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* Added solution for Project Euler problem 121 * Updated typing for 3.9 * updating DIRECTORY.md Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
65 lines
2.0 KiB
Python
65 lines
2.0 KiB
Python
"""
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A bag contains one red disc and one blue disc. In a game of chance a player takes a
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disc at random and its colour is noted. After each turn the disc is returned to the
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bag, an extra red disc is added, and another disc is taken at random.
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The player pays £1 to play and wins if they have taken more blue discs than red
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discs at the end of the game.
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If the game is played for four turns, the probability of a player winning is exactly
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11/120, and so the maximum prize fund the banker should allocate for winning in this
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game would be £10 before they would expect to incur a loss. Note that any payout will
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be a whole number of pounds and also includes the original £1 paid to play the game,
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so in the example given the player actually wins £9.
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Find the maximum prize fund that should be allocated to a single game in which
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fifteen turns are played.
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Solution:
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For each 15-disc sequence of red and blue for which there are more red than blue,
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we calculate the probability of that sequence and add it to the total probability
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of the player winning. The inverse of this probability gives an upper bound for
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the prize if the banker wants to avoid an expected loss.
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"""
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from itertools import product
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def solution(num_turns: int = 15) -> int:
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"""
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Find the maximum prize fund that should be allocated to a single game in which
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fifteen turns are played.
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>>> solution(4)
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10
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>>> solution(10)
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225
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"""
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total_prob: float = 0.0
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prob: float
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num_blue: int
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num_red: int
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ind: int
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col: int
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series: tuple[int, ...]
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for series in product(range(2), repeat=num_turns):
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num_blue = series.count(1)
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num_red = num_turns - num_blue
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if num_red >= num_blue:
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continue
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prob = 1.0
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for ind, col in enumerate(series, 2):
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if col == 0:
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prob *= (ind - 1) / ind
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else:
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prob *= 1 / ind
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total_prob += prob
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return int(1 / total_prob)
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if __name__ == "__main__":
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print(f"{solution() = }")
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