How Low (or High) Can You Go?

In “high-low” you see a sequence of independent uniform $[0,1]$ numbers and, before each new one, guess whether it will be higher or lower than the previous. The optimal rule is to guess “high” when the last number is below $\tfrac12$ and “low” when it is above. Playing this way, what is the probability you earn at least two points, that is, you correctly compare $x_2$ to $x_1$ and then $x_3$ to $x_2$?

The Fiddler, Zach Wissner-Gross, September 5, 2025(original post)

Solution

Condition on the middle value $x_2$. Given $x_2$, the two comparisons are independent: the second is correct with probability $B(x_2)=\max(x_2,1-x_2)$, and integrating the first over $x_1$ gives $A(x_2)=x_2+\tfrac12$ for $x_2<\tfrac12$ and $\tfrac32-x_2$ for $x_2>\tfrac12$. Hence $$\begin{aligned} P(\text{2 points})&=\int_0^{1}A(x_2)\,B(x_2)\,dx_2\\ &=2\int_0^{1/2}\!\bigl(x_2+\tfrac12\bigr)(1-x_2)\,dx_2 =\frac{13}{24}\approx\boxed{54.2\%}. \end{aligned}$$

The computation

Play the game: draw three numbers, apply the optimal rule (guess high if the previous is below $\tfrac12$, else low) on both comparisons, and take the fraction of rounds scoring both.

import numpy as np
rng = np.random.default_rng(0); M = 8_000_000
x1, x2, x3 = (rng.random(M) for _ in range(3))
ok = lambda p, n: np.where(p < .5, n > p, n < p)         # high if prev<.5 else low
print(round((ok(x1, x2) & ok(x2, x3)).mean(), 5))        # 0.5417 = 13/24

Extra Credit

A friend has been playing forever and amassed a huge score. Given only that, what is the probability they win the next round?

Solution

A long survivor sits at a value drawn from the game’s quasi-stationary distribution. The kernel that carries a surviving value $v$ to the next is $K(v,u)=\mathbf 1[v<\tfrac12,\,u>v]+\mathbf 1[v>\tfrac12,\,u<v]$; its leading eigenfunction is $\varphi(v)\propto e^{v/\lambda}$ on $(0,\tfrac12)$ (symmetric about $\tfrac12$), and substituting back forces $e^{1/(2\lambda)}=2$. The per-round survival probability, which is exactly the chance of winning the next round, is the eigenvalue $$\lambda=\frac{1}{2\ln 2}\approx\boxed{0.7213}.$$ (The source value is paywalled; this closed form is my own.)

The computation

Discretise the survival kernel $K$ on a fine grid and power-iterate to its leading eigenvalue: that eigenvalue is the long-run per-round survival probability, the chance the veteran wins again.

n = 2000; xs = (np.arange(n) + .5) / n; dx = 1 / n
K = np.zeros((n, n))
for i, v in enumerate(xs): K[i, (xs > v) if v < .5 else (xs < v)] = 1.0
w = np.ones(n)
for _ in range(20000):
    w2 = (K.T @ w) * dx; lam = w2.sum() / w.sum(); w = w2 / np.linalg.norm(w2)
print(round(lam, 4), round(1/(2*np.log(2)), 4))          # 0.7211 (n=2000) -> 0.7213