Itertools 创建列表并计算概率

Question

我正在计算“Susie”赢得比赛的概率。

“Susie”赢得比赛的概率 = 0.837
“鲍勃”赢得比赛的概率 = 0.163

如果第一个赢得 n 场比赛的人赢得比赛，那么 n 的最小值是多少，这样 Susie 赢得比赛的机会大于 0.9？

到目前为止，我有这个代码：

import itertools

W = 0.837
L = 0.163
for product in itertools.product(['W','L'], repeat=3): #3=number of games
    print product

哪些打印：

('W', 'W', 'W')
('W', 'W', 'L')
('W', 'L', 'W')
('W', 'L', 'L')
('L', 'W', 'W')
('L', 'W', 'L')
('L', 'L', 'W')
('L', 'L', 'L')

然后我想使用这些结果来计算“Susie”赢得比赛的总体概率。

我已经在纸上解决了这个问题，玩的游戏越多，“Susie”赢得比赛的机会就越大。

Answer 1

您可以使用字典来获取概率：

import itertools
import operator

probabilities = {'W':0.837, 'L':0.163}

for product in itertools.product(['W','L'], repeat=3): #3=number of games
    p = reduce(operator.mul,
               [probabilities[p] for p in product])
    print product, ":", p

reduce 函数使用第一个参数中给出的函数累加列表的所有元素 - 这里我们通过乘法累加它们。

这为您提供了每个事件序列的概率。从中您可以轻松选择哪个是“Susie 赢得比赛”，并对概率求和。这样做的一种可能性是：

import itertools
import operator

probabilities = {'W':0.837, 'L':0.163}

winProbability = 0
for product in itertools.product(['W','L'], repeat=3): #3=number of games
    p = reduce(operator.mul,
               [probabilities[p] for p in product])

    if product.count('W') > 1: #works only for 3 games
        winProbability += p
        print "Susie wins:", product, "with probability:", p
    else:
        print "Susie looses:", product, "with probability:", p

print "Total probability of Susie winning:", winProbability 

该条件仅适用于 3 游戏，但我真的要把这个留给你 - 很容易将它推广到 n 游戏:)

Answer 2

您还需要循环遍历n 的值。另请注意，'first to n'与'best out of 2n-1'相同。所以我们可以说m = 2 * n - 1，看看谁在该盘中赢得了最多的比赛。 max(set(product), key=product.count) 是一种简短但不透明的方式来计算谁赢得了最多的比赛。另外，当您可以将值直接存储在元组中时，为什么还要费心用字符串表示概率然后使用字典来读取它们。

import itertools

pWin = 0 #the probability susie wins the match
n = 0
while pWin<0.9:
    n += 1
    m = 2 * n - 1
    pWin = 0
    for prod in itertools.product([0.837,0.163], repeat=m):
        #test who wins the match
        if max(set(prod), key=prod.count) == 0.837:
            pWin += reduce(lambda total,current: total * current, prod)
print '{} probability that Susie wins the match, with {} games'.format(pWin, n)

Answer 3

我对@desired_login 的观点很感兴趣，但我想我会尝试计算排列而不是遍历它们：

import sys
if sys.hexversion >= 0x3000000:
    rng = range     # Python 3.x
else:
    rng = xrange    # Python 2.x

def P(n, k):
    """
    Calculate permutations of (n choose k) items
    """
    if 2*k > n:
        k = n - k
    res = 1
    for i in rng(k):
        res = res * (n-i) // (i+1)
    return res

Ps = 0.837    # Probability of Susie winning one match
Px = 0.980    # Target probability

# Probability of Susie winning exactly k of n matches
win_k         = lambda n,k: P(n, k) * Ps**k * (1.0-Ps)**(n-k)
# Probability of Susie winning k or more of n matches
win_k_or_more = lambda n,k: sum(win_k(n, i) for i in rng(k, n+1))

def main():
    # Find lowest k such that the probability of Susie winning k or more of 2*k - 1 matches is at least Px
    k = 0
    while True:
        k += 1
        n = 2*k - 1
        prob = win_k_or_more(n, k)
        print('Susie wins {} or more of {} matches: {}'.format(k, n, prob))
        if prob >= Px:
            print('At first to {} wins, Susie has >= {} chance of winning the match.'.format(k, Px))
            break

if __name__=="__main__":
    main()

对于 Px=0.98，这会导致

Susie wins 1 or more of 1 matches: 0.837
Susie wins 2 or more of 3 matches: 0.9289544940000001
Susie wins 3 or more of 5 matches: 0.9665908247127419
Susie wins 4 or more of 7 matches: 0.9837066988309756
At first to 4 wins, Susie has >= 0.98 chance of winning the match.

此算法的运行时间类似于 O(n^3)，而其他算法的运行时间为 O(2^n)。