Non-Transitive Dice
Three fair 6-sided dice $A$, $B$, and $C$ have the following faces:
- Die $A$: $\{2, 2, 6, 6, 7, 7\}$
- Die $B$: $\{1, 1, 5, 5, 9, 9\}$
- Die $C$: $\{3, 3, 4, 4, 8, 8\}$
Each value appears on exactly two faces, so each die shows each of its three values with probability $1/3$.
Find the probability that die $A$ shows a strictly higher value than die $B$.
Hints
- Each die has only 3 distinct values, each with probability $1/3$. This gives you a $3 \times 3$ grid of equally likely outcomes to check.
- Condition on the value of $B$. For each value $B$ can show, count how many of $A$'s values strictly exceed it.
- Use the law of total probability: $P(A > B) = \sum_{b} P(A > b) \cdot P(B = b)$ where the sum is over $B$'s three values $\{1, 5, 9\}$.
Worked Solution
How to Think About It: Each die has three equally likely outcomes (each with probability $1/3$), so this is a $3 \times 3$ grid of cases. Condition on what $B$ rolls and count how often $A$ beats it. This is a brute-force-friendly problem -- there are only 9 cases to check.
Quick Estimate: Die $A$ has values $\{2, 6, 7\}$ and die $B$ has values $\{1, 5, 9\}$. On average, $A$ has mean $5$ and $B$ has mean $5$, so the dice are roughly evenly matched in expectation. But the question is about $P(A > B)$, not about means. Let's count: $B$ rolls low (1) a third of the time, and $A$ always beats 1. $B$ rolls medium (5) a third of the time, and $A$ beats it with 6 or 7 (2 out of 3). $B$ rolls high (9) a third of the time, and $A$ never beats it. So roughly $P \approx (1/3)(1) + (1/3)(2/3) + (1/3)(0) = 5/9 \approx 0.556$.
Approach: Enumerate all 9 equally likely pairs and count victories.
Formal Solution:
Since each die shows each of its three distinct values with probability $1/3$, we have 9 equally likely outcome pairs. Condition on the value of $B$:
| $B$ rolls | Prob | $A$ values that beat $B$ | $P(A > B \mid B)$ | |-----------|------|--------------------------|--------------------| | $1$ | $1/3$ | $2, 6, 7$ (all three) | $3/3 = 1$ | | $5$ | $1/3$ | $6, 7$ (two of three) | $2/3$ | | $9$ | $1/3$ | none | $0$ |
By the law of total probability:
$$P(A > B) = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{2}{3} + \frac{1}{3} \cdot 0 = \frac{1}{3} + \frac{2}{9} = \frac{5}{9}$$
Answer: $P(A > B) = \dfrac{5}{9} \approx 0.556$.
Note: This problem is part of the "non-transitive dice" family. One can verify that $P(B > C) > 1/2$ and $P(C > A) > 1/2$ as well, creating a cycle: $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$ -- each with probability $5/9$. This violates transitivity, which is the surprising punch line.
Intuition
Non-transitive dice are a beautiful example of how pairwise comparisons can violate transitivity -- something that deeply contradicts most people's intuition. We are used to thinking "if A beats B and B beats C, then A beats C," but that logic fails for random variables compared by $P(X > Y)$.
This has real applications in finance and decision theory. When comparing strategies, assets, or portfolios pairwise by $P(\text{outperformance})$, you can get cycles. Strategy A outperforms B most of the time, B outperforms C most of the time, yet C outperforms A most of the time. This is why ranking by pairwise win rates can be misleading -- it's not a total order. In tournament design, voting theory, and even sports analytics, non-transitivity is a well-known pitfall. The takeaway: always check whether your comparison metric is transitive before using it to rank alternatives.