骰子游戏 1

首先，我们首先考虑用两个骰子掷出总和为 6 的概率以及用两个骰子掷出总和为 7 的概率。

我们可以通过以下方式掷出 6：1&5、5&1、2&4、4&2、3&3。由于掷两个骰子的方式共有 36 种，因此掷出总和为 6 的概率为 $\frac{5}{36}$。我们可以对 7 的和进行同样的操作。 7 可以通过以下方式进行滚动：1&6、6&1、2&5、5&2、3&4、4&3。这给了我们滚动总和为 7 作为 $\frac{6}{36} = \frac{1}{6}$ 的概率。

根据此信息，你可能倾向于相信玩家 B 将赢得游戏，因为掷出 7 的概率大于掷出 6 的概率。但是，你必须记住此问题的一个重要部分 - 玩家 A 先走。因此，我们想要研究先行的优势是否抵消了与滚动总和为 6 相关的较低概率。

玩家A可以通过以下方式赢得比赛：

1. 玩家 A 第一次掷骰子的总和为 6。
2. 玩家 A 在第一次掷骰时未能掷出 6，然后玩家 B 未能掷出 7，然后玩家 A 掷出 6。

请注意，只要玩家 A 在玩家 B 掷出 7 之前最终掷出 6，模式 #2 就可以无限期地继续下去。本质上，存在一系列事件，会发生一些 $k$ 次数的失败，直到我们遇到玩家 A 掷出 6 的第一次成功。我们刚才描述的过程可以通过几何分布进行建模。

设 $p$ 为用两个骰子掷出总和为 6 的概率，设 $r$ 为用两个骰子掷出总和为 7 的概率。那么我们可以将事件的概率表示为：$(1-p)^k(1-r)^kp$

其中 $k = 0,1,2,3....$。我们可以将这些概率相加，得到玩家 A 获胜的概率：

由于 $p=\frac{5}{36}$ 和 $r=\frac{1}{6}$（我们在上面计算了这些值），这等于 $\frac{30}{61} \approx 0.49$。因此，本场比赛只是稍微对玩家B有利。

这是我们刚刚观察到的结果的 python 模拟：

import random
from typing import Callable

random.seed(42)

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7, 1)

def play_game() -> bool:
    """Returns a boolean indicating whether player A won"""
    pa_turn = True
    while True:
        dice_roll_sum = roll_dice() + roll_dice()
        if pa_turn and dice_roll_sum == 6:
            return True
        if not pa_turn and dice_roll_sum == 7:
            return False
        pa_turn = not pa_turn

iterations = 10_000
pa_win_prob = sum([play_game() for _ in range(iterations)]) / iterations
print(f"The probability of Player A winning is {pa_win_prob}")

Original Explanation

To begin, let's start by considering the probability of rolling a sum of 6 with two dice and the probability of rolling a sum of 7 with two dice.

We can roll a six in the following ways: 1&5, 5&1, 2&4, 4&2, 3&3. Since there are 36 total ways to roll two dice, the probability of rolling a sum of 6 is $\frac{5}{36}$. We can do the same for rolling a sum of 7. A seven can be rolled in the following ways: 1&6, 6&1, 2&5, 5&2, 3&4, 4&3. This gives us the probability of rolling a sum of 7 as $\frac{6}{36} = \frac{1}{6}$.

Based on this information, you may be inclined to believe that player B will win the game because the probability of rolling a 7 is larger than the probability of rolling a 6. However, there's one important part of this problem that you must remember - player A goes first. Therefore, we want to investigate whether the advantage of going first offsets the lower probability associated with rolling a sum of 6.

Player A can win the game in the following ways:

1. Player A rolls a sum of 6 on their first roll.
2. Player A fails to roll a sum of 6 on their first roll, then player B fails to roll a sum of 7, and then player A rolls a sum of 6.

Note that pattern #2 can continue indefinitely as long as player A eventually rolls a sum of 6 before player B rolls a sum of 7. In essence, there is a sequence of events that take place with some $k$ number of failures until we encouter the first success of player A rolling a 6. The process we just described can be modeled by the geometric distribution.

Let $p$ be the probability of rolling a sum of 6 with two dice and let $r$ be the probability of rolling a sum of 7 with two dice. Then we can express the probability of the event as: $(1-p)^k(1-r)^kp$

where $k = 0,1,2,3....$. We can sum these probabilities to get the probability that player A wins:

Since $p=\frac{5}{36}$ and $r=\frac{1}{6}$ (we calculated these values above), this equals $\frac{30}{61} \approx 0.49$. Therefore, this game just slightly favors player B.

Here is a python simulation of the result we just observed:

import random
from typing import Callable

random.seed(42)

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7, 1)

def play_game() -> bool:
    """Returns a boolean indicating whether player A won"""
    pa_turn = True
    while True:
        dice_roll_sum = roll_dice() + roll_dice()
        if pa_turn and dice_roll_sum == 6:
            return True
        if not pa_turn and dice_roll_sum == 7:
            return False
        pa_turn = not pa_turn

iterations = 10_000
pa_win_prob = sum([play_game() for _ in range(iterations)]) / iterations
print(f"The probability of Player A winning is {pa_win_prob}")