In order to find the starting position of the tug-of-war to make a fair fight, we define the function f on [-0.5,0.5] as

f(x) =: Prob(Player 1 wins at a starting position of x).

The symmetry of the game implies

f(x) = Prob(Player 1 wins in the first move) + ∫_x^½ Prob(Player 2 wins starting at position (-y)) dy

= (½ + x) + ∫_x^½(1 - f(-y))dy.

Differentiating and applying the fundamental theorem of calculus, we get

f’(x) = 1 - (1 - f(-x)) = f(-x).

Differentiating again and employing the chain rule, and then substituting the equation above, we get

f’’(x) = f’(-x)·(-1) = -f’(-x) = -f(x). The general solution to this differential equation is

f(x) = Asin(x) + Bcos(x).

We need two boundary conditions to determine f exactly. First, it’s clear we must have f(½) = 1. Second, from the above formula, f’(0) = f(0)

Therefore, Asin(½) + Bcos(½) = 1, and Acos(0) - B sin(0) = Asin(0) + Bcos(0).

The second equation implies A = B, and the first equation then gives

A = B = 1/(sin(½) + cos(½)), so f(x) = (sin(x) + cos(x))/(sin(½) + cos(½)).

Therefore the answer is the solution to (sin(x) + cos(x))/(sin(½) + cos(½)) = ½ on [-½,0]. A calculator can give this to the desired accuracy, -0.2850001…, or by using that

sin(x) + cos(x) = √(2)·sin(x + 𝜋/4)

we can exactly solve that the answer is arcsin(sin(½ + 𝜋/4)/2) - 𝜋/4.

Congrats to this month’s solvers who successfully found the answer for running an honest robot tug-of-war!