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我有以下代码试图解决以下问题：

掷 n 次骰子 m 次，计算得到至少一个 6 的概率。

我知道掷 2 个骰子时至少得到 1 个 6 的确切概率是 11/36。

我下面的程序似乎希望概率为 0.333，这很接近，但应该是 11/36 对吧？

如果这些建议可以在我制作的标准代码上继续，那就太好了，但也感谢矢量化代码。

import random
from sys import argv

m = int(argv[1]) # performing the experiment with m dice n times
n = int(argv[2]) # Throwing m dice n times
s = 0            # Counts the number of times m dies shows at least one 6

print '%.g dice are thrown %.g times' % (m, n)

for i in xrange(n):
    list = []    # used to clear the list for new die count
    for q in xrange(m):
        r = random.randint(1,6)#Picks a random integer on interval [1,6]
        list.append(r)         #appends integer value
        if len(list) == m:     #when list is full, that is when m dice has been thrown
            for i in xrange(len(list)):
                #print list
                if list[i] == 6: #if the list of elements has a six add to the counter
                    s += 1
                    pass #I want the loop to exit when it finds an element = 6

print 'Number of times one of the n dice show at least one 6: %.g' % s  
print 'Probability of at least 1 six from %.g dice is = %2.3f' % (m,s/float(n))

如果不清楚，我将编辑代码和问题。

输出示例：

Terminal > python one6_ndice.py 2 1000000
2 dice are thrown 1e+06 times
Number of times one of the n dice show atleast one 6: 3e+05
Probability of atleast 1 six from 2 dice is = 0.333
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2 回答 2

2

我认为问题出在这里：

 pass #I want the loop to exit when it finds an element = 6

pass不会退出循环。 pass是无操作命令；它什么都不做。您可能想要break（退出循环）。

顺便说一句，不要调用你的列表list——这会破坏内置的list.

对于更紧凑的表达式，您可以考虑

sum(any(random.randint(1,6) == 6 for die in xrange(n)) for trial in xrange(m))

或者

sum(6 in (random.randint(1,6) for die in range(n)) for trial in range(m))
于 2012-11-10T15:05:03.703 回答
1

您不必在列表上循环，也不必检查它的长度。只需喂它并检查其中是否有 6：

for i in xrange(n):
    list = []
    for q in xrange(m):
        r = random.randint(1, 6)
        list.append(r)
    if 6 in list:
        s += 1

如果您希望您的程序更紧凑并且不想每次都提供一个列表，您可以break在获得“6”时停止生成：

for i in xrange(n):
    for q in xrange(m):
        if random.randint(1, 6) == 6:
            s += 1
            break
于 2012-11-10T15:05:17.197 回答