Got the closed-form solution: P_i = a^i (1/a - 1) where a is the real-valued solution of a^3 + a^2 + a - 1 = 0 . a = 1/3 (-1 - 2/(17 + 3 sqrt(33))^(1/3) + (1

Got the closed-form solution: P_i = a^i (1/a - 1) where a is the real-valued solution of a^3 + a^2 + a - 1 = 0 . a = 1/3 (-1 - 2/(17 + 3 sqrt(33))^(1/3) + (17 + 3 sqrt(33))^(1/3)) edit: The probability to win comes out as a^2 ~= 0.2955977425220847709809965928515386138989754484466083115379546015... edit2: To arrive to this solution I expressed the equations in terms of R_i = P(choosing a number larger than i), then substitute P_i = R_{i-1} - R_i . This gets rid of the sums when you evaluate Q_{j+1} - Q_j , and arrives at R_{j+1}^2 = R_{j-1}(2 R_j - R_{j-1}). We know that R_0 = 1, and lim R_i = 0. I just tried R_i = a^i, and it works, with the value of "a" calculated above. edit3: To name names: the solution is a geometric distribution [1] What I ended up reducing the algebra problem to is a homogeneous recurrence relation, albeit non-linear. As it has constant coefficients, a^i is the obvious candidate to test it with. I got successfully nerd-sniped, this is a fun problem. Next time with 4 players... [1] https://en.wikipedia.org/wiki/Geometric_distribution https://en.wikipedia.org/wiki/Geometric_distribution