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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))