Solution for fifty question 9: craps (#69)

* solution: analysis

* add(fifty): 9 craps solutions

docs/fifty/9_craps.md

@@ -0,0 +1,205 @@
+# Craps
+
+## Problem
+
+???+ question "Question"
+
+    === "English"
+
+        The game of craps, played with two dice, is one of America's fastest and most popular gambling games.
+        Calculating the odds associated with it is an instructive exercise.
+        The rules are these. Only totals for the two dice count.
+        The player throws the dice and wins at once if the total for the first throw is 7 or 11,
+        loses at once if it is 2, 3, or 12. Any other throw is called his "point."
+        If the first throw is a point, the player throws the dice repeatedly
+        until he either wins by throwing his point again or loses by throwing 7.
+        What is the player's chance to win?
+
+    === "中文"
+
+        掷骰子赌博游戏是美国最迅速和最受欢迎的赌博游戏之一，使用两颗骰子进行。计算相关的赔率是一项有益的练习。
+        规则如下：只计算两颗骰子的总和。
+        玩家掷骰子，如果第一次掷出的总点数为7或11，则立刻获胜；如果是2、3或12，则立刻输掉。其他点数称为玩家的“点数”。
+        如果第一次掷出的点数是一个点数，玩家将反复掷骰子，直到再次掷出点数获胜，或通过掷出7输掉。
+        玩家获胜的概率是多少？
+
+## Tip
+
+??? tip "Tip"
+
+    === "English"
+
+        Focus on breaking down the game into three possible outcomes on the first roll: win immediately, lose immediately, or get a point number. Then calculate the probabilities for each scenario.
+
+    === "中文"
+
+        将游戏分解为第一次掷骰子的三种可能结果：立即获胜、立即失败或得到一个点数。然后计算每种情况的概率。
+
+
+## Solutions
+
+### Solution1: Analysis
+
+??? success "Solution1: Analysis"
+
+    === "English"
+
+        First, we can list all possible point totals and their corresponding probabilities:
+
+        |    | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
+        | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
+        | Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
+
+        Let's denote winning on the first roll as $W_1$, losing on the first roll as $L_1$, and getting a point number as $P_1$.
+
+        The probability of winning on the first roll is:
+
+        $$
+        P(W_1) = P(7, 11) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9}
+        $$
+
+        Similarly, the probability of losing on the first roll is:
+
+        $$
+        P(L_1) = P(2, 3, 12) = \frac{1}{36} + \frac{2}{36} + \frac{1}{36} = \frac{4}{36} = \frac{1}{9}
+        $$
+
+        If a point number is rolled on the first roll, it could be 4, 5, 6, 8, 9, or 10.
+        Let's take the case of rolling a 4 as an example and calculate the probability of the player losing:
+
+        $$
+        \begin{align}
+        P(L_{p4}) &= P(7) + (1 - P(4) - P(7)) \cdot P(7) + (1 - P(4) - P(7))^2 \cdot P(7) + \cdots \\
+                  &= P(7) \cdot (1 + (1 - P(4) - P(7)) + (1 - P(4) - P(7))^2 + \cdots) \\
+                  &= P(7) \cdot \frac{1}{1 - (1 - P(4) - P(7))} \\
+                  &= P(7) \cdot \frac{1}{P(4) + P(7)} \\
+                  &= \frac{6}{36} \cdot \frac{1}{\frac{3}{36} + \frac{6}{36}} \\
+                  &= \frac{6}{36} \cdot \frac{1}{\frac{9}{36}} \\
+                  &= \frac{2}{3}
+        \end{align}
+        $$
+
+        Similarly, we can calculate the probabilities of losing when the point number is 5, 6, 8, 9, or 10:
+
+        $$
+        P(L_{p5}) = P(L_{p9}) = \frac{3}{5}
+        $$
+
+        $$
+        P(L_{p6}) = P(L_{p8}) = \frac{6}{11}
+        $$
+
+        $$
+        P(L_{p10}) = P(L_{p4}) = \frac{2}{3}
+        $$
+
+        From this, we can calculate the probability of the player winning after rolling a point number on the first roll:
+
+        $$
+        \begin{align}
+        P(L_p)  &= P(4) \cdot (1-P(L_{p4})) + P(5) \cdot (1-P(L_{p5})) + P(6) \cdot (1-P(L_{p6})) + P(8) \cdot (1-P(L_{p8})) + P(9) \cdot (1-P(L_{p9})) + P(10) (1-\cdot P(L_{p10})) \\
+                &= \frac{3}{36} \cdot (1-\frac{2}{3}) + \frac{4}{36} \cdot (1-\frac{3}{5}) + \frac{5}{36} \cdot (1-\frac{6}{11}) + \frac{5}{36} \cdot (1-\frac{6}{11}) + \frac{4}{36} \cdot (1-\frac{3}{5}) + \frac{3}{36} \cdot (1-\frac{2}{3}) \\
+                &= \frac{1}{36} + \frac{2}{45} + \frac{25}{396} + \frac{25}{396} + \frac{2}{45} + \frac{1}{36} \\
+                &= \frac{134}{495}
+        \end{align}
+        $$
+
+        Therefore, the total probability of the player winning is:
+
+        $$
+        P(W) = P(W_1) + P(W_p) = \frac{2}{9} + \frac{134}{495} = \frac{244}{495} \approx 0.49293
+        $$
+
+        So the probability of the player winning is approximately 49.293%.
+
+    === "中文"
+
+        首先，我们可以列出所有可能的点数和对应的概率：
+
+        |    | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
+        | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
+        | 概率 | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
+
+        我们将第一次掷骰子获胜的情况记为 $W_1$，输掉的情况记为 $L_1$，得到点数的情况记为 $P_1$。
+
+        那么，首次掷骰子获胜的概率为：
+
+        $$
+        P(W_1) = P(7, 11) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9}
+        $$
+
+        同理，首次掷骰子输掉的概率为：
+
+        $$
+        P(L_1) = P(2, 3, 12) = \frac{1}{36} + \frac{2}{36} + \frac{1}{36} = \frac{4}{36} = \frac{1}{9}
+        $$
+
+        如果首次掷骰子得到点数，那么点数的值可能为4、5、6、8、9、10中的任意一个。
+        我们以点数为4的情况为例，计算此时玩家输掉的概率：
+
+        $$
+        \begin{align}
+        P(L_{p4}) &= P(7) + (1 - P(4) - P(7)) \cdot P(7) + (1 - P(4) - P(7))^2 \cdot P(7) + \cdots \\
+                  &= P(7) \cdot (1 + (1 - P(4) - P(7)) + (1 - P(4) - P(7))^2 + \cdots) \\
+                  &= P(7) \cdot \frac{1}{1 - (1 - P(4) - P(7))} \\
+                  &= P(7) \cdot \frac{1}{P(4) + P(7)} \\
+                  &= \frac{6}{36} \cdot \frac{1}{\frac{3}{36} + \frac{6}{36}} \\
+                  &= \frac{6}{36} \cdot \frac{1}{\frac{9}{36}} \\
+                  &= \frac{2}{3}
+        \end{align}
+        $$
+
+        同理，我们可以计算得到点数为5、6、8、9、10时玩家输掉的概率分别为：
+
+        $$
+        P(L_{p5}) = P(L_{p9}) = \frac{3}{5}
+        $$
+
+        $$
+        P(L_{p6}) = P(L_{p8}) = \frac{6}{11}
+        $$
+
+        $$
+        P(L_{p10}) = P(L_{p4}) = \frac{2}{3}
+        $$
+
+        由此得到首次掷骰子得到点数且玩家赢的概率为：
+
+        $$
+        \begin{align}
+        P(L_p)  &= P(4) \cdot (1-P(L_{p4})) + P(5) \cdot (1-P(L_{p5})) + P(6) \cdot (1-P(L_{p6})) + P(8) \cdot (1-P(L_{p8})) + P(9) \cdot (1-P(L_{p9})) + P(10) (1-\cdot P(L_{p10})) \\
+                &= \frac{3}{36} \cdot (1-\frac{2}{3}) + \frac{4}{36} \cdot (1-\frac{3}{5}) + \frac{5}{36} \cdot (1-\frac{6}{11}) + \frac{5}{36} \cdot (1-\frac{6}{11}) + \frac{4}{36} \cdot (1-\frac{3}{5}) + \frac{3}{36} \cdot (1-\frac{2}{3}) \\
+                &= \frac{1}{36} + \frac{2}{45} + \frac{25}{396} + \frac{25}{396} + \frac{2}{45} + \frac{1}{36} \\
+                &= \frac{134}{495}
+        \end{align}
+        $$
+
+        所以得到玩家总的获胜概率为：
+
+        $$
+        P(W) = P(W_1) + P(W_p) = \frac{2}{9} + \frac{134}{495} = \frac{244}{495} \approx 0.49293
+        $$
+
+        所以玩家获胜的概率为约为49.293%。
+
+
+
+### Solution2: Simulation
+
+??? success "Solution2: Simulation"
+
+    === "English"
+
+        The overall process of the game is relatively simple, and we can simulate it through the following code:
+
+        ```python exec="true" source="material-block" session="fifty-9"
+        --8<-- "docs/fifty/snippet/9_craps.py:solution2"
+        ```
+
+    === "中文"
+
+        游戏的过程总体比较简单，我们可以通过下面的代码来模拟:
+
+        ```python exec="true" source="material-block" session="fifty-9"
+        --8<-- "docs/fifty/snippet/9_craps.py:solution2"
+        ```

