二项分布模拟

Question

假设A队和B队正在进行一系列游戏，第一个 赢得4场比赛的球队赢得了系列赛。假设A队有55% 赢得每场比赛的机会以及每场比赛的结果 独立的

（a）a队赢得系列赛的概率是多少？发表 准确的结果并通过仿真验证

（b）预计的比赛次数是多少？确切地说 通过仿真验证了该方法的有效性

（c）A队预计的比赛次数是多少 赢得系列赛？给出了精确的结果，并通过仿真进行了验证

（d）现在假设我们只知道A队更有可能获胜 每个游戏，但不知道确切的概率。如果最 可能玩的游戏数是5，这意味着什么 A队赢得每场比赛的概率是多少

这就是我所做的，但没有得到它…需要一些输入。多谢各位

import numpy as np

probs = np.array([.55 ,.45])
nsims = 500000

chance = np.random.uniform(size=(nsims, 7))

teamAWins = (chance > probs[None, :]).astype('i4')
teamBWins = 1 - teamAWins

teamAwincount = {}
teamBwincount = {}
for ngames in range(4, 8):
    afilt = teamAWins[:, :ngames].sum(axis=1) == 4
    bfilt = teamBWins[:, :ngames].sum(axis=1) == 4

    teamAwincount[ngames] = afilt.sum()
    teamBwincount[ngames] = bfilt.sum()

    teamAWins = teamAWins[~afilt]
    teamBWins = teamBWins[~bfilt]

teamAwinprops = {k : 1. * count/nsims for k, count in teamAwincount.iteritems()}
teamBwinprops = {k : 1. * count/nsims for k, count in teamBwincount.iteritems()}

Answer 1

好的，这是让你开始的想法和代码

这是我认为是Negative Binomial Distribution，这是很容易实现的，并计算出最受欢迎和处于劣势的概率

有了这些代码，就定义了一整套事件，概率之和正好为1。因此，您可以：

1. 得到准确的答案
2. 根据概率检查模拟

模拟代码为多个事件和单事件模拟器添加了计数器。到目前为止，它的概率似乎与 负二项公式

代码，Python 3.8 x64 Win10

import numpy as np
import scipy.special

# Negative Binomial as defined in
# https://mathworld.wolfram.com/NegativeBinomialDistribution.html
def P(x, r, p):
    return scipy.special.comb(x+r-1, r-1)*p**r*(1.0-p)**x

def single_event(p, rng):
    """
    Simulates single up-to-4-wins event,
    returns who won and how many opponent got
    """
    f = 0
    u = 0
    while True:
        if rng.random() < p:
            f += 1
            if f == 4:
                return (True,u) # favorite won
        else:
            u += 1
            if u == 4:
                return (False,f) # underdog won

def sample(p, rng, N):
    """
    Simulate N events and count all possible outcomes
    """

    f = np.array([0, 0, 0, 0], dtype=np.float64) # favorite counter
    u = np.array([0, 0, 0, 0], dtype=np.float64) # underdog counter

    for _ in range(0, N):
        w, i = single_event(p, rng)
        if w:
            f[i] += 1
        else:
            u[i] += 1

    return (f/float(N), u/float(N)) # normalization

def expected_nof_games(p, rng, N):
    """
    Simulate N events and computes expected number of games
    """

    Ng = 0
    for _ in range(0, N):
        w, i = single_event(p, rng)

        Ng += 4 + i # 4 games won by winner and i by loser

    return float(Ng)/float(N)


p = 0.55

# favorite
p04 = P(0, 4, p)
p14 = P(1, 4, p)
p24 = P(2, 4, p)
p34 = P(3, 4, p)

print(p04, p14, p24, p34, p04+p14+p24+p34)

# underdog
x04 = P(0, 4, 1.0-p)
x14 = P(1, 4, 1.0-p)
x24 = P(2, 4, 1.0-p)
x34 = P(3, 4, 1.0-p)
print(x04, x14, x24, x34, x04+x14+x24+x34)

# total probability
print(p04+p14+p24+p34+x04+x14+x24+x34)

# simulation of the games
rng = np.random.default_rng()
f, u = sample(p, rng, 200000)
print(f)
print(u)

# compute expected number of games

print("expected number of games")
E_ng = 4*p04 + 5*p14 + 6*p24 + 7*p34 + 4*x04 + 5*x14 + 6*x24 + 7*x34
print(E_ng)
# same result from Monte Carlo
print(expected_nof_games(p, rng, 200000))