Python 如何获得变量输出的平均值？

所以，我有一个变量，输出0或1。现在，我想运行10000次，得到平均值

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() 

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所以基本上，我想知道掷骰子返回7的几率是多少

要执行10000次滚动，可以使用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：刚刚意识到您还有一个名为“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，请尝试重命名变量。

您可以使用创建10000个六面骰子两次，- 他们，每个人，把它喂进 数一数。请参见骰子示例，以获取对代码进行了一些解释的代码注释

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硬币投掷（0,1）2枚硬币的10k投掷示例和总价值： 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：

0可以获得25%，2可以获得25%，1可以获得50%

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6面骰子示例，共10k，共2个骰子，滚动并求和： 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：要对齐输出中的数字（

{k:>2}

，

{v:>5}

，

{v/sumall*100:2.2f}

）

整个过程看起来如何？for循环将从roll_dice函数的开头开始，一直运行到count=0的位置。如果你使用我的代码，平均值=。。。和return语句应该在for循环之外。然而，Patrick比我有更多的经验，而且可能有比我更好的解决方案。只是一个提示。为什么不

x，y=random。示例（可用，2）

？“示例”确保不会绘制两次相同的示例。或者直接使用

z=sum（random.sample（可用，2））

？您的骰子掷骰返回0或1，返回7的几率为零，您不需要任何平均值即可知道这一点。

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% 

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}%")

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%