Fair Dice Comparison: d20 vs. 3d6
Two players each roll dice independently and repeatedly. Player A rolls a single fair 20-sided die (faces 1 through 20). Player B rolls three fair 6-sided dice and takes their sum. Each round, both players roll simultaneously. If their results are equal, they roll again. The game ends the first time the results differ, and whichever player has the higher number wins.
What is the probability that Player A wins?
Hints
- Don't try to compute the full PMF of 3d6 -- look for a symmetry property instead.
- If $(a, b, c)$ sums to $k$, what does $(7-a, 7-b, 7-c)$ sum to? This bijection implies $P(Y = k) = P(Y = 21-k)$.
- Condition on Player A's roll and use the PMF symmetry to show $P(A > B) = P(B > A)$. Since ties are re-rolled, the game is fair.
Worked Solution
How to Think About It: At first glance this looks messy -- you would need the full PMF of the sum of three d6s, which ranges from 3 to 18. But the question is really asking about a comparison game conditioned on disagreement. The key observation is a symmetry property of the 3d6 distribution: the PMF is symmetric around 10.5, meaning $P(Y = k) = P(Y = 21 - k)$ for all $k$. If you can prove that, the rest falls out cleanly.
Quick Estimate: Player A's die covers 1-20 uniformly, while Player B's sum covers 3-18. Player A auto-loses on rolls of 1 or 2 (B can never roll that low), but auto-wins on 19 or 20 (B can never roll that high). That is 2 values each way out of 20, so those cancel. The interior overlap region (3-18) has the symmetry of the 3d6 sum, which suggests the game might be perfectly fair. Gut estimate: $P(A \text{ wins}) = 1/2$.
Approach: Prove the PMF symmetry of 3d6, then use the law of total probability conditioning on Player A's roll.
Formal Solution:
Let $X$ be Player A's roll (uniform on $\{1, \ldots, 20\}$) and $Y$ be Player B's sum of three d6s. Since the game continues until $X \neq Y$, we need $P(X > Y \mid X \neq Y)$. By symmetry of the stopping rule, this equals $P(X > Y) / P(X \neq Y)$, and similarly for $P(X < Y)$. So it suffices to show $P(X > Y) = P(X < Y)$.
Step 1 -- PMF symmetry of 3d6. For any outcome $(a, b, c)$ with $a + b + c = k$, the complementary outcome $(7 - a, 7 - b, 7 - c)$ sums to $21 - k$. Since $1 \le a, b, c \le 6$ implies $1 \le 7 - a, 7 - b, 7 - c \le 6$, this is a valid outcome. The map $(a,b,c) \mapsto (7-a, 7-b, 7-c)$ is a bijection between outcomes summing to $k$ and outcomes summing to $21-k$. Since all $6^3 = 216$ outcomes are equally likely:
$$P(Y = k) = P(Y = 21 - k) \quad \text{for all } k.$$
Step 2 -- Symmetry of winning probabilities. Condition on $X = k$:
$$P(X > Y) = \sum_{k=1}^{20} P(X > Y \mid X = k) \cdot P(X = k) = \frac{1}{20} \sum_{k=1}^{20} P(Y < k).$$
Using the PMF symmetry, $P(Y < k) = P(Y > 21 - k)$. Substituting and re-indexing ($k \to 21 - k$):
$$P(X > Y) = \frac{1}{20} \sum_{k=1}^{20} P(Y > 21 - k) = \frac{1}{20} \sum_{k=1}^{20} P(Y > k) = P(X < Y).$$
Step 3 -- Conclusion. Since the game only ends when $X \neq Y$, and $P(X > Y) = P(X < Y)$, we have:
$$P(A \text{ wins}) = \frac{P(X > Y)}{P(X > Y) + P(X < Y)} = \frac{1}{2}.$$
Answer: The probability that Player A wins is $\dfrac{1}{2}$. The game is perfectly fair.
Intuition
The deep insight here is that the sum of dice whose faces are $\{1, 2, \ldots, m\}$ has a PMF symmetric around $(m+1)/2$ per die, or $(m+1)d/2$ for $d$ dice. This follows from the "complement" bijection: replacing each face $a$ with $m+1-a$ preserves validity and maps sums of $k$ to sums of $d(m+1)-k$. For 3d6, symmetry is around 10.5, meaning sums of $k$ and $21-k$ are equally likely. Once you have that, the uniform d20 treats high and low values of $Y$ identically, making the comparison symmetric.
This symmetry trick appears constantly in dice problems and discrete probability. Whenever you see a sum of i.i.d. discrete uniform variables, check for this complement symmetry before grinding through case analysis -- it often collapses the problem entirely.