我該如何簡化此代碼

Question

我是一名初學者，我想知道是否有更簡單的方式在Python中編寫此代碼。我假設某種類型的字典，但我不知道如何寫出來。 幾天前我正在巡航，我玩胡扯。我想知道賠率是否有點正確。所以，我寫了這個，但我知道有一個更簡單的方法。

import random 

dice2 = 0 
dice3 = 0 
dice4 = 0 
dice5 = 0 
dice6 = 0 
dice7 = 0 
dice8 = 0 
dice9 = 0 
dice10 = 0 
dice11 = 0 
dice12 = 0 

for i in range(100000): 
      dice1 = random.randint(1,6) 
      dice2 = random.randint(1,6) 
      number = dice1 + dice2 
      #print(dice1) 
      if number == 2: 
         dice2 +=1 
      elif number == 3: 
         dice3 += 1 
      elif number == 4: 
         dice4 += 1 
      elif number == 5: 
         dice5 += 1 
      elif number == 6: 
         dice6 += 1 
      elif number == 7: 
         dice7 += 1 
      elif number == 8: 
         dice8 += 1 
      elif number == 9: 
         dice9 += 1 
      elif number == 10: 
         dice10 += 1 
      elif number == 11: 
         dice11 += 1 
      elif number == 12: 
         dice12 += 1 

total = dice2+dice3+dice4+dice5+dice6+dice7+dice8+dice9+dice10+dice11+dice12 

在此結束時，它只打印出2-12數字的命中百分比。

Answer 1

from random import randint 

dice = [0]*11 
for i in range(100000): 
    dice[randint(1,6)+randint(1,6)-2] += 1 

total = sum(dice) #it is 100000, of course 
for i, v in enumerate(dice, 2): 
    print('{0}: {1}%'.format(i, v*100.0/total)) 

Answer 2

我會用Counter，因爲這就是它被用於製作：

from random import randint 
from collections import Counter 

counts = Counter(randint(1, 6) + randint(1, 6) for i in range(100000)) 
total = sum(counts.values()) 
number_of_tens = counts[10] 

Answer 3

import random 

def roll(n=6): 
    return random.randint(1, n) 

dice = dict.fromkeys(range(2, 13), 0) 

for i in range(100000): 
    number = roll() + roll() 
    dice[number] += 1 

total = float(sum(dice.values())) 
for k,v in dice.items(): 
    print "{}, {:.2%}".format(k, v/total)