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*) = *A*sin(*x*) + *B*cos(*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, *A*sin(½) + *B*cos(½) = 1, and *A*cos(0) - *B* sin(0) = *A*sin(0) + *B*cos(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!**