Question

因此，我有一个输出0或1的变量。现在，我要运行10,000次并获得其平均值。

import random

def roll_dice():

    available = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6]

    x = random.sample(available, 1)
    del available[x[0]]
    y = random.sample(available, 1)
    z = x[0] + y[0]

    if z == 7:
        count = 1
    else:
        count = 0

    print(z)
    print(count)
    return count

roll_dice()

所以基本上，我想知道骰子掷回7的机会是什么。

Answer 1

您可以使用random.choices()两次创建10000个6面骰子，-zip() 它们，每个sum() tuple，然后将其送入collections.Counter 数一下请参阅骰子示例，以获取有关代码解释的代码注释。

掷硬币2枚硬币和总价值的掷硬币（0,1）示例：

from collections import Counter
import random

random.seed(42)
r = range(2)   
c = Counter( (map(sum, ( zip(random.choices(r,k=10000),random.choices(r,k=10000))))))
sumall = sum(c.values())
for k,v in c.most_common():
    print(f"Chance for {k:>2}: {v:>5} out of {sumall} = {v / sumall * 100:2.2f}%")

输出：

Chance for  1:  4989 out of 10000 = 49.89%   # about 50%
Chance for  2:  2540 out of 10000 = 25.40%   # about 25%
Chance for  0:  2471 out of 10000 = 24.71%   # about 25%

数学：

A    B           # for summed values:
0    0    25%    
1    0    25%    # combine it with the one below
0    1    25%    # combine it with the one above
1    1    25%

0可获得25％，2可获得25％，1可获得50％。

6面骰子示例，将2个骰子的10k滚动并求和：

from collections import Counter
import random

random.seed(42)

r = range(1,7)  
c = Counter( (map(sum, ( zip(random.choices(r,k=10000),random.choices(r,k=10000))))))
# explanation of the last code line:
#   random.choices(r,k=10000) creates 10000 random numbers between 1 and 6
#     [1,2,4,...]   and [6,1,6,...]
#   zip takes 2 such 10k lists and makes 10k tuples 
#     [ (1,6),(2,1),(4,6) ... ]
#   map( sum, zip( ...) ) applies sum() to all 2-tuples
#     [7,3,10,...]
#   Counter creates a dict with the sum als key and counts how often it occures

# the rest is just pretty printing:
print(c)
sumall = sum(c.values())

for k,v in c.most_common():
    print(f"Chance for {k:>2}: {v:>5} out of {sumall} = {v / sumall * 100:2.2f}%")

输出：

Counter({ 7: 1673, 8: 1406, 6: 1372,  5: 1090, 9: 1089, 10: 823, 4: 821, 
         11:  591, 3:  570, 2:  291, 12:  274})

Chance for  7:  1673 out of 10000 = 16.73%   # thats about the % of your dice/binary logic
Chance for  8:  1406 out of 10000 = 14.06%
Chance for  6:  1372 out of 10000 = 13.72%
Chance for  5:  1090 out of 10000 = 10.90%
Chance for  9:  1089 out of 10000 = 10.89%
Chance for 10:   823 out of 10000 = 8.23%
Chance for  4:   821 out of 10000 = 8.21%
Chance for 11:   591 out of 10000 = 5.91%
Chance for  3:   570 out of 10000 = 5.70%
Chance for  2:   291 out of 10000 = 2.91%
Chance for 12:   274 out of 10000 = 2.74%

Doku：

格式化：format mini language以对齐输出中的数字（{k:>2}，{v:>5}，{v / sumall * 100:2.2f}）

Answer 2

好吧，要执行10,000次滚动，您可以这样使用for循环：

for count in range(0,10001):
    # random script here

找到平均值的一种方法是在所述循环中包含if语句，例如：

avg = 0
if z == 7:
    count = 1
    if count == 1:
        avg += 1
else:
    count = 0
avg = (avg // 10000)
return avg

希望有帮助。

编辑：刚刚意识到您也有一个名为“ count”的变量。我不确定这是否会干扰循环，因此如果遇到任何问题，请尝试重命名该变量。

如何获得变量输出的平均值？

2 个答案:

掷硬币2枚硬币和总价值的掷硬币（0,1）示例：

6面骰子示例，将2个骰子的10k滚动并求和：