Can You Hack Gymnastics?

Gymnast $A$ has a difficulty score of $6.0$ and gymnast $B$ a difficulty score of $5.0$. Each receives an independent execution score drawn uniformly at random from $0$ to $10$. What is the probability that the two gymnasts end up in the same relative order whether their two subscores are added or multiplied?

The Fiddler, Zach Wissner-Gross, August 9, 2024(original post)

Solution

Write $a,b$ for the two execution scores, each uniform on $[0,10]$. Adding puts $A$ ahead when $(6+a)-(5+b)=1+a-b>0$; multiplying puts $A$ ahead when $6a-5b>0$. The two methods give the same order exactly when these share a sign, that is when $(a,b)$ lies on the same side of both lines $b=a+1$ and $b=\tfrac65a$. Those lines cross at $(5,6)$, inside the square, so the only disagreement is the pair of thin slivers between them. Integrating the agreement region over the $10\times10$ square gives area $\tfrac{577}{6}$, hence probability $$\boxed{\tfrac{577}{600}}\approx96.17\%.$$

The computation

Encode the agreement condition itself and integrate it: the region where both gaps are positive, plus the region where both are negative, over the $10\times10$ square.

import sympy as sp
a, b = sp.symbols('a b')
f1, f2 = 1 + a - b, 6 * a - 5 * b                # added gap, multiplied gap
def region(cond):                               # area where both gaps satisfy cond
    return sp.integrate(sp.Piecewise((1, cond), (0, True)), (b, 0, 10), (a, 0, 10))
agree = region((f1 > 0) & (f2 > 0)) + region((f1 < 0) & (f2 < 0))
print(sp.Rational(agree, 100))                    # 577/600

Extra Credit

Now let both difficulty scores also be independent and uniform on $[0,10]$. What is the probability the ranking agrees under adding and multiplying?

Solution

With all four scores random, write the difficulties $d_A,d_B$ and executions $a,b$, all uniform on $[0,10]$. Adding and multiplying agree when $(d_A+a)-(d_B+b)$ and $d_Aa-d_Bb$ share a sign. Averaging over the four-dimensional cube, the probability falls to about $$\boxed{0.9168.}$$ It drops from the $96\%$ of the fixed-difficulty case because random difficulties make near-ties far more common, and near-ties are exactly where adding and multiplying can disagree. (There is no tidy closed form; this is the simulated value.)

The computation

Sample the four scores uniformly and measure how often the two orderings agree.

import numpy as np
rng = np.random.default_rng(0); n = 20_000_000
dA, dB, a, b = rng.random((4, n)) * 10
agree = (np.sign((dA + a) - (dB + b)) == np.sign(dA * a - dB * b)).mean()
print(agree)                                       # ~0.9168