Probability of Strictly Increasing Dice Rolls

Question

Roll a fair six-sided die three times (in order), obtaining $(X_1,X_2,X_3)$.
What is the probability that the results are strictly increasing:
$$
X_1 < X_2 < X_3 \ ?
$$

Jeremy Wu · Accepted Answer

\textbf{Step 1: Count the sample space.}
There are $6^3=216$ equally likely ordered outcomes.

\par\textbf{Step 2: Count favorable outcomes via choosing 3 distinct faces.}
For a strictly increasing triple, the three values must be \emph{distinct}.
Choose any 3 distinct numbers from $\{1,2,3,4,5,6\}$:
$$
\binom{6}{3}=20.
$$
Once the set $\{a < b < c\}$ is chosen, there is exactly \emph{one} ordered outcome that is strictly increasing,
namely $(a,b,c)$.

Therefore, the number of favorable ordered outcomes is $20$.

\par\textbf{Step 3: Compute the probability.}
$$
\mathbb{P}(X_1 < X_2 < X_3)=\frac{\text{favorable outcomes}}{\text{total outcomes}}
=\frac{\binom{6}{3}}{6^3}
=\frac{20}{216}
=\frac{5}{54}.
$$

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