如何解决二项分布模拟python
假设A和B两支球队进行了一系列比赛,第一支球队 球队赢得4连胜。假设A队有55% 赢得每场比赛的机会,并且每场比赛的结果都是 独立的。
(a)A队赢得系列赛的概率是多少?给一个 准确的结果,并通过仿真进行确认。
(b)预期打多少场比赛?给出确切的 结果并通过仿真进行确认。
(c)鉴于甲队,预计将打多少场比赛 赢得系列赛?给出准确的结果并通过仿真确认。
(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 甲队赢得每场比赛的几率?
这是我所做的,但是没有得到..需要一些输入。谢谢
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()}
解决方法
好的,这是使您前进的想法和代码。
我相信这是Negative Binomial Distribution,它很容易实现,并且可以计算出喜欢和失败者的概率。
使用该代码,您可以定义整个事件集,其概率正确地总计为1。由此可以:
- 获取确切答案
- 根据概率检查模拟
仿真代码为许多事件和单个事件模拟器添加了计数器。到目前为止,看起来它的概率与 负二项式
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],dtype=np.float64) # favorite counter
u = np.array([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,N):
"""
Simulate N events and computes expected number of games
"""
Ng = 0
for _ in range(0,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,p)
p24 = P(2,p)
p34 = P(3,p)
print(p04,p14,p24,p34,p04+p14+p24+p34)
# underdog
x04 = P(0,1.0-p)
x14 = P(1,1.0-p)
x24 = P(2,1.0-p)
x34 = P(3,1.0-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,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,200000))
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