{-

- 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

main = putStrLn (show ans)

ans = div (product [1 .. turns+1]) (sum [ways turns i | i <- [(div turns 2)+1 .. turns]])

ways :: Integer -> Integer -> Integer

ways 0 0 = 1

ways 0 _ = 0

ways k b = k * (ways (k - 1) b) + (ways (k - 1) (b - 1))