A probability game

Assume that two positive numbers are given, $X$ and $Y$, with unknown joint probability distribution $P$, and $X\ne Y$ a.s.

A player draws randomly one of the numbers and has to guess if the number is smaller or larger than the other unrevealed number, i.e., let $U\sim Bernoulli(\frac{1}{2})$ independent of $X,Y$, then the player sees $Z_1=UX+(1-U)Y,$ and $Z_2=(1-U)X+UY$ remains unseen.

A random guess (coin-flip) would due to the sampling $U$, indepedently of $F$, have probability $\frac{1}{2}$ of correct guessing. The question is if we can find a better strategy?

Assume we can sample $\tilde{Z}$ from a distribution with support covering $X$ and $Y$. Then the player can instead guess that $Z_1>Z_2$ iff $Z_1>\tilde{Z}$. In this case the player guesses correct with probability

$$\begin{array}{rl}& \mathbb{P}(Z_1>\tilde{Z},Z_1>Z_2)+\mathbb{P}(Z_1<\tilde{Z},Z_2>Z_1)=\\ & \mathbb{P}(Z_1>\tilde{Z},Z_1>Z_2,U=1)+\mathbb{P}(Z_1>\tilde{Z},Z_1>Z_2,U=0)\\ & +\mathbb{P}(Z_1<\tilde{Z},Z_2>Z_1,U=1)+\mathbb{P}(Z_1<\tilde{Z},Z_2>Z_1,U=0)=\\ & \frac{1}{2}\mathbb{P}(X>\tilde{Z},X>Y)+\frac{1}{2}\mathbb{P}(Y>\tilde{Z},Y>X)+\frac{1}{2}\mathbb{P}(X<\tilde{Z},X<Y)+\frac{1}{2}\mathbb{P}(Y<\tilde{Z},Y<X)\end{array}$$

Applying $\mathbb{P}(A)+\mathbb{P}(B)=\mathbb{P}(A\cup B)+\mathbb{P}(A\cap B)$, we have

$$\begin{array}{rl}& \frac{1}{2}[\mathbb{P}(X>Y)+\mathbb{P}(X>Y,X>\tilde{Z}>Y)+\\ & \mathbb{P}(X<Y)+\mathbb{P}(X<Y,X<\tilde{Z}<Y)]\\ & =\frac{1}{2}\{1+\mathbb{P}(\tilde{Z}\text{ between }X\text{ and }Y)\},\end{array}$$

which under the given assumptions is strictly larger than $\frac{1}{2}$. The player should intuitively try to to choose the distribution of $\tilde{Z}$ that “separates” $X$ and $Y$.

Example in python

import sys
import numpy as np
import scipy.linalg as linalg
from   scipy.stats import norm, bernoulli, uniform

if sys.version_info.major < 3 or sys.version_info.minor<6:
      raise Exception("Must be using Python >= 3.6")

def rexp(n: int): return np.exp(-uniform.rvs(size=n))

class game:
      """Simulation of simple probability game"""
      def __init__(self,
		 n: int,
		 gen=lambda n: -np.log(uniform.rvs(size=n)),
		 rho=.5):
	    R = np.matrix([[1, rho], [rho, 1]]);
	    L = linalg.cholesky(R)
	    xy = np.matrix(np.resize(norm.rvs(size=2*n), (n,2)))*L
	    self.u = bernoulli.rvs(0.5, size=n)
	    self.w = np.array(xy[range(len(xy)), self.u])
	    self.z = np.array(xy[range(len(xy)), 1-self.u])
	    self.x = np.array(xy[:,[0]])
	    self.y = np.array(xy[:,[1]])
	    self.g = gen(n)
	    self.guess = self.w>self.g
	    self.true = self.w>self.z

n = int(1e5)
sim = game(n)
print(np.mean(sim.true==sim.guess))

0.60851

Example in R

sim <- function(n, delta=0, ...) {
  xy <- exp(mets::rmvn(n, ...)) + delta
  u <- rbinom(n, 1, 0.5)
  w <- xy[cbind(seq(n), u+1)]
  z <- xy[cbind(seq(n), 2-u)]
  wmax <- w>z
  cbind(w=w, z=z, wmax=wmax, x=xy[,1], y=xy[,2])
}

n <- 1e5
val <- sim(n=n, rho=0.5)
z0 <- rexp(n, 1)
guess <- val[, "w"]>z0
mean(guess==val[, "wmax"])

[1] 0.6037

val <- sim(n=1e5, rho=0.5, mu=c(-3,3))
z0 <- rexp(n, 1)
guess <- val[, "w"]>z0
mean(guess==val[, "wmax"])

[1] 0.96077

val <- sim(n=1e5, rho=0.5, mu=c(10,10))
z0 <- rexp(n, 1)
guess <- val[, "w"]>z0
mean(guess==val[, "wmax"])

[1] 0.49956