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Craps

Problem

Question

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?

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

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.

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

Solution1: Analysis

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%.

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

Solution2: Simulation

The overall process of the game is relatively simple, and we can simulate it through the following code:

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()

Simulated probability of winning: 0.49313

Theoretical probability of winning: 244/495 ≈ 0.49293

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:

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()

Simulated probability of winning: 0.49259

Theoretical probability of winning: 244/495 ≈ 0.49293