Problem 121

DIRECTORY.md

@@ -771,6 +771,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_119/sol1.py)
   * Problem 120
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_120/sol1.py)
+  * Problem 121
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_121/sol1.py)
   * Problem 123
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_123/sol1.py)
   * Problem 125

project_euler/problem_121/sol1.py

@@ -0,0 +1,64 @@
+"""
+A bag contains one red disc and one blue disc. In a game of chance a player takes a
+disc at random and its colour is noted. After each turn the disc is returned to the
+bag, an extra red disc is added, and another disc is taken at random.
+
+The player pays £1 to play and wins if they have taken more blue discs than red
+discs at the end of the game.
+
+If the game is played for four turns, the probability of a player winning is exactly
+11/120, and so the maximum prize fund the banker should allocate for winning in this
+game would be £10 before they would expect to incur a loss. Note that any payout will
+be a whole number of pounds and also includes the original £1 paid to play the game,
+so in the example given the player actually wins £9.
+
+Find the maximum prize fund that should be allocated to a single game in which
+fifteen turns are played.
+
+
+Solution:
+    For each 15-disc sequence of red and blue for which there are more red than blue,
+    we calculate the probability of that sequence and add it to the total probability
+    of the player winning. The inverse of this probability gives an upper bound for
+    the prize if the banker wants to avoid an expected loss.
+"""
+
+from itertools import product
+
+
+def solution(num_turns: int = 15) -> int:
+    """
+    Find the maximum prize fund that should be allocated to a single game in which
+    fifteen turns are played.
+    >>> solution(4)
+    10
+    >>> solution(10)
+    225
+    """
+    total_prob: float = 0.0
+    prob: float
+    num_blue: int
+    num_red: int
+    ind: int
+    col: int
+    series: tuple[int, ...]
+
+    for series in product(range(2), repeat=num_turns):
+        num_blue = series.count(1)
+        num_red = num_turns - num_blue
+        if num_red >= num_blue:
+            continue
+        prob = 1.0
+        for ind, col in enumerate(series, 2):
+            if col == 0:
+                prob *= (ind - 1) / ind
+            else:
+                prob *= 1 / ind
+
+        total_prob += prob
+
+    return int(1 / total_prob)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")