Dice Order

We roll three fair dice one at a time, giving an ordered triple. What is the probability the three values come out strictly increasing?

Reveal solutionHide solution

#Count the favourable rolls

Strictly increasing forces all three faces distinct, and any three distinct faces sort into exactly one increasing order. So the favourable rolls are the three-element subsets of $\{1, \dots, 6\}$ , of which there are $\binom{6}{3} = 20$ .

#Divide by everything

Three dice give $6^3 = 216$ equally likely ordered rolls, so

\PP(\text{strictly increasing}) = \frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54}. \tag{1}

The tempting answer $1/6$ , the chance that a fixed distinct triple happens to land sorted, forgets that the dice can also tie.

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Each set of three distinct faces sorts into exactly one increasing roll. The 20 such sets out of 216 ordered rolls give a probability of 5/54.