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Roll three fair dice. What is the probability that all three show different values?

Accepted Answer

Sequence the rolls. The first die is free. The second must avoid one value, the third must avoid two: P = 1 cdot frac56 cdot frac46 = frac2036 = frac59 approx 0.556. Equivalently by counting: frac6 cdot 5 cdot 46^3 = frac120216 = frac59. This "sequential avoidance" pattern, multiply survival probabilities one object at a time, is the same engine behind the birthday paradox, so make it automatic here first.

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