Can You Sprint to the Finish?

You and one rival approach the finish line. Each of you privately assesses your legs as a number uniform on $[0,1]$, then decides whether to sprint. A sprinter always beats a non-sprinter; if both sprint, the fresher legs win; if neither sprints, it is a coin flip. Your rival’s manager has announced that the rival will sprint whenever their legs feel at least $0.5$. Playing optimally, what is your chance of winning?

The Fiddler, Zach Wissner-Gross, July 25, 2025(original post)

Solution

Sprinting dominates: against a non-sprinter you win for certain (versus a coin flip if you also hold back), and against a sprinter you win with probability equal to your chance of fresher legs (versus a sure loss if you hold back). So the optimal rule is to always sprint, whatever your legs. With your legs $y\sim U[0,1]$ and the rival sprinting only when their legs $o\ge\tfrac12$, $$\begin{aligned} P(\text{win})&=\underbrace{\tfrac12}_{\text{rival holds back}}\cdot 1 +\tfrac12\!\!\int_{1/2}^{1}\!\!2\,(1-o)\,do\\ &=\tfrac12+\tfrac12\cdot\tfrac14=\frac58\approx\boxed{62.5\%}. \end{aligned}$$

The computation

Run the race: you always sprint, the rival sprints only on legs $\ge\tfrac12$. Against a holding rival you win; against a sprinting one the fresher legs win. The win rate lands on $\tfrac58$.

import numpy as np
rng = np.random.default_rng(0); M = 8_000_000
y = rng.random(M); o = rng.random(M); op = o >= .5        # you always sprint
win = np.where(~op, 1.0, (y > o)*1.0)                     # vs non-sprinter win; else fresher legs
print(round(win.mean(), 5))                              # 0.625 = 5/8

Extra Credit

Now three riders race. Both rivals sprint at $0.5$ or better. Your manager fills the blank in “sprint if your legs feel at least ” with a number chosen uniformly at random in $[0,1]$. Before you draw that threshold and test your legs, what is your chance of winning?

Solution

Average over the random threshold $t$, your legs $y$, and the rivals’ legs. Since $t\sim U[0,1]$, you sprint with probability $y$; conditioning on the number of sprinting rivals ($0,1,2$ with probabilities $\tfrac14,\tfrac12,\tfrac14$, each with legs $U[\tfrac12,1]$), $$\begin{aligned} P(\text{win})&=\tfrac14\!\int_0^1\!\Bigl(y+\tfrac{1-y}{3}\Bigr)dy +\tfrac12\!\int_{1/2}^1\! y(2y-1)\,dy\\ &\quad+\tfrac14\!\int_{1/2}^1\! y(2y-1)^2 dy\\ &=\frac16+\frac{5}{48}+\frac{7}{192}=\frac{59}{192}\approx\boxed{30.7\%}. \end{aligned}$$ (The source value is paywalled; this closed form is my own.)

The computation

Simulate the three-rider race directly: draw your random threshold and everyone’s legs, decide who sprints, and award the win to the sprinter with the freshest legs (a three-way coin flip when nobody sprints).

rng = np.random.default_rng(1); M = 8_000_000
t, y, o1, o2 = (rng.random(M) for _ in range(4))
ys = y >= t; s1 = o1 >= .5; s2 = o2 >= .5
legs = np.stack([y, o1, o2]); spr = np.stack([ys, s1, s2])
big = np.where(spr, legs, -1.0); anyone = spr.any(0)
youmax = (big[0] >= big[1]) & (big[0] >= big[2]) & ys
print(round(np.where(anyone, youmax*1.0, 1/3).mean(), 5))   # 0.3073 = 59/192