How it works
Multiply each possible value by its probability, then add:
E = (value₁ × probability₁) + (value₂ × probability₂) + …
A fair six-sided die has expected value (1 + 2 + … + 6) ÷ 6 = 3.5. The expected value does not have to be a possible outcome.
The most powerful shortcut is linearity of expectation: the expected value of a sum is the sum of the expected values, even when the parts depend on each other. Doubling a payout doubles its expected value. For two coin flips the expected number of heads is ½ + ½ = 1.
Expected Value practice problems with solutions
These problems come from Mathyessy's question bank. Try each one before opening the solution.
A game pays 10 dollars with probability 0.5 and 0 otherwise. What is the expected value in dollars?
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Answer: $5
10 × 0.5 + 0 × 0.5 = 5.
A die pays double the number rolled in dollars. What is the expected payout in dollars?
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Answer: $7
The average roll is 3.5, and doubling every payout doubles the average: 2 × 3.5 = 7.
A game pays 16 dollars for three heads, 4 dollars for exactly two heads, and 0 otherwise when flipping three coins. What is the expected value in dollars?
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Answer: $3.50
P(three heads) = 1/8 and P(exactly two heads) = 3/8. E = 16 × 1/8 + 4 × 3/8 = 2 + 1.5 = 3.5.
A game pays 8 dollars for each head in two flips plus the number rolled on a fair die in dollars. What is the expected value in dollars?
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Answer: $11.50
By linearity, add the parts. Expected heads in two flips is 1, worth 8 × 1 = 8 dollars. The expected die roll is 3.5. Total 8 + 3.5 = 11.5.
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