🎰Expected Value

A fair stall rolls a die: roll a 6 and win 6 yuan, anything else wins nothing. One play "might" earn you 6 yuan — but what if you play all afternoon? How much is each turn worth on average? This question has an exact answer, called the expected value.

Probabilities as weights

For one fair die, the expected value of the roll is $(1+2+3+4+5+6) \div 6 = 3.5$. The standard recipe multiplies each outcome by its probability and adds everything up:

$1 \times \frac{1}{6} + 2 \times \frac{1}{6} + \cdots + 6 \times \frac{1}{6} = \frac{21}{6} = 3.5$

Probable outcomes carry more say — the same idea as the weighted averages of mean, median and mode, with probabilities wearing the weights. Notice: no die face shows 3.5. It is not the result of any single roll but the value the long-run average settles toward.

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Heads rate

Flip onceFlip 100×Flip 1000×↺

The more you flip, the closer the heads rate gets to 50% — that's the law of large numbers.

Use it on the simplest wager of all: heads wins 1 yuan, tails loses 1 yuan. The expectation is $1 \times \frac{1}{2} + (-1) \times \frac{1}{2} = 0$ — an even break, a "fair game". Flip 10 times and the heads ratio may swing wildly; flip 1000 times and it hugs 50%. Expectation is not a promise for one round — it is the long-run average that the law of large numbers delivers.

Work out a lottery's expectation

Expectation is not a single result

Buy one ticket and you will never hold "2 yuan": you get 100, or 25, or 0. Expectation is a weighted average level, not a guarantee for one draw — a single outcome can land miles away from it. But as the count grows, the average clings ever closer to the expectation: in the coin experiment, after 1000 flips the heads ratio has stopped wandering away from 50%. One bet rides on luck; ten thousand bets ride on expectation.

Why the house always wins

In a casino, every game's expectation is negative for the player. Take roulette: the European wheel has 37 pockets — 18 red, 18 black, 1 zero. Bet 1 yuan on red: winning probability $\frac{18}{37}$, losing probability $\frac{19}{37}$, so the expectation is

$1 \times \frac{18}{37} + (-1) \times \frac{19}{37} = -\frac{1}{37} \approx -0.027$

Losing under 3 cents per yuan wagered looks harmless? Millions of players place billions of bets, and about 2.7 cents leave every yuan — the house sleeps soundly. The raffle's −1 yuan and the wheel's −0.027 yuan share one moral: when the expectation is negative, the longer you play, the more punctually you lose. To keep exploring probability, visit probability; combined events wait in compound probability.

Check yourself

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Quick quiz

↺ Try again

1. 1. One roll of a fair die: what is the expected value of the number shown?

   3.534

2. 2. 10 straws hide one 9-yuan prize, the rest are blanks. You draw one for free — what is the expected prize?

   0.9 yuan9 yuan4.5 yuan

3. 3. A game has an expectation of −0.5 yuan per play. After about 1000 plays you will have...

   lost around 500 yuanbroken evengained around 500 yuan