(*

* Solution to Project Euler problem 121

* Copyright (c) Project Nayuki. All rights reserved.

*

* https://www.nayuki.io/page/project-euler-solutions

* https://github.com/nayuki/Project-Euler-solutions

*)

(*

* At the beginning of turn number k (0-based), there are k + 2 discs to choose from.

* Hence a game that has n turns has (n+1) * n * ... * 1 = (n + 1)! outcomes.

*

* Let f(k, b) be the number of ways to accumulate exactly b blue discs after k turns.

* We can see that:

* - f(0, 0) = 1.

* - f(0, b) = 0, for b > 0.

* - f(k, 0) = k * f(k - 1, 0), for k > 0.

* (Add a red disc, where there are k ways)

* - f(k, b) = f(k - 1, b - 1) + k * f(k - 1, b), for k > 0, b > 0.

* (Add a blue disc (1 way) or add a red disc (k ways))

*

* Next, we calculate the sum f(n, j) + f(n, j+1) + ... + f(n, n),

* where j is the smallest number of blue discs accumulated that exceeds

* the number of red discs accumulated (which is n - j). So j = ceil((n + 1) / 2).

*

* Finally, the probability of winning is that sum divided by (n + 1)!.

* For any game where the cost of playing is 1 and the probability of winning is p,

* the maximum sustainable prize is 1 / p, therefore the maximum sustainable integer prize is floor(1 / p).

*)

turns = 15;

Ways[0, 0] = 1;

Ways[0, _] = 0;

Ways[k_, b_] := k * Ways[k - 1, b] + Ways[k - 1, b - 1]

Floor[(turns + 1)! / Sum[Ways[turns, i], {i, Ceiling[(turns + 1) / 2], turns}]]