Python 程序，计算掷公平骰子 12 次时至少两次获得 1 的概率

Question

import random
import sys

bestcounter1 = 0
bestcounter2=0
get_sample = int(sys.argv[1])

for i in range(get_sample):
    for i in range(12):
        if (random.randint(1,6)==1):
            bestcounter1+=1
            bestcounter2+=1

oneatleasttwice = (bestcounter2*1.0)/(2*(get_sample))

#Divide by 2 to make both comparable. Otherwise 2 will always be greater than 1 !
print("One atleast twice in 12 rolls: ", oneatleasttwice)

谁能解释一下这里使用的逻辑是否正确？我得到的输出总是在 1 左右。

谢谢

Answer 1

您必须将计数器放置在正确的位置。假设 bestcounter1 用于在每次运行(12 次)期间对 1 的值进行计数，而 bestcounter2 用于在您获得 2 个或多个 1 值时对运行进行计数。然后您的 main for 循环应如下所示:

for i in range(get_sample):
    # reset before every run
    bestcounter1 = 0
    for i in range(12):
        if random.randint(1, 6) == 1:
            # count values of 1
            bestcounter1 += 1
        # check if we got 2 or more values of 1
        if bestcounter1 >= 2:
            # count proper cases
            bestcounter2 += 1
            break

oneatleasttwice = bestcounter2 / get_sample

我运行一百万次，结果为 61.9%。