在N次成功之前，我该如何模拟翻转？

你和我决定玩一个游戏，大家轮流掷硬币。第一位总共翻转10个头的玩家将赢得比赛。自然，关于谁应该先走有一个争论。

此游戏的模拟结果显示，前一个掷骰的玩家比第二个掷骰的玩家赢6％（第一个掷骰的玩家大约有53％的时间获胜）。我有兴趣对此进行建模分析。

这不是二项式随机变量，因为没有固定的试验次数（直到有人得到10个脑袋时才翻转）。我该如何建模？它是负二项式分布吗？

为了能够重新创建我的结果，这是我的python代码：

import numpy as np
from numba import jit


@jit
def sim(N):

    P1_wins = 0
    P2_wins = 0

    for i in range(N):

        P1_heads = 0
        P2_heads = 0
        while True:

            P1_heads += np.random.randint(0,2)

            if P1_heads == 10:
                P1_wins+=1
                break

            P2_heads+= np.random.randint(0,2)
            if P2_heads==10:
                P2_wins+=1
                break
    return P1_wins/N, P2_wins/N


a,b = sim(1000000)

尾的数量的实现之前的分布的头是负二项与参数10和1 / 2。让˚F是概率函数和摹生存函数：每个ñ ≥ 0，˚F （ñ ）是玩家的机会ň前尾巴10头和g ^ （ñ ）是玩家的机会ñ之前或更多的尾巴10头。$10$$10$$1/2$$f$$G$$n\ge 0$$f(n)$$n$$10$$G(n)$$n$$10$

因为玩家独立地滚动，机会第一玩家获胜与恰好滚动尾部被由第二玩家辊上的机会，机会相乘得到Ñ或多个尾部，等于˚F （Ñ ）ģ （Ñ ）。$n$$n$$f(n)G(n)$

求和所有可能给人的第一玩家的获胜几率为$n$

大约是一半时间的。$3\%$

通常，用任何正整数m代替，答案可以根据超几何函数给出：它等于$10$$m$

当使用偏头硬币的机会为下，这可以概括为$p$

这是R一百万个此类游戏的模拟。报告的估计值为。将其与理论结果进行比较的二项式假设检验的Z得分为− 0.843，差异不明显。$0.5325$$-0.843$

n.sim <- 1e6
set.seed(17)
xy <- matrix(rnbinom(2*n.sim, 10, 1/2), nrow=2)
p <- mean(xy[1,] <= xy[2,])
cat("Estimate:", signif(p, 4), 
    "Z-score:", signif((p - 0.532909774) / sqrt(p*(1-p)) * sqrt(n.sim), 3))

我们可以像这样对游戏进行建模：

• 玩家A反复翻转硬币，得到结果$A_1, A_2, \dots$，直到他们得到共计10头。让10头时指数是随机变量$X$。
• 玩家B也是一样。设第10个磁头的时间索引为随机变量$Y$，它是$X$的iid副本。
• 如果$X \le Y$，玩家A获胜; 否则，玩家B获胜。即， $$\begin{align} \Pr(A\text{ wins})&= \Pr(X \ge Y) = \Pr(X > Y) + \Pr(X = Y)\\ \Pr(B\text{ wins})&= \Pr(Y > X) = \Pr(X > Y). \end{align}$$

因此，胜率的差距为

正如您所怀疑的那样，$X$（和$Y$）基本上是根据负二项分布分布的。表示法各不相同，但在Wikipedia的参数化中，正面为“失败”，背面为“成功”。在实验停止之前，我们需要$r = 10$ “失败”（正面），成功概率$p = \tfrac12$。然后，“成功”的数量为$X - 10$，具有

Thus Player B's win rate is $\Pr(Y > X) \approx 46.7\%$, and Player A's is $\frac{619\,380\,496}{1\,162\,261\,467} \approx 53.3\%$.

Let $E_{ij}$ be the event that the player on roll flips i heads before the other player flips j heads, and let $X$ be the first two flips having sample space $\{ hh,ht,th,tt\}$ where h means heads and t tails, and let $p_{ij} \equiv Pr(E_{ij})$.

Then $p_{ij}=Pr(E_{i-1j-1}|X=hh)*Pr(X=hh)+Pr(E_{i-1j}|X=ht)*Pr(X=ht)+Pr(E_{ij-1}|X=th)*Pr(X=th)+Pr(E_{ij}|X=tt)*Pr(X=tt)$

Assuming a standard coin $Pr(X=*)=1/4$ means that $p_{ij}=1/4*[p_{i-1j-1}+p_{i-1j}+p_{ij-1}+p_{ij}]$

solving for $p_{ij}$, $= 1/3*[p_{i-1j-1}+p_{i-1j}+p_{ij-1}]$

But $p_{0j}=p_{00}=1$ and $p_{i0}=0$, implying that the recursion fully terminates. However, a direct naive recursive implementation will yield poor performance because the branches intersect.

An efficient implementation will have complexity $O(i*j)$ and memory complexity $O(min(i,j))$. Here's a simple fold implemented in Haskell:

Prelude> let p i j = last. head. drop j $ iterate ((1:).(f 1)) start where
  start = 1 : replicate i 0;
  f c v = case v of (a:[]) -> [];
                    (a:b:rest) -> sum : f sum (b:rest) where
                     sum = (a+b+c)/3 
Prelude> p 0 0
1.0
Prelude> p 1 0
0.0
Prelude> p 10 10
0.5329097742513388
Prelude> 

UPDATE: Someone in the comments above asked whether one was suppose to roll 10 heads in a row or not. So let $E_{kl}$ be the event that the player on roll flips i heads in a row before the other player flips i heads in a row, given that they already flipped k and l consecutive heads respectively.

Proceeding as before above, but this time conditioning on the first flip only, $p_{k,l} = 1-1/2*[p_{l,k+1}+p_{l,0}]$ where $p_{il}=p_{ii}=1, p_{ki}=0$

This is a linear system with $i^2$ unknowns and one unique solution.

To convert it into an iterative scheme, simply add an iterate number $n$ and a sensitivity factor $\epsilon$:

$p_{k,l,n+1} = 1/(1+\epsilon)*[\epsilon*p_{k,l,n} +1-1/2*(p_{l,k+1,n}+p_{l,0,n})]$

Choose $\epsilon$ and $p_{k,l,0}$ wisely and run the iteration for a few steps and monitor the correction term.