[Above is a Mathematica-derived solution to the systems of equations described in the 4-player tournament described below]

Without loss of generality we can assume an arrow’s distance to the center of the target is U[0,1] distributed. Let *P _{j,k}*(

*x*) denote the probability that Player

*j*out of

*k*in the

**current**line will eventually win the game, given that the current best dart is at distance

*x*.

In a 2-player archery tournament, we would have *P*_{1,2}(*x*) + *P*_{2,2}(*x*) = 1, *P*_{1,2}(0) = 0, and *P*_{1,2}(*x*) = ∫_{0}^{x} *P*_{2,2}(*u*) d*u*. Taking derivatives of both sides in that last equation and substituing gives *P*‘_{1,2}(*x*) = −*P*_{1,2}(x). Solving gives *P*_{1,2}(*x*) = 1−e^{−x}.

In the 3-player version of the game, we get the following system of equations:

*P*_{1,3}(*x*) = ∫_{0}^{x} *P*_{3,3}(*u*) d*u*

*P*_{2,3}(*x*) = (1 - *x*)*P*_{1,2}(*x*) + ∫_{0}^{x} *P*_{1,3}(*u*) d*u*

*P*_{1,3}(*x*) + *P*_{2,3}(*x*) + *P*_{3,3}(*x*) = 1

*P*_{1,3}(0) = *P*_{2,3}(0) = 0.

And for the 4-player version we have the following:

*P*_{1,4}(*x*) = ∫_{0}^{x} *P*_{4,4}(*u*) d*u*

*P*_{2,4}(*x*) = (1 - *x*)*P*_{1,3}(*x*) + ∫_{0}^{x} *P*_{1,4}(*u*) d*u*

*P*_{3,4}(*x*) = (1 - *x*)*P*_{2,3}(*x*) + ∫_{0}^{x} *P*_{2,4}(*u*) d*u*

*P*_{1,4}(*x*) + *P*_{2,4}(*x*) + *P*_{3,4}(*x*) + *P*_{4,4}(*x*) = 1

*P*_{1,4}(0) = *P*_{2,4}(0) = *P*_{3,4}(0) = 0.

In particular we have *P*_{4,4}(1) = (-5/4)*(cos(1) + sin(1)) + (1/2)**e*^{-1} + (*e*^{-1/2})*(cos(sqrt(3)/2) + 5/sqrt(3)*sin(sqrt(3)/2)), or approximately 0.18343765086.

**Congrats to this month’s solvers!**