Strictly Increasing Dice Rolls
You roll three fair six-sided dice one at a time. What is the probability that the three results come out in strictly increasing order -- that is, the first roll is less than the second, which is less than the third?
Hints
- Think about splitting the event into two parts: first, that all three dice show different values, and second, that those values happen to be in the right order.
- Given three distinct values, how many orderings are there, and how many are strictly increasing? Use symmetry.
- Compute $P(\text{all distinct}) = 1 \cdot \frac{5}{6} \cdot \frac{4}{6}$, then multiply by $\frac{1}{3!}$ since only one of the $3!$ orderings of distinct values is strictly increasing.
Worked Solution
How to Think About It: There are $6^3 = 216$ equally likely ordered outcomes. The question is: out of all those triples, how many satisfy $d_1 < d_2 < d_3$? The slick move is to notice that any set of three distinct values can be arranged in exactly $3! = 6$ orderings, and exactly one of those is strictly increasing. So the problem reduces to: how often do we get three distinct values? Multiply that by
Quick Estimate: Roughly, two out of six faces collide on the second roll ($P(\text{distinct so far}) \approx 5/6$), and about two more on the third ($4/6$). So the fraction with all distinct values is around $(5/6)(4/6) \approx 56\%$. One-sixth of those are in increasing order: $0.56/6 \approx 0.093$. The exact answer should be close to $5/54 \approx 0.093$.
Approach: Decompose via $P(\text{strictly increasing}) = P(\text{all distinct}) \times P(\text{increasing} \mid \text{all distinct})$.
Formal Solution:
Let $A$ be the event that all three dice show different values, and $B$ the event that they appear in strictly increasing order. Note that $B \subseteq A$ (you cannot have a strict increase with a repeated value), so:
$P(B) = P(A) \cdot P(B \mid A)$
Computing $P(A)$: Roll sequentially. The first die can be anything. The second must differ from the first -- probability $5/6$. The third must differ from both -- probability $4/6$. So:
$P(A) = 1 \cdot \frac{5}{6} \cdot \frac{4}{6} = \frac{20}{36} = \frac{5}{9}$
Computing $P(B \mid A)$: Given all three values are distinct, there are $3! = 6$ equally likely orderings of those values (by symmetry of the dice). Exactly one ordering is strictly increasing. So:
$P(B \mid A) = \frac{1}{6}$
Combining:
$P(B) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54}$
Answer: $\dfrac{5}{54} \approx 9.26\%$
Intuition
The key principle here is symmetry over orderings. Whenever you have $k$ distinct values drawn from any symmetric distribution (dice, uniform draws, etc.), all $k!$ orderings are equally likely. So the probability of any specific ordering -- strictly increasing, strictly decreasing, alternating -- is exactly
This shows up constantly in quant interviews and in real work. Order statistics problems -- 'what is the probability the minimum exceeds $x$?', 'what is the distribution of the median?' -- often reduce to exactly this symmetry argument. Any time you see 'strictly ordered' combined with 'independent symmetric random variables,' your first move should be: find the probability of no ties, then use