diff --git a/docs/fifty/9_craps.md b/docs/fifty/9_craps.md new file mode 100644 index 0000000..fa8b484 --- /dev/null +++ b/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" + ``` diff --git a/docs/fifty/snippet/9_craps.py b/docs/fifty/snippet/9_craps.py new file mode 100644 index 0000000..9b9e0d2 --- /dev/null +++ b/docs/fifty/snippet/9_craps.py @@ -0,0 +1,42 @@ +# --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) diff --git a/mkdocs.yml b/mkdocs.yml index 8331db1..b84d155 100644 --- a/mkdocs.yml +++ b/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