Expected Value

A game with three prizes

A spinner at a fair has 10 equal sectors. Five pay nothing, four pay $10 and one pays $50. Let X be the prize from one spin, in dollars.

The distribution of X comes from counting sectors: $P\left(X=0\right)=\frac{5}{10}=\frac{1}{2}$P(X = 0) = 5/10 = 1/2, $P\left(X=10\right)=\frac{4}{10}=\frac{2}{5}$P(X = 10) = 4/10 = 2/5 and $P\left(X=50\right)=\frac{1}{10}$P(X = 50) = 1/10. Check the total: $\frac{5}{10}+\frac{4}{10}+\frac{1}{10}=1$5/10 + 4/10 + 1/10 = 1.

Ten equal sectors: five pay $0, four pay $10, and the colored one pays $50.

Weight each value by its probability

Suppose 10 spins land exactly in the long-run proportions: five land on $0, four on $10 and one on $50. The prizes add up to 5 × 0 + 4 × 10 + 1 × 50 = 0 + 40 + 50 = 90 dollars, which is 90 ÷ 10 = 9 dollars a spin.

Dividing each count by 10 turns it into a probability. So the same 9 comes from multiplying each value by its probability and adding: $0\times \frac{1}{2}+10\times \frac{2}{5}+50\times \frac{1}{10}=0+4+5=9$0 × 1/2 + 10 × 2/5 + 50 × 1/10 = 0 + 4 + 5 = 9.

This number is called the expected value of X, written E(X). For any discrete random variable, E(X) is the sum of each value times its probability. Over 1000 spins the prizes come to about 500 × 0 + 400 × 10 + 100 × 50 = 9000 dollars, still 9 dollars a spin: E(X) is the long-run average prize per spin.

Ten spins in the long-run proportions, as weights on a beam from 0 to 50: five at 0, four at 10 and one at 50. The beam balances at 9, which is E(X).

Nobody wins $9

E(X) = 9 is not a prize on the spinner. No single spin pays $9: each spin pays $0, $10 or $50. The expected value is an average over many spins, not a prediction for one.

It lands between the prizes, just below 10, and well to the right of 0, the most likely prize. The one $50 sector pulls it there: that sector alone adds $50\times \frac{1}{10}=5$50 × 1/10 = 5 to E(X). If it paid nothing, E(X) would be 0 + 4 = 4.

The plain arrows point at the three prizes, 0, 10 and 50. The colored one points at E(X) = 9, between 0 and 10, which is not one of the prizes.

A fair game

Now charge $9 a spin. A player's gain is the prize minus the fee, X − 9, which is −9, 1 or 41 dollars with the same probabilities $\frac{1}{2}$1/2, $\frac{2}{5}$2/5 and $\frac{1}{10}$1/10. Its expected value is $-9\times \frac{1}{2}+1\times \frac{2}{5}+41\times \frac{1}{10}=-4.5+0.4+4.1=0$−9 × 1/2 + 1 × 2/5 + 41 × 1/10 = −4.5 + 0.4 + 4.1 = 0.

A quicker way: every gain is 9 less than its prize, so the average gain is 9 less than the average prize, E(X − 9) = 9 − 9 = 0. A game whose expected gain is 0 is called fair: over many spins, neither the player nor the stall comes out ahead.

At a fee of $10 the expected gain is 9 − 10 = −1 dollar a spin. The stall then expects to make about $1 on every spin, or about $100 over 100 spins.

The usual mistakes

Averaging the values without their probabilities. (0 + 10 + 50) ÷ 3 = 20 treats the three prizes as equally likely, but five sectors pay $0 and only one pays $50.

Taking the larger value as the answer. If X is 0 or 14 with probability $\frac{1}{2}$1/2 each, 14 happens only half the time: $E\left(X\right)=0\times \frac{1}{2}+14\times \frac{1}{2}=7$E(X) = 0 × 1/2 + 14 × 1/2 = 7.

Expecting E(X) to be a possible value. An expected number of heads of 1.5 is correct, even though no toss gives 1.5 heads.