Probability That a Larger Die Wins

Probability · Medium · Free problem

You have two fair dice: one $m$-sided with faces

, 2, \ldots, m$ and one $n$-sided with faces
, 2, \ldots, n$, where $n > m$. The dice are rolled independently.

Find the probability that the $n$-sided die shows a strictly larger value than the $m$-sided die.

Then compute the answer for $m = 20$, $n = 30$.

Hints

  1. Split the event based on whether the $n$-sided die shows a value above $m$ or not -- in one case the result is immediate, in the other you have a symmetric contest.
  2. When both dice are effectively in the same range $\{1, \ldots, m\}$, use the symmetry argument: by equal probability, ties aside, each die is equally likely to be larger.
  3. Apply the law of total probability: $P(X > Y) = P(X > Y \mid X > m)\,P(X > m) + P(X > Y \mid X \leq m)\,P(X \leq m)$, and show the conditional win probability in the second term is $(m-1)/(2m)$.

Worked Solution

How to Think About It: The $n$-sided die has a home-field advantage -- it has more faces, so it can reach values that the $m$-sided die simply cannot. Any time the $n$-sided die lands above $m$, it wins outright, no matter what the other die shows. That chunk alone gives it a $(n - m)/n$ edge. The only real contest happens when both dice fall in the shared range

$ to $m$, and there the situation is symmetric. So partition the event space on whether the $n$-sided die exceeds $m$ or not, handle each case cleanly, and you are done in a few lines.

Quick Estimate: For $m = 20$, $n = 30$: the $n$-sided die lands above 20 with probability

0/30 = 1/3$ -- automatic wins. When it lands in
$-
0$ (probability /3$), we have two effectively uniform dice on the same range. By symmetry, the $n$-sided die is strictly larger half the time minus the ties. Ties happen with probability
/m = 1/20$ conditional on both being in range. So the conditional win probability is roughly $(1 - 1/20)/2 \approx 0.475$. Total estimate:
/3 + (2/3)(0.475) \approx 0.333 + 0.317 = 0.65$. The exact answer of
3/20 = 0.65$ matches perfectly.

Approach: Condition on whether the $n$-sided die lands above $m$ (guaranteed win) or within

$-$m$ (symmetric contest). Apply the law of total probability.

Formal Solution:

Let $X$ be the outcome of the $n$-sided die and $Y$ the outcome of the $m$-sided die. Define the event $G = \{X > m\}$, i.e., the $n$-sided die lands on a face that the $m$-sided die cannot reach.

Step 1: Partition. By the law of total probability,

$P(X > Y) = P(X > Y \mid G)\,P(G) + P(X > Y \mid G^c)\,P(G^c)$

Step 2: Evaluate $P(G)$. The $n$-sided die takes values $m+1, m+2, \ldots, n$, giving $n - m$ winning faces out of $n$:

$P(G) = \frac{n - m}{n}, \qquad P(G^c) = \frac{m}{n}$

Step 3: Conditional on $G$. If $X > m$, then $X > Y$ with certainty (since $Y \leq m$), so $P(X > Y \mid G) = 1$.

Step 4: Conditional on $G^c$. Given $X \leq m$, both $X$ and $Y$ are uniform on $\{1, 2, \ldots, m\}$ and independent. There are $m^2$ equally likely pairs. Ties $(X = Y)$ occur in exactly $m$ of them. The remaining $m^2 - m = m(m-1)$ pairs have $X \neq Y$. By symmetry between $X$ and $Y$ (both uniform on the same range), exactly half of these have $X > Y$:

$P(X > Y \mid G^c) = \frac{m(m-1)/2}{m^2} = \frac{m-1}{2m}$

Step 5: Combine.

$P(X > Y) = 1 \cdot \frac{n-m}{n} + \frac{m-1}{2m} \cdot \frac{m}{n} = \frac{n-m}{n} + \frac{m-1}{2n} = \frac{2(n-m) + (m-1)}{2n} = \frac{2n - m - 1}{2n}$

Alternative (direct counting): Count the pairs $(X, Y)$ with $X > Y$ directly. For each value $Y = k$ (with $k = 1, \ldots, m$), $X$ can be any of $k+1, k+2, \ldots, n$, giving $n - k$ winning outcomes. Summing over all $k$:

$\sum_{k=1}^{m}(n - k) = mn - \sum_{k=1}^{m} k = mn - \frac{m(m+1)}{2}$

Dividing by total outcomes $mn$:

$P(X > Y) = \frac{mn - m(m+1)/2}{mn} = 1 - \frac{m+1}{2n} = \frac{2n - m - 1}{2n}$

Both methods agree.

Answer: For general $m$, $n$ with $n > m$:

$P(X > Y) = \frac{2n - m - 1}{2n}$

For $m = 20$, $n = 30$:

$P(X > Y) = \frac{60 - 20 - 1}{60} = \frac{39}{60} = \frac{13}{20} = 0.65$

Intuition

The key structural insight is to decompose the probability space into a region where the outcome is obvious and a region where it requires calculation. Whenever one variable has a strictly larger range, there is always a guaranteed-win zone -- the values only it can take. That contribution is $(n - m)/n$ and requires no calculation at all. The remaining probability mass sits in the shared range where, by symmetry, the contest is nearly 50-50 (discounting ties). This decomposition pattern -- partition into 'trivial' and 'nontrivial' zones, handle each cleanly -- comes up constantly in discrete probability problems.

The direct counting approach is equally clean and worth knowing: when you want $P(X > Y)$ for discrete uniform variables with different ranges, just count favorable pairs by fixing $Y$ and summing over valid $X$ values. The resulting arithmetic series ($mn - m(m+1)/2$ divided by $mn$) collapses to the same formula. In interviews, being able to reach the same answer two independent ways -- conditioning and direct counting -- is a strong signal. It also confirms there are no off-by-one errors, which are the most common mistake when counting discrete outcomes.

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