The Lazy Duelist: Optimal Shooting Distance

Two duelists, A and B, walk toward each other. At each step, each can choose to shoot. Duelist A hits with probability $p_A = 1/2$ if they shoot at the current moment, and duelist B hits with probability $p_B = 1/3$ . They take turns stepping closer (A shoots first on each round, then B if A misses).

But here is the puzzle, stripped to its probabilistic core:

A and B simultaneously decide (without knowing the other's choice) whether to shoot NOW or wait one more round. If they wait, both hit probabilities double (up to certainty). A shoots first. Should A shoot now or wait?

Actually, let's pose the clean classic version:

Problem (Mosteller, Problem 20 — The Duel):

Two men, $A$ and $B$ , take turns shooting at each other. $A$ shoots first. At each turn, each man has probability $\frac{1}{2}$ of hitting the other. What is the probability that $A$ wins (i.e., hits $B$ before $B$ hits $A$ )?

Now the twist: what if A is a worse shot — A hits with probability $p$ and B hits with probability $q$ , with $p < q$ ? Surprisingly, there is a clean closed form. Find it, and determine: for what values of $p$ is it better to go second (i.e., $P(A \text{ wins}) < 1/2$ )?

geometric seriesfirst-mover advantageduelprobabilityclassic

Answer: The Duelist's Dilemma: Who Wins When A Shoots First?

Key Idea / Intuition

Going first sounds like an advantage — but it only helps if you actually hit. Each round, A shoots first; if A misses (probability $1-p$ ), then B gets a shot. The game is memoryless, so the probability of A winning satisfies a clean geometric series. The surprise: A wins more than half the time if and only if $p > q(1-p)$ , i.e., $p > q/(1+q)$ . Going first is not always enough to overcome a skill disadvantage.

Formal Proof / Solution

Setup. Each round proceeds:

A shoots and hits B with probability $p$ . → A wins.
If A misses (probability $1-p$ ), B shoots and hits A with probability $q$ . → B wins.
If both miss (probability $(1-p)(1-q)$ ), the round resets.

Computing $P(A \text{ wins})$ .

Let $W$ = event A wins. Within one round:

A wins with probability $p$ .
B wins with probability $(1-p)q$ .
Round resets with probability $(1-p)(1-q)$ .

Since the game resets to the same state, we get the equation:

$P(A \text{ wins}) = p + (1-p)(1-q) \cdot P(A \text{ wins})$

Let $\alpha = P(A \text{ wins})$ . Then:

$\alpha = p + (1-p)(1-q)\,\alpha$

$\alpha \left[1 - (1-p)(1-q)\right] = p$

$\boxed{\alpha = \frac{p}{1 - (1-p)(1-q)} = \frac{p}{p + q - pq}}$

Sanity check: If $p = q = 1/2$ : $\alpha = \frac{1/2}{1/2 + 1/2 - 1/4} = \frac{1/2}{3/4} = \frac{2}{3}$

So A wins with probability $2/3$ when both are equally skilled — going first is a big advantage when you have a 50% hit rate!

When does A win with probability less than $1/2$ ?

We need $\alpha < 1/2$ : $\frac{p}{p + q - pq} < \frac{1}{2}$ $2p < p + q - pq$ $p < q - pq = q(1-p)$ $\frac{p}{1-p} < q$

So A is the underdog (despite going first) exactly when:

$q > \frac{p}{1-p}$

For example, if $p = 1/3$ , then A is an underdog whenever $q > (1/3)/(2/3) = 1/2$ .

Interpretation. Going first gives A a guaranteed first shot, but if A's hit probability is low, the advantage is small. A skilled B can overcome the order disadvantage as long as B's skill $q$ exceeds the threshold $p/(1-p)$ .

Summary Table for $p = q = 1/2$ :

| Event | Probability | |-------|-------------| | A wins in round 1 | $1/2$ | | B wins in round 1 | $1/4$ | | Game continues | $1/4$ | | A wins overall | $\mathbf{2/3}$ |

The clean formula $\alpha = \dfrac{p}{p + q - pq}$ is the heart of the problem.

Written to question file and answer included below.

Source: Fifty Challenging Problems in Probability with Solutions, Frederick Mosteller, Problem 20

Type: ProbabilitySource: Fifty Challenging Problems in Probability with Solutions, Frederick Mosteller, Problem 20Edit on GitHub ↗