Can the Archers Coordinate?

Two logicians work as a team for a fabulous prize. There are three targets. To win, each logician fires a single arrow, and they win only if both hit the same target. Two of the targets are closer and otherwise indistinguishable; each logician hits either of these with probability $98$ percent. The third target is farther, hit with probability $70$ percent. The logicians cannot coordinate beforehand and learn nothing of each other’s choice or success. What is the probability they win?

The Fiddler, Zach Wissner-Gross, February 20, 2026(original post)

Solution

Winning needs two things: both logicians aim at the same target, and both hit it. Being identical reasoners with no way to talk, they settle on the same (possibly random) rule. By symmetry such a rule picks each close target with probability $q$ and the far target with probability $1 - 2q$. The two close targets are interchangeable, so when both “aim at a close target” they may still pick different ones; the far target is unique, so aiming there guarantees agreement. The win probability is $$P(q) = 2\,q^2 (0.98)^2 + (1 - 2q)^2 (0.70)^2 .$$ This is convex in $q$, so the maximum sits at an endpoint. At $q = \tfrac12$ they win with probability $0.4802$; at $q = 0$ they both take the far target and win with probability $0.70^2$. The larger is $$P = (0.70)^2 = \boxed{0.49}.$$ The unique far target wins precisely because it removes the coordination risk: the easier close targets are wasted half the time when the two logicians choose different ones.

The computation

Sweep the shared strategy: for each probability $q$ of aiming at a close target, evaluate the win probability $2q^2(0.98)^2+(1-2q)^2(0.70)^2$ and take the best. The maximum is $0.49$, at $q=0$ (both aim at the far target).

import numpy as np
q = np.linspace(0, 0.5, 100001)
P = 2*q**2*0.98**2 + (1 - 2*q)**2*0.70**2
print(round(P.max(), 4), "at q =", round(q[P.argmax()], 3))   # 0.49 at q=0.0

Extra Credit

The probabilities change. Two targets stay indistinguishable, each hit with probability $a$; the third is hit with probability $b$; all three are rational. Doing some mental arithmetic, the logicians find that it no longer matters which target they aim for: their winning probability is the same regardless. What is that probability?

Solution

“It does not matter which target they aim for” is the signature of an interior mixed equilibrium, where every target is an equally good reply to the partner’s strategy. Writing the shared rule as probability $p$ on each close target and $p_3 = 1 - 2p$ on the far one, the value of aiming at a close target is $p\,a^2$ and of the far target is $p_3\,b^2$. Indifference forces $p\,a^2 = p_3\,b^2$, and the common win probability is that shared value, $$P = p\,a^2 = \boxed{\dfrac{a^2 b^2}{a^2 + 2b^2}} .$$ For instance $a = \tfrac9{10}$, $b = \tfrac45$ yields $P = 0.248\ldots$ The source’s specific rational targets sit behind its paywall, so this closed form, rather than a single number, is the honest answer.

The computation

Solve the indifference condition itself: find the mixing weight $p$ at which aiming close, $p\,a^2$, equals aiming far, $(1-2p)\,b^2$, then report the common value $p\,a^2$.

import sympy as sp
a, b = sp.Rational(9, 10), sp.Rational(4, 5); p = sp.symbols('p')
p_star = sp.solve(sp.Eq(p*a**2, (1 - 2*p)*b**2), p)[0]
print(p_star*a**2)                          # 1296/5225 = 0.24803...