一个Python程序，计算当掷一个公平骰子12次时，至少两次得到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)

有人能解释一下这里使用的逻辑是否正确吗？我得到的输出总是在1左右。在

谢谢

Answer 1

你必须把你的柜台放在正确的地方。假设bestcounter1用于在每次运行（12次转鼓）期间计数1的值，而{}用于在获得2个或更多值1时计算运行次数。那么主for循环应该如下所示：

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

我得到了61.9%的结果100万次。在