令牌之争

让我们稍微概括一下，假设每个玩家一开始都有 $a$ 代币，而玩家 $A$ 一回合获得代币的概率是 $p$。

令 $x$ 为 $A$ 的代币数量减去 $a$。因此游戏从 $x=0$ 开始，玩家 $A$ 需要在达到 $x = -a$ 之前达到 $x = a$ 才能获胜。

我们可以将游戏视为有偏差的、吸引人的随机游走：$x$从零开始，在每一步，$x$增加1（概率为$p$）或减少1（概率为$1-p$），直到$x = a$或$x = -a$。

当$A$的代币数量减去$a$为$x$时，令$R_x$为$A$的获胜概率。然后我们知道： $$R_a = 1 \ \text{and} \ R_{-a} = 0$$ 和 $$R_n = pR_{n+1} + (1-p)R_n - 1$$ 对于$-a < n < a$。上面的方程是线性递推关系的一个例子，我们可以按如下方式求解 $R_n$： $$pR_{n+1} = R_n - (1-p)R_{n-1}$$ 这有特征方程 $$px^2 - x + 1 - p = 0$$ 我们假设 $p > \frac{1}{2}$。那么这个方程的两个根就是$a_1 = 1$和$a_2 = \frac{1-p}{p}$。那么存在常量 $c_1$ 和 $c_2$ 使得： $$R_n = c_1a_1^n + c2a^n_2$$ 与 $0 = R_{-a}$ 和 $1 = R_a$。求解，我们找到$c_1 = \frac{a_2^a}{a^{2a}_{2} - a^{2a}_{1}}$和$c_2 = \frac{-a_1^a}{a^{2a}_2 - a^{2a}_{1}}$。我们对 $R_0$ 感兴趣，我们发现： $$R_0 = c_1 + c_2 = \frac{1}{a_1^a + a_2^a}$$ 对于我们的特定问题，$a=12$ 和 $p = \frac{9}{14}$（因为用三个骰子掷出总和为 11 和 14 的概率分别为 $\frac{1}{8}$ 和 $\frac{5}{72}$），因此 $a_2 = \frac{5}{9}$ 和 $A$ 玩家获胜的概率为： $$R_0 = \frac{1}{1 + (\frac{5}{9})^{12}} \approx 0.99913631$$

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Original Explanation

Let's generalize slightly and suppose that each player starts with $a$ tokens and the probability of player $A$ gaining a token on one turn is $p$.

Let $x$ be the number of $A$'s tokens minus $a$. So the game begins with $x=0$, and player $A$ needs to reach $x = a$ before reaching $x = -a$ in order to win.

We can view the game as a biased, absorbing random walk: $x$ begins at zero, and at each step, $x$ increases by one (with probability $p$) or decreases by 1 (with probability $1-p$) until $x = a$ or $x = -a$.

Let $R_x$ be $A$'s probability of winning when the number of $A$'s tokens minus $a$ is $x$. Then we know:

$$R_a = 1 \ \text{and} \ R_{-a} = 0$$

and

$$R_n = pR_{n+1} + (1-p)R_n - 1$$

for $-a < n < a$. The equation above is an example of a linear recurrence relation, and we can solve for $R_n$ as follows:

$$pR_{n+1} = R_n - (1-p)R_{n-1}$$

and this has characteristic equation

$$px^2 - x + 1 - p = 0$$

Let's assume $p > \frac{1}{2}$. Then the two roots of this equation are $a_1 = 1$ and $a_2 = \frac{1-p}{p}$. Then there exist constants $c_1$ and $c_2$ such that:

$$R_n = c_1a_1^n + c2a^n_2$$

with $0 = R_{-a}$ and $1 = R_a$. Solving, we find $c_1 = \frac{a_2^a}{a^{2a}_{2} - a^{2a}_{1}}$ and $c_2 = \frac{-a_1^a}{a^{2a}_2 - a^{2a}_{1}}$. We are interested in $R_0$ and we find:

$$R_0 = c_1 + c_2 = \frac{1}{a_1^a + a_2^a}$$

For our particular problem, $a=12$ and $p = \frac{9}{14}$ (since the probabilities of throwing a sum of 11 and 14 with three dice are $\frac{1}{8}$ and $\frac{5}{72}$, respectively), so $a_2 = \frac{5}{9}$ and so the probability of player $A$ winning is:

$$R_0 = \frac{1}{1 + (\frac{5}{9})^{12}} \approx 0.99913631$$