How to (Maybe) Survive a Math Gulag: A Guide

06 Nov, 2025

You're sent to a gulag¹.
The one way you get out is if you beat your opponent in the following game: There is a button RND in many old-school calculators. It generates a random number between $(0, 1)$ , assumed to follow the uniform distribution. To allow players some control, you may both choose a threshold $τ$ . When the first drawn number is less than $τ$ , it is discarded and replaced by a new random number.
Whoever has the greater number wins.
What threshold do you choose to maximize your chance to survive?

Probability of Winning if You Keep Your Score $x$

We compute $P (youwin ∣ you keep x)$ given that the opponent follows the same threshold $τ$ .

Case 1: $x \leq τ$ (low draw)

If your score $x$ is $\leq τ$ , you only win if the opponent also redraws (i.e., their first score $\leq τ$ ).

P (opponent redraws) = τ

After redrawing, the opponent’s new score $y$ is uniformly distributed in $[0, 1]$ . You win if $y \leq x$ , which occurs with probability $x$ .

Hence,

P (win ∣ x \leq τ) = τ \cdot x = τ x .

Case 2: $x \geq τ$ (high draw)

There are two possibilities:

Opponent redraws: This happens with probability $τ$ , and you win if $x > y$ (probability $x$ ). Thus, contribution: $τ \cdot x$ .
Opponent keeps: This happens with probability $(1 - τ)$ . You win if the opponent's score lies in $(τ, x]$ , which has probability $(x - τ)$ .

Adding these contributions gives

P (win ∣ x \geq τ) = τ x + (x - τ) .

To Summarize

P (win if you keep x) = {\begin{matrix} τ x, & x \leq τ, \ τ x + x - τ, & x \geq τ . \end{matrix}

Probability of Winning if You Redraw

Let your redraw be $y ~ U [0, 1]$ .

Case A: Opponent redraws

This occurs with probability $τ$ . You win if $y > y_{op}$ , where both are independent uniforms.

P (y > y_{op}) = \frac{1}{2} .

Thus, the contribution is $τ \cdot \frac{1}{2}$ .

Case B: Opponent keeps

This occurs with probability $(1 - τ)$ . Given that the opponent’s score $x_{op} > τ$ , we have

P (y > x_{op} ∣ x_{op} > τ) = \int_{τ}^{1} P (y > x_{op}), \frac{d x_{op}}{1 - τ} = \int_{τ}^{1} (1 - x_{op}), \frac{d x_{op}}{1 - τ} = \frac{1 - τ}{2} .

Thus, the contribution is $(1 - τ) \cdot \frac{1 - τ}{2}$ .

Total

P (win if you redraw) = τ \cdot \frac{1}{2} + (1 - τ) \cdot \frac{1 - τ}{2} = \frac{1 - τ + τ^{2}}{2} .

Nash Equilibrium

At equilibrium, a player must be indifferent between keeping or redrawing when their score equals the threshold:

P (win if you keep) = P (win if you redraw)

For $x = τ$ , both cases yield

P (win if you keep) = τ^{2} .

And,

P (win if you redraw) = \frac{1 - τ + τ^{2}}{2} .

Equating these:

τ^{2} = \frac{1 - τ + τ^{2}}{2} .

Solving for $τ$

Simplify:

2 τ^{2} = 1 - τ + τ^{2} \Rightarrow τ^{2} + τ - 1 = 0 .

Solving using the quadratic formula:

τ = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 1)}}{2 (1)} = \frac{- 1 \pm \sqrt{5}}{2} .

Since $τ \in [0, 1]$ , we take the positive root:

τ^{*} = \frac{- 1 + \sqrt{5}}{2} \approx 0.618 .

Which happens to be the golden ratio $ϕ$ . Why? I don't know. I will update this or make a new post if I ever figure it out. #StayTuned²

So in conclusion, if you're opponent is as logical and well-read in math as you are, then the survival rate is 50% for you again. Which means you may or may not die (?), so it really doesn't change much of your life trajectory if you've ridden a citibike in nyc at least once.
Thank you for reading this semi-informative and probably incomplete proof. Always welcome to email me for errors and improvements here.

Not the Soviet Union gulag ↩
Or maybe I will quit math before reaching that point :0↩

#math #problems