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06 · Expected value

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Probability, lesson 6

Expected value

The average outcome, and why it decides almost every bet.

A chance says how often something happens. To judge a bet you also need how much it pays, and expected value puts the two into one number. You need chances of outcomes, and the rule that they add up to 1.

By the end, you can:

  1. Find a bet's expected value, and the fair price to play it.
  2. Add expected values, even when the parts depend on each other.
  3. Choose by comparing what you hold with what carrying on is worth.

Part 1. The average of a bet

Find a bet's expected value, and the fair price to play it.

Roll a die and win the number it shows. One game tells you almost nothing, since you might win 1 or 6. Ask instead what you win per game, over many games.

Example 1.1

The die game

Roll a fair die and win the number it shows. On average, how much do you win per game?

  1. Picture 6,000 games6,000 × 1/6 = 1,000

    Each face comes up about 1,000 times.

  2. Add up the winnings1,000 × 21 = 21,000

    The faces add up to 1 + 2 + 3 + 4 + 5 + 6 = 21, and each is won about 1,000 times.

  3. Divide by the games21,000 / 6,000 = 3.5

    The win per game.

Answer3.5 a game, a number the die can't even show.

The 6,000 cancels out. What's left is each outcome times its chance, added up: 1 × 1/6, plus 2 × 1/6, and so on up to 6 × 1/6. Any number of games gives the same 3.5. It's the long-run average, just as a chance is the long-run share.

Expected value

The expected value of a random amount X, written E[X], is each value times its chance, added up. Σ, read "sum", means add this up over every value. For the die, E[X] = 21/6 = 3.5. A bet's fair price is its expected value: pay that, and neither side has an edge.

E[X] = Σ x × P(x)

Outcomes needn't be equally likely: weight each by its chance. A bet that wins 10 with chance 0.1 and loses 1 otherwise is worth 10 × 0.1 − 1 × 0.9 = +0.1 a play. It's a good bet, yet it loses 9 plays in 10. Expected value isn't what usually happens. It's what happens on average.

Die game: 1/6 on each face 1 2 3 4 5 6 balances at 3.5 Win 10 (chance 0.1), else lose 1 balances at 0.1 0.9 on −1 0.1 on 10
Die game: 1/6 on each face 1 2 3 4 5 6 balances at 3.5 Win 10 (chance 0.1), else lose 1 balances at 0.1 0.9 on −1 0.1 on 10
Figure 1Put a weight on each value, as heavy as its chance. The expected value is where the beam balances: 3.5 for the die, 0.1 for the bet.

Your turn

It costs 4 to play the die game. Should you play?

You pay 4 and win 3.5 on average, so each game is worth −0.50 to you. You'll win some games, but over many the average settles near −0.50 a game. The fair price, where neither side has an edge, is 3.50.

Part 2. Expected values add

Add expected values, even when the parts depend on each other.

Keep going with Pro

You've done the first part. Pro opens the rest of this lesson, and every other one.

Still to come
  1. Expected values add
  2. Keep or roll again
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Next lessonVariance and riskSame average, different ride: why spread matters as much as the mean. Start lesson 7

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