Probability distributions, E(X), and why the expected value is usually a number the variable can never be.
Here is a distribution on 1, 2, 3, 4. The triangle underneath is the balance point, and it sits at E(X). Shift the probability and watch where it has to move to keep the bars balanced.
Notice how rarely the pivot lands underneath a bar. That is the whole point of the word "expected".
E(X) is not the value you expect. It is the long-run average, so it is a balance point, not a prediction. The most likely value is the tallest bar, and the two are usually different numbers. A family cannot have 2.4 children and a die cannot land on 3.5, and both are perfectly good expected values.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | 0.1 | 0.2 | 0.3 | 0.4 |
| x × P | 0.1 | 0.4 | 0.9 | 1.6 |
P(X = x) = 4 + x18 for x ∈ {1, 2, 3}.
A game is fair when the expected gain is zero. Not when the chances are equal, and not when it feels reasonable. Work out E(gain) including the stake, and see whether it comes to 0.
1. For the table above, find E(X).
2. For P(X = x) = 4 + x18 on {1, 2, 3}, find E(X) to 3 decimal places.
3. A game pays out 5 with probability 0.3. What stake makes it fair?
Checking that the probabilities sum to 1, and using that to find an unknown. It is often the first mark and it is free.
Writing ΣxP(X = x) out, not just the answer. The method mark survives an arithmetic slip; a bare number does not.
On a fair game, include the stake. The expected winnings and the expected gain are different quantities, and only one of them is zero in a fair game.
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