Question

我正试图找出'苏西'赢得比赛的可能性。

'苏西'赢得比赛的可能性= 0.837
“鲍勃”赢得比赛的可能性= 0.163

如果第一个赢得n场比赛的人赢了一场比赛，那么苏的最小值是多少，以便有超过0.9的机会赢得比赛？

到目前为止，我有这段代码：

import itertools

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

打印哪些：

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

然后我想用这些结果来计算'苏西'赢得整场比赛的概率。

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

您可以使用词典来表示概率：

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积列表的所有元素 - 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。从这里你可以很容易地选择哪一个是“苏西赢得比赛”，并总结概率。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

这个条件仅适用于3个游戏，但我真的把这个游戏留给你了 - 很容易将其归结为n个游戏：）

Answer 2

您还需要循环n的值。另请注意，“首先n”与“2n-1最佳”相同。所以我们可以说m = 2 * n - 1并看看谁赢得了那场比赛中的大多数比赛。 max(set(product), key=product.count)是一种简短而不透明的方法，可以解决谁赢得了最多的比赛。此外，当您可以直接将值存储在元组中时，为什么还要用字符串表示概率，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

我被@ desired_login的观点所吸引，但我想我会尝试计算排列而不是迭代它们：

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.

此算法的运行时类似于O（n ^ 3），而其他算法则为O（2 ^ n）。

Itertools创建一个列表并计算出概率

3 个答案: