MISSION · Data and probability
Expected Value
The average you'd get per go if you played forever
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The idea in one sentence
Multiply each outcome by its probability and add them all up; the result is what you’d average per attempt over a very long run.
\[E(X) = \sum x_i \, P(x_i)\]A fair die
The outcomes are 1 to 6, each with probability 1/6:
\[E(X) = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = 3.5\]You will never roll a 3.5. That’s not a flaw in the calculation — the expected value is the average of the outcomes, not a prediction of any single one. Roll a thousand times and the mean will sit very close to 3.5.
When all outcomes are equally likely, expected value is just the ordinary mean. What makes it more than that is unequal probabilities: it’s a weighted average, where each outcome’s weight is how often it happens.
A game with money
Win £5 with probability 0.2, otherwise lose £2.
\[E(X) = 5(0.2) + (-2)(0.8) = 1 - 1.6 = -£0.60\]Sixty pence lost per go, on average. You might win the first three times — the expected value says nothing about that. It says that over a few hundred goes you’re down about £60, and that number is remarkably reliable.
A game with \(E(X) = 0\) is called fair. Commercial games are never fair: the expected value is negative for the player by design, and that gap is the operator’s income.
The lottery
A £2 ticket with a 1-in-14-million chance of £10 million:
\[E(X) = 10{,}000{,}000 \times \frac{1}{14{,}000{,}000} - 2 \approx 0.71 - 2 = -£1.29\]Even a jackpot that vast doesn’t rescue it, because the probability is smaller than the prize is big. That comparison — how big against how unlikely — is the only thing expected value ever measures.
Where the answer isn’t the whole story
Expected value ignores spread entirely, and sometimes spread is what matters.
Compare a certain £50 with a coin flip for £0 or £100. Both have \(E(X) = 50\), and almost nobody treats them the same. Insurance has negative expected value for the buyer, and buying it is still sensible, because the worst case is worse than the average is good.
So expected value answers “what happens over many repetitions?”, and that’s the wrong question when you only get one go and the downside is severe.
Other things it computes
Take the expected value of \((X - \mu)^2\) and you get the variance. Expected value is the machine underneath most of statistics, not just gambling problems: a binomial distribution has \(E(X) = np\) by exactly this calculation, and that’s where the familiar formula comes from.
Common mix-ups
- Expecting the expected value. 3.5 is not a face on the die.
- Forgetting the negative outcomes, or the cost of playing.
- Probabilities not summing to 1, which means an outcome is missing.
- Using it for one-off decisions with a catastrophic downside, where the long-run average never gets a chance to apply.
Check yourself
Find the expected value of one roll of a fair die. Then: you win £5 with probability 0.2 and lose £2 otherwise — what’s the expected value per go?
The die averages 21/6 = 3.5. The game gives 5(0.2) − 2(0.8) = 1 − 1.6 = −0.6, so it costs about 60p a go in the long run.
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