I'm trying to model a series of plays on a game of American roulette. This is where you can't bet on two numbers (the zeros) rather than European Roulette where you can't bet on one.

If you bet on even odds for example red, there is a 47% (16/38) chance of winning 1. This means a 53% chance of losing 1. So it is likely that you will lose (47% * 1 + 57% * -1) or 0.05.

How do I calculate how much you will lose on 2+ subsequent plays? I'm assuming that at each subsequent play you will lose more because the odds are stacked against you.

I'm not sure how to do this calculation. Any help will be much appreciated.

  • $\begingroup$ You add the losses. $\endgroup$
    – whuber
    Commented Oct 16, 2020 at 17:56

1 Answer 1


What you are looking for is the expected profit from the bet.

It's denoted $E(X)$ where $X$ is a "placeholder" for the random profit you would get from a game. Notice that the profit can also be negative. For the bet on red your expected profit is, as you correctly calculated,

\begin{align} E(X) &= P(X=-1)\times(-1) + P(X=1)\times1\\ &= 0.53\times(-1) + 0.47\times1\\ &= -0.06 \end{align}

meaning that you will on average lose 6 cents for each dollar you play.

Now lets say you play two games on red, with a random profit of $X_1$ and $X_2$. Now we are interested in the expected value of the total profit $X_1 + X_2$, which is $E(X_1 + X_2)$. It turns out that you can calculate the expected value for each variable separately. So, for two games we get,

\begin{align} E(X_1 + X_2) &= E(X_1) + E(X_2)\\ &= (-0.06) + (-0.06) = -0.12 \end{align}

which is just twice the loss from one game. For $n$ games the loss will simply be $n\times(-0.06)$, so the number of games multiplied by the expected loss in each game.

  • $\begingroup$ Thanks for the great explanation. I understood it now. $\endgroup$ Commented Oct 17, 2020 at 10:33

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