Probability of Three Dice Being Strictly Increasing

You roll three fair six-sided dice in sequence. We want the probability that the outcomes satisfy first < second < third.

How to think about it

  1. There are 6 × 6 × 6 = 216 total ordered outcomes.
  2. A strictly increasing triple must use 3 distinct numbers.
  3. Choose any 3 different faces from {1,2,3,4,5,6}. The number of choices is C(6,3) = 20.
  4. For each chosen set, there is exactly one strictly increasing order.
P(first < second < third) = C(6,3) / 6³
= 20 / 216 = 5 / 54 ≈ 0.09259
Favorable outcomes
20
Total outcomes
216
Exact probability
5 / 54
Decimal
9.26%

Another viewpoint: if three rolls are all distinct, then their values can be arranged in 3! = 6 possible orders, and exactly one of those is increasing.

Try a random roll

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Simulation estimate

All 20 favorable outcomes

Each one is a strictly increasing triple: