What is the expected value of rolling a standard six-sided die?

Question

Accepted Answer

3.5. The expected value is the weighted average of all possible outcomes. For a standard die: E(X) = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21/6 = 3.5. Each outcome has probability 1/6, so E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5. Note that the expected value does not need to be a value that can actually be rolled.

The value 3 is the median (or one of the middle values) of the possible outcomes rather than the expected value; some students select 3 thinking it is the "middle" of 1 through 6, but the true middle (average) is 3.5. The value 4 overestimates the expected value, perhaps from rounding up or from an incorrect calculation. The value 21 is the sum of all outcomes (1+2+3+4+5+6 = 21) before dividing by the number of outcomes; this step of division by 6 is essential.

Expected value is a fundamental probability concept on the DSST exam. Students should understand that it represents the long-run average outcome of a random process and can calculate it for various probability distributions. Expected value is used in decision-making, games of chance, insurance, and many other practical contexts.

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