Python 二项分布模拟
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) 预计玩的游戏数是多少?确切地说 通过仿真验证了该方法的有效性 (c) 考虑到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Python 二项分布模拟,python,random,probability,montecarlo,Python,Random,Probability,Montecarlo,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) 预计玩的游戏数是多少?确切地说 通过仿真验证了该方法的有效性 (c) 考虑到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
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()}
好的,这是让你开始的想法和代码 这是我认为是,这是很容易实现,并计算概率为最喜欢的和落后的 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:
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))
将numpy导入为np
进口特殊商品
#中定义的负二项式
# https://mathworld.wolfram.com/NegativeBinomialDistribution.html
def P(x,r,P):
返回scipy特殊梳(x+r-1,r-1)*p**r*(1.0-p)**x
def单_事件(p、rng):
"""
模拟单个最多4胜事件,
返回谁赢了以及有多少对手赢了
"""
f=0
u=0
尽管如此:
如果rng.random()
好的,下面是让您开始的想法和代码
这是我认为是,这是很容易实现,并计算概率为最喜欢的和落后的
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:
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))
将numpy导入为np
进口特殊商品
#中定义的负二项式
# https://mathworld.wolfram.com/NegativeBinomialDistribution.html
def P(x,r,P):
返回scipy特殊梳(x+r-1,r-1)*p**r*(1.0-p)**x
def单_事件(p、rng):
"""
模拟单个最多4胜事件,
返回谁赢了以及有多少对手赢了
"""
f=0
u=0
尽管如此:
如果rng.random()