六的倍数

假设 $s$ 是 $n$ 掷骰后我们的骰子的总和。 $6$ 的可能值与 $s \mod 6$ 的值相同。 乍一看，这可能表明 $n$ 掷骰后 $s$ 可能存在 $6$ 的情况，但请注意，无论 $s$ 是什么，都存在 $1$ 掷骰，使总和达到 6 的倍数。（即，如果 $2 \equiv s\mod6$，则必须掷出 $4$ 才能达到 $6$ 的倍数）

这意味着你下一次掷骰子达到 6 倍数的概率将始终为 $\frac16$，从而导致 $p = \frac16$ 的几何分布。这种随机变量的期望由 $\mu = \frac1p= 6$ 给出。

我们也可以在不知道几何分布行为的情况下得到 $6$。假设我们预期的掷骰数是 $x$。我们可以说 $x = \frac16(1) + \frac56(x+1)$ $\\$（有 $\frac16$ 机会我们得到 $6$ 的倍数，$\frac56$ 有机会我们在额外掷骰后回到第一个方块）。求解该系统我们还得到$x = 6$。

import random
from typing import Callable

roll_dice: Callable [[None], int] = lambda: random.randrange(1,7)

total_rolls = 0
num_iters = 10_000
#simulates rolling until we get a multiple of 6 num_iters times
for i in range(num_iters):

   run_sum = roll_dice()
   rolls = 1

   #reroll until the sum is a multiple of 6
   while run_sum % 6 != 0:
      run_sum += roll_dice()
      rolls += 1

   total_rolls += rolls

#expecting value close to 6.00
print(total_rolls/num_iters)

Original Explanation

Suppose $$s$$ is the sum of our die after $$n$$ rolls. There are $$6$$ possible values for what $$s \mod 6$$ can be. At first glance, that might suggest there are $$6$$ possible cases for $$s$$ after $$n$$ rolls, but notice that regardless of what $$s$$ is, there exists exactly $$1$$ roll that brings the sum to a multiple of 6. (ie. If $$2 \equiv s\mod6$$ you must roll a $$4$$ to reach a multiple of $$6$$)

This means the probability of reaching a multiple of 6 on your next roll will always be $$\frac16$$, resulting in a geometric distribution with $$p = \frac16$$. The expectation of such a random variable is given by $$\mu = \frac1p= 6$$.

We can also get $$6$$ without knowing the behavior of a geometric distribution. Suppose our number of expected rolls is $$x$$. We can say $$x = \frac16(1) + \frac56(x+1)$$ $$\\$$ (there is a $$\frac16$$ chance we get a multiple of $$6$$ and a $$\frac56$$ chance we go back to square one after an additional roll). Solving this system we also get $$x = 6$$.

import random
from typing import Callable

roll_dice: Callable [[None], int] = lambda: random.randrange(1,7)

total_rolls = 0
num_iters = 10_000
#simulates rolling until we get a multiple of 6 num_iters times
for i in range(num_iters):

   run_sum = roll_dice()
   rolls = 1

   #reroll until the sum is a multiple of 6
   while run_sum % 6 != 0:
      run_sum += roll_dice()
      rolls += 1

   total_rolls += rolls

#expecting value close to 6.00
print(total_rolls/num_iters)