# nayuki/Project-Euler-solutions

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 {- - 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))