Question

我希望通过以下功能进行1000次互动，以了解您是否在此游戏中获胜或放松。

游戏的设计是这样的，你扔掉一对骰子，然后拿回钱或者放钱。假设我们从5个硬币开始。

投出 12 会产生 1.5 个硬币。

投出 11 会产生 1 个硬币。

投出 10 会产生 0.5 硬币。

投出 9,8或7 不会产生任何结果。

从您的硬币数量中投掷 6,5,4,3,2或1 扣除 0.5 硬币。

到目前为止，这是我的实现：

def luckCalc():

    amount = 5

    # if 12 then 1/36 chance

    if random.randrange(1,7) == 6 and random.randrange(1,7) == 6:
        amount = amount + 1.5

    # if 11 then 2/36 chance
    elif  (random.randrange(1,7) == 5 and random.randrange(1,7) == 6) or (random.randrange(1,7) == 6 and random.randrange(1,7) == 5):
        amount = amount + 1

   # if 10 then 3/36 chance

   elif (random.randrange(1,7) == 5 and random.randrange(1,7) == 5) or (random.randrange(1,7) == 4 and random.randrange(1,7) == 6) or (random.randrange(1,7) == 6 and random.randrange(1,7) == 4):
       amount = amount + 0.5

   # if 9,8,7
   # 4/36 + 5/36 + 6/36 chance
   # 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 chance
   # 2+6, 3+5, 4+4, 5+3, 6+2 chance
   # 3+6, 4+5, 5+4, 6+3 chance
   # then no change in amount 

   # if 6,5,4,3,2,1
   # chances...
   # then amount -0.5

return amount 

# Iterate over the dice throwing simulator and calculate total

total = 0.0
for a in range(1000):
    total = total + luckCalc()

print (total)

我在功能结束时停止编码，因为我认识到必须有一个更优雅的解决方案来实现这一目标。任何有趣的建议，我一直听到的蒙特卡洛是什么？

Answer 1

每次拨打random.randrange(1,7)时，都会生成新随机数。由于您正在测试单个“转弯”，因此请滚动两次：

def roll_die():
    return random.randrange(1, 7)

total = roll_die() + roll_die()

看看总和是否在一个范围内：

def play_turn():
    total = roll_die() + roll_die()

    if total == 12:
        return 1.5
    elif total == 11:
        return 1.0
    elif total == 10:
        return 0.5
    elif total <= 6:
        return -0.5
    else:  # total is 7, 8, or 9
        return 0

以下是100,000轮的结果：

>>> from collections import Counter
>>> counts = Counter(play_turn() for i in xrange(100000))
>>> counts
    Counter({-0.5: 41823, 0: 41545, 0.5: 8361, 1.0: 5521, 1.5: 2750})
>>> probabilities = {score: count / 100000.0 for score, count in counts.items()}
>>> probabilities
    {-0.5: 0.41823, 0: 0.41545, 0.5: 0.08361, 1.0: 0.05521, 1.5: 0.0275}

Answer 2

你实际上可以将你正在做的所有事情都滚动到一个函数中：

from random import randrange

def play_game(rolls=1000, amount=5, n=6):
    """Play game 'rolls' times, starting with 'amount' on 'n'-sided dice.""" 
    for i in range(rolls):
        roll = randrange(1, n+1) + randrange(1, n+1)
        if roll == 12:
            amount += 1.5
        elif roll == 11:
            amount += 1
        elif roll == 10:
            amount += 0.5
        elif roll < 7:
            amount -= 0.5
    return amount

Answer 3

我注意到代码中的一些内容。首先，对于6-1个案例，你实际上并没有从金额中减去0.5。其次，既然你没有传递每个循环的初始数量，那么你的总数就会增加5到6.5之间，这使得总数毫无意义。

更有效的总数是每次传递金额：

def luckCalc( amount ):

然后为你的循环：

total = 5.0
for a in range(1000):
    total = luckCalc(total)

Blender的答案刚刚在我写这篇文章时发布，是简化主要功能的好方法。

Answer 4

我个人喜欢将结果表设置为数组（或字典，但这更符合我的目的，因为每个结果都是少数可能的整数之一），每个骰子的索引都设置为值由此产生的变化。见下文。

import random

def luckCalc(coins=5):

  diceroll = random.randint(1,6)+random.randint(1,6) #roll them bones

  #here's that table I was talking about....
  results_table = ['index 0 is blank',"you can't roll a one on two dice",-.5,-.5,-.5,-.5,-.5,0,0,0,.5,1,1.5]

  coins += results_table[diceroll] #changes your coins value based on your roll (as an index of results_table)

  if results_table[diceroll] > 0: #change the string if your result was + or -
    result = "gained {}".format(results_table[diceroll])
  else:
    result = "lost {}".format(results_table[diceroll]*-1)

  print("You {} coins, putting you at {}".format(result,coins)) #report back to the user
  return coins #this is how you save your output


#CONSTANTS GO HERE -- YOU CAN CHANGE THESE TO CHANGE YOUR PROGRAM
STARTING_COINS = 5
HOW_MANY_ITERATIONS = 1000

#this way we don't modify a constant
coins = STARTING_COINS

#do it how many times?
for _ in range(HOW_MANY_ITERATIONS): #oh yeah that many times
  coins = luckCalc(coins) #runs the function and saves the result back to coins

#report to the user your final result.
print("After {} rolls, your final total is {}".format(HOW_MANY_ITERATIONS,coins))

Python中的概率骰子游戏有两个骰子

4 个答案: