将2个骰子滚动1000次，计算两次命中之和的次数

Question

作业要求我们通过实验找出2到12的概率作为总和。我们想将两个骰子滚动1000次并计算总和为2、3，…和12的次数。到目前为止，我已经将其作为代码，但是我无法获得教授要求的输出。Output that the professor is looking for.我到目前为止所拥有的：

import random as r

die_1 = r.randint(1,6)
die_2 = r.randint(1,6)


print('For-Loop')
for i in range(2,13):
    r.seed(1)
    counter = 0
    for j in range(1000):
        if i == r.randint(2,12):
            counter = counter + 1
    print("sum = ", i,  " count = ",  counter)

Answer 1

from random import randint

rolls = [sum([randint(1, 6), randint(1, 6)]) for i in range(1000)]

for i in range(2, 13):
    print(f'Sum of {i} was rolled {rolls.count(i)} times')

Answer 2

我试图解释评论中发生的一切：

from collections import defaultdict
from random import randint

# Roll the two dice how many times?
n = 1000

# Create a dictionary to store the results
results = defaultdict(int)

# Loop n times
for _ in range(n):
    # Get random numbers for the two dice
    die_1 = randint(1, 6)
    die_2 = randint(1, 6)
    # Increase the corresponding result by 1
    results[die_1 + die_2] += 1

# Print results
print(results)

可能打印出以下内容：

defaultdict(<class 'int'>, {7: 160, 8: 134, 6: 145, 9: 107, 3: 50, 10: 76, 12: 26, 4: 86, 5: 128, 2: 37, 11: 51})

您还可以通过绘图轻松地说明结果：

import matplotlib.pyplot as plt
plt.bar(results.keys(), results.values())
plt.show()

Answer 3

您没有正确考虑概率。 r.randint(2, 12)与独立滚动两个骰子不同（对于某些值，它们是多个两个骰子的总和为相同值）。

import collections
import random

print("For Loop")

occurrences = []
for trial in range(1000):
    die1 = random.randint(1, 6)
    die2 = random.randing(1, 6)
    occurrences.append(die1 + die2)
counter = collections.Counter(occurrences)
for roll, count in counter.items():
    print(f"sum = {roll} count = {count}") 

如果您不想导入标准库的其他部分，则可以自己制作计数器。

import random

print("For Loop")
occurrences = {}
for trial in range(1000):
    die1 = random.randint(1, 6)
    die2 = random.randing(1, 6)
    roll = die1 + die2
    current = occurrences.setdefault(roll, 0)
    occurrences[roll] = current + 1

for roll, count in occurrences.items():
    print(f"sum = {roll} count = {count}")

请注意，输出将因所涉及的随机性而稍有不同。