Three Dice in Strictly Increasing Order
A fair six-sided die is rolled three times in sequence, giving results $X_1, X_2, X_3$. What is the probability that the results are strictly increasing, that is, $X_1 < X_2 < X_3$?
Hints
- Strictly increasing requires all three values to be different. Start by computing the probability that no two rolls agree.
- Given three distinct values, the three rolls landed on them in one of 3! equally likely orders. How many of those orders are increasing?
- $P(\text{all distinct}) = 1 \cdot \tfrac{5}{6} \cdot \tfrac{4}{6}$, and only $1$ of the $3! = 6$ orderings is increasing.
Worked Solution
How to Think About It: There are two independent conditions hiding here: the three rolls must be distinct, and, given distinct values, they must come out in ascending order. Symmetry handles the second condition: once you know the three distinct values, every one of their $3! = 6$ arrangements is equally likely.
Quick Estimate: Rolls are distinct about $56\%$ of the time ($\tfrac{5}{6} \cdot \tfrac{4}{6}$), and one in six of those is increasing: roughly $9\%$.
Formal Solution:
*Step 1 -- All three distinct.* The second roll differs from the first with probability $5/6$, and the third differs from both with probability $4/6$, so
$$P(\text{all distinct}) = \frac{5}{6} \cdot \frac{4}{6} = \frac{20}{36} = \frac{5}{9}.$$
*Step 2 -- Order given distinct.* Conditional on three distinct values, the $3!$ orderings are equally likely (the rolls are exchangeable), and exactly one ordering is increasing:
$$P(X_1 < X_2 < X_3 \mid \text{all distinct}) = \frac{1}{6}.$$
*Step 3 -- Multiply.*
$$P(X_1 < X_2 < X_3) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54} \approx 0.0926.$$
*Direct count.* An increasing triple is the same as choosing 3 different faces out of 6: $\binom{6}{3} = 20$ outcomes out of $6^3 = 216$, and $20/216 = 5/54$.
Answer: $P(X_1 < X_2 < X_3) = \dfrac{5}{54} \approx 0.0926$.
Intuition
Any strictly increasing outcome is just a choice of three distinct faces, and by symmetry each of the $3!$ orders of three distinct faces is equally likely, so exactly one sixth of the "all distinct" outcomes are increasing. Counting $\binom{6}{3} = 20$ favorable triples out of 216 says the same thing. Splitting "distinct values" from "their order" is the standard move for any pattern-in-a-sequence question, from runs in coin flips to the probability that successive trade prices are monotone.