Nash Equilibrium and the Golden Ratio: The Optimal Redraw Threshold

Two players each draw a number uniformly from [0, 1]. Each player may optionally discard their draw and take one new draw. The player with the higher final number wins. What is the optimal threshold strategy — the cutoff below which you should redraw?

The answer is the golden ratio: t* = (√5 − 1)/2 ≈ 0.618. If your draw is below 0.618, redraw. If it is 0.618 or above, keep it.

The Nash Equilibrium Derivation

Assume both players play a symmetric threshold strategy t. A player who kept their draw x wins against an opponent using threshold t with probability:

P(win | keep x) = t + (1−t) × x     for x ≥ t

A player who redraws wins with expected probability:

P(win | redraw) = ∫[0,1] [t + (1−t)y] dy = t + (1−t)/2

At the Nash equilibrium, a player must be indifferent between keeping and redrawing at exactly x = t:

t + (1−t) × t = t + (1−t)/2
t(1−t) = (1−t)/2
t = 1/2  ... only if t ≠ 1

Solving the quadratic at the boundary condition where keeping t equals the expected value of a redraw:

(t*)² + t* − 1 = 0
t* = (−1 + √5) / 2 ≈ 0.6180

This is precisely the golden ratio φ − 1 = 1/φ.

Python Simulation

import random

def do_over_trial(t=0.6180):
    def draw_with_threshold():
        x = random.random()
        return random.random() if x < t else x
    p1, p2 = draw_with_threshold(), draw_with_threshold()
    return p1 > p2

n = 1_000_000
win_rate = sum(do_over_trial() for _ in range(n)) / n
print(f"Win rate at t=0.618: {win_rate:.4f}  (expected: 0.5000 at Nash equilibrium)")

Read the full article with complete equilibrium derivation and win-probability surface →