Student Question

In the game of roulette, a player can place a
$6 bet on the number 10 and have a 1 / 38
probability of winning. If the metal ball lands on
10, the player gets to keep the $6
paid to play the game and the player is awarded an additional $210.
Otherwise, the player is awarded nothing, and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. What is the expected value?

Quick answer:

The expected value of a game where losing costs $6 with a probability of 37/38 and winning yields $210 with a probability of 1/38 is approximately -$0.32.

Expert Answers

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We are told that the cost of a bet is $6. The bettor wins with a probability of 1/38, and the yield is $210 (they get their $6 entry back). The bettor loses with a probability 37/38 and loses their $6 entry.

We have a probability distribution:


Note that the probabilities sum to 1 and that all possibilities have been covered (as the bettor either wins or loses at each play).

We can calculate the mean and standard deviation of a probability distribution. We essentially treat it as a weighted mean of a distribution.


The key is to understand that the expected value (or expectation) of the game is the mean of the underlying probability distribution.

Thus, `E(x)=mu~~-0.3158`

Obviously, a player will never lose 32 cents at one game. Either the player loses $6 or wins $210. But over time, the average loss (in this case) is between 31 and 32 cents. If the player played 100 times, they should expect to lose about $31.58.

In a calculator that has tables, you can enter the possible wins/losses in one column and their corresponding probabilities in another column and compute the two-variable statistics on those columns. In a calculator without tables, you can enter the numerators as the frequencies (assuming the denominators of the probabilities are the same) and run two-variable statistics.

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