在python中实现涉及重组合的概率函数答案

【问题标题】：Implementing probability function involving heavy combinatorics in python在python中实现涉及重组合的概率函数
【发布时间】：2021-03-12 19:16:04
【问题描述】：

这是关于我在math stack 上提出的问题的答案。

我希望将此问题/解决方案转换为 python，但我无法解释此处使用的所有符号。

我意识到这篇文章有点过于“给我代码”而不是一个好问题，但我问这个问题的目的是为了理解这里所涉及的数学。我不太了解这里使用的数学符号的语言，但如果我看到它，我可以很好地解释 python 来概念化答案。

问题可以这样设置

import numpy as np

bag = np.hstack((
    np.repeat(0, 80),
    np.repeat(1, 21),
    np.repeat(3, 5),
    np.repeat(7,1)
    ))

【问题讨论】：

标签： python probability combinatorics

【解决方案1】：

我不确定这是否正是您所追求的，但这就是我要计算的方式，例如，得到总和 == 6 的概率。它比数学更实用，只是解决了这个特殊问题，所以我不确定它是否能帮助你理解数学。


import numpy as np
import itertools
from collections import Counter
import pandas as pd


bag = np.hstack((
    np.repeat(0, 80),
    np.repeat(1, 21),
    np.repeat(3, 5),
    np.repeat(7,1)
    ))



#107*106*105*104*103*102*101*100*99*98
#Out[176]: 127506499163211168000 #Permutations

##Need to reduce the number to sample from without changing the number of possible combinations
reduced_bag = np.hstack((
    np.repeat(0, 10), ## 0 can be chosen all 10 times
    np.repeat(1, 10), ## 1 can be chosen all 10 times
    np.repeat(3, 5), ## 3 can be chosen up to 5 times
    np.repeat(7,1) ## 7 can be chosen once
    ))


## There are 96 unique combinations
number_unique_combinations = len(set(list(itertools.combinations(reduced_bag,10))))

### sorted list of all combinations
unique_combinations = sorted(list(set(list(itertools.combinations(reduced_bag,10)))))

### sum of each unique combination
sums_list = [sum(uc) for uc in unique_combinations]

### probability for each unique combination

probability_dict = {0:80, 1:21, 3:5, 7:1} ##Dictionary to refer to 
n = 107 ##Number in the bag

probability_list = []


##This part is VERY slow to run because of the itertools.permutations
for x in unique_combinations:
    print(x)
    p = 1 ##Start with the probability again
    n = 107 ##Start with a full bag for each combination
    count_x = Counter(x)
    for i in x:
        i_left = probability_dict[i] - (Counter(x)[i] - count_x[i]) ##Number of that type left in bag
        p *= i_left/n ##Multiply the probability 
        n = n-1 # non replacement
        count_x[i] = count_x[i] - 1 ##Reduce the number in the bag
    p *= len(set(list(itertools.permutations(x,10)))) ##Multiply by the number of permutations per combination
    probability_list.append(p)
        
##sum(probability_list) ## Has a rounding error
##Out[57]: 1.0000000000000002
##

##Put the combinations into dataframe
ar = np.array((unique_combinations,sums_list,probability_list))
df = pd.DataFrame(ar).T
##Name the columns
df.columns = ["combination", "sum","probability"]

## probability that sum is >= 6
df[df["sum"] >= 6]['probability'].sum()
## 0.24139909236232826
## probability that sum is == 6
df[df["sum"] == 6]['probability'].sum()
## 0.06756408790812335

【讨论】：