Expectations Galore: Roulette

Brian Alspach

Poker Digest Vol. 2, No. 21, October 8 - 21, 1999

Roulette is easily analyzed with respect to expectation so we begin our analysis of various games with roulette. In spite of the simplicity of the calculations, we shall find there are some interesting aspects. We shall assume we are dealing with a roulette wheel having both 0 and 00.

If we bet on a single number, the payout is 35-to-1 when we win. The probability of winning is 1/38 and the probability of losing is 37/38. Thus, using the formula for expectation given in the previous article, the expectation is 35(1/38) + (-1)(37/38) = -1/19. If we bet a single chip on two numbers, the payout is 17-to-1 when we win. The probability of winning this bet is 1/19 and the probability of losing is 18/19. The formula for expectation yields 17(1/19) + (-1)(18/19) = -1/19.

If we bet a single chip on three numbers, the payout is 11-to-1 for a winner. The probability of winning is 3/38 and the probability of losing is 35/38. The expectation is given by 11(3/38) + (-1)(35/38) = -1/19.

You may also bet a single chip on four numbers, six numbers, 12 numbers or 18 numbers for respective payouts of 8-to-1, 5-to-1, 2-to-1 and 1-to-1. Upon calculating the expectation for each of these possibilities, you will find the answer always is -1/19. The fact the answer in all cases is -1/19 means that all of these types of bets are equivalent to each other in terms of expectation. The bettor is giving away 1/19 of each bet on average no matter what kind of bet is placed.

The fact the expectation is the same for each type of bet should induce a curious person to ask whether or not something is going on mathematically. I don't suppose anyone will be surprised to learn the answer is yes.

With a single chip the bettor can bet on 1,2,3,4,6,12, or 18 numbers at a time. Note that all of these numbers divide 36. Secondly, the payout scheme is based on 36 as follows. Divide 36 by the number of numbers on which you are betting and subtract one from the answer. This is the payout you will receive if you win the bet. For example, when betting a single chip on four numbers, you divide 36 by four to obtain nine. Subtracting one from nine gives eight and this bet pays 8-to-1 for a winner.

If a roulette wheel had only 36 numbers, the preceding calculation of the payout would produce a game for which the expectation is zero. The house edge then comes from the introduction of extra positions. If there is a single extra position added (a so-called European roulette wheel), the house advantage is 1/(36+1) = 1/37 or slightly less than three percent. If two extra positions are added, the house advantage becomes 2/(36+2) = 1/19 as calculated earlier.

Let us now see how to extend these ideas to a general roulette-like game. Choose a positive integer n and let $d(1),d(2),\ldots,d(t)$be the positive divisors of n. Let $w(i) = n/d(i)-1, i = 1,\ldots,t$. If a bettor wagers one chip on d(i) numbers, pay w(i)-to-1 for a win. Finally, introduce x extra positions into the game which will provide the house edge. (Roulette is the special case of n=36 and x=2.)

We claim the general roulette-like game described above has a house edge of x/(n+x) for all possible bets. To verify the claim we calculate the expectation for this general roulette-like game. Suppose we are betting one chip on d(i) numbers. The probability of winning is d(i)/(n+x) and the probability of losing is (n+x-d(i))/(n+x). The expectation is then


which simplifies to x/(n+x) upon substituting the expression for w(i) given earlier.

Therefore, we see the expectation and payouts for roulette or any roulette-like game are functions of the number of positions on which the payouts are based, divisors of this number, and the number of extra positions introduced. It simply is a function of elementary arithmetic.

The above indicates there is great difficulty in escaping the expectation present in roulette. Can a bettor do anything? If you stop and watch a roulette game in action, you seldomly see a player betting a single chip at a time. It is common for players to bet many chips and several different types of bets for each spin. Let's take a look at one such scheme.

I know a person who bets nine squares at a time with a single chip on each square. The probability of winning is 9/38 and the probability of losing is 29/38. If the player wins, he is paid 35-to-1 for the winning position, but loses the other eight chips for a total profit of 27. If he loses, all nine chips bite the dust. The expectation is then 27(9/38) + (-9)(29/38) = -9/19 but this wager risks nine chips so the expectation per chip bet is still -1/19. We can see that the -1/19 sits there as barrier. Some people would have us believe there is a way around this barrier. We shall discuss this in the future.

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