docs/fifty/snippet/9_craps.py

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+# --8<-- [start:solution2]
+import random
+
+
+def roll_dice():
+    """Roll two dice and return the sum."""
+    return random.randint(1, 6) + random.randint(1, 6)
+
+
+def play_craps():
+    """Play one game of Craps."""
+    first_roll = roll_dice()
+
+    if first_roll in [7, 11]:
+        return 1  # Win
+    elif first_roll in [2, 3, 12]:
+        return 0  # Lose
+    else:
+        point = first_roll
+        while True:
+            next_roll = roll_dice()
+            if next_roll == point:
+                return 1  # Win
+            elif next_roll == 7:
+                return 0  # Lose
+
+
+def simulation(num_games=1000000):
+    """Run a simulation of Craps games."""
+    wins = sum(play_craps() for _ in range(num_games))
+    win_probability = wins / num_games
+    print(f"Simulated probability of winning: {win_probability:.5f}\n")
+    print(f"Theoretical probability of winning: 244/495 ≈ {244 / 495:.5f}\n")
+
+
+# Run the simulation
+simulation()
+# --8<-- [end:solution2]
+
+
+if __name__ == "__main__":
+    simulation(num_games=1000000)

mkdocs.yml

@@ -20,6 +20,7 @@ nav:
     - "6. Chuck a Luck": fifty/6_chuck_a_luck.md
     - "7. Curing Compulsive Gambler": fifty/7_curing_compulsive_gambler.md
     - "8. Perfect Bridge Hand": fifty/8_perfect_bridge_hand.md
+    - "9. Craps": fifty/9_craps.md
   - Mystery:
     - mystery/index.md
     - "3. Fertility Policy": mystery/3_fertility_policy.md