如何将变量中的偏差值合并到函数中？[python]

Question

我在模拟一个场景，一个自助洗衣店里有十双袜子，还有六只随机丢失的袜子。在此之后，可以至少剩下四对，最多七对。我编写了一个代码，它运行并模拟概率，但是当机会不相等时，它将所有的概率视为平等的机会。我的问题是，现在如何将偏见纳入这段代码：

import random #random number generator
# variables
number_of_pairs = 10 # how many pairs of socks there are originally
total_socks = number_of_pairs * 2 # how many total socks there are
socks_lost = 6 # how many socks were lost
number_of_simulations = 1000 # total number of trials being done
prob_of_7_pairs = (1/6) # probability that there are still seven pairs of socks left
prob_of_6_pairs = (2/9) # probability that there are still six pairs of socks left
prob_of_5_pairs = (5/18) # probability that there are still five pairs of socks left
prob_of_4_pairs = (1/3) # probability that there are still four pairs of socks left
four_pairs = 0 # how many times out of a thousand trials that there are still four pairs left
five_pairs = 0 # how many times out of a thousand trials that there are still five pairs left
six_pairs = 0 # how many times out of a thousand trials that there are still six pairs left
seven_pairs = 0 # how many times out of a thousand trials that there are still seven pairs left
# function
for i in range(0, number_of_simulations): # i is the trial runs, range is from 0 to how many trials there are
    possible_pairs = random.randint(4,7) # random number generator will randomly select between four and seven pairs of socks
    if possible_pairs == 4: 
        four_pairs += 1 # if there are only four possible pairs of socks in one of the trials, the program will add one to the total
    elif possible_pairs == 5:
        five_pairs += 1 # if there are only five possible pairs of socks in one of the trials, the program will add one to the total
    elif possible_pairs == 6:
        six_pairs += 1 # if there are only six possible pairs of socks in one of the trials, the program will add one to the total
    elif possible_pairs == 7:
        seven_pairs += 1 # if there are only seven possible pairs of socks in one of the trials, the program will add one to the total
# results
print('There were four possible pairs of socks', four_pairs, 'of times.') # prints out how many times 1 was rolled out of a thousand rolls
print('There were five possible pairs of socks', five_pairs, 'of times.') # prints out how many times 2 was rolled out of a thousand rolls
print('There were six possible pairs of socks', six_pairs, 'of times.') # prints out how many times 3 was rolled out of a thousand rolls
print('There were seven possible pairs of socks', seven_pairs, 'of times.') # prints out how many times 4 was rolled out of a thousand rolls
print('The total number of trials was', number_of_simulations,'.') # makes sure the program does a thousand trials

Answer 1

假设每只股票都用0- 19 (含)之间的数字表示，这两对是股票0& 1，股票2和3，等等。然后你从0- 19 (包括)中随机选择6个数字，代表丢失的6个股票，并计算出剩余的对数。这样做了很多次(大约100,000次)，并找出每一个数字的次数作为剩余的对数。

import random

# variables
number_of_pairs = 10
total_stocks = number_of_pairs * 2
stocks_lost = 6
number_of_simulations = 100_000
prob_of_7_pairs = (1/6)
prob_of_6_pairs = (2/9)
prob_of_5_pairs = (5/18)
prob_of_4_pairs = (1/3)
four_pairs = 0
five_pairs = 0
six_pairs = 0
seven_pairs = 0

# function
for _ in range(number_of_simulations):
    missing_stocks = set(random.sample(range(total_stocks), stocks_lost)) # generate stacks_lost unique random numbers
    possible_pairs = 0
    # loop through the pairs
    for i in range(0, total_stocks, 2):
        # if both stocks of a pair aren't missing, increment possible_pairs
        if i not in missing_stocks and i + 1 not in missing_stocks:
            possible_pairs += 1
    if possible_pairs == 4:
        four_pairs += 1
    elif possible_pairs == 5:
        five_pairs += 1
    elif possible_pairs == 6:
        six_pairs += 1
    elif possible_pairs == 7:
        seven_pairs += 1
    else:
        raise Exception("not supposed to happen")

# results
print(f'There were four possible pairs of socks {four_pairs / number_of_simulations:.2%} of the times.')
print(f'There were five possible pairs of socks {five_pairs / number_of_simulations:.2%} of the times.')
print(f'There were six possible pairs of socks {six_pairs / number_of_simulations:.2%} of the times.')
print(f'There were seven possible pairs of socks {seven_pairs / number_of_simulations:.2%} of the times.')

输出

有4对可能的袜子，34.71%的次数。

有5双可能的袜子，占51.87%。

有6双可能的袜子，占13.11%。

有7双可能的袜子，占0.31%。

我不知道你从哪里得到你的概率，也不知道如何自己计算，但他们似乎错了。

编辑：添加了一个工作示例，请记住我更改了一些变量名。

如果您对此代码有任何疑问，请不要犹豫。