calculation of expected winnings but rest assured it's wrong.

The usual caveat to mention with this startegy is that it actually requires infinitely deep pockets and is thus not realistic. If you have any loss limit where you have to walk away, no matter how large, your expected value is negative cause of the remote chance of losing a huge amount. This is a consequence of the optional stopping theorem.