条件骰子

Conditional Die Rolls

专题: Probability / 概率
难度: L4
来源: OpenQuant

第 1 小问

题目详情

Dan 掷骰子，直到获得 $6$ 。鉴于他“没有”看到 $5$ ，Dan 掷骰子的预期次数是多少？

Dan rolls a dice until he gets a $6$ . Given that he did not see a $5$ , what is the expected number of times Dan rolled his die?

解析

如果 Dan 没有看到 $5$ ，则意味着条件是 $6$ 在 $5$ 之前出现。当掷一个公平的骰子时，这种情况会发生一半，因此该情况的概率为 $\frac12$ 。

给定条件下滚动 $6$ 的概率为 $\frac{\frac16}{\frac12} = \frac13$ 。因为卷是彼此独立的，所以该分布是 $p = \frac13$ 的几何分布，结果是： $\mu = \frac1p = \boxed3$ （请注意，如果他没有看到 $5$ ，则滚动 $6$ 的概率不是 $\frac15$ 。这表明 Dan 滚动 $1,2,3,4,$ 和 $6$ 的可能性相同，这不可能是真的，因为你知道 $6$ 出现在 $5$ 之前，而其他数字不一定是这种情况）

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000
for i in range(num_iters):

    sequence = []
    while(True):

        sequence.append(roll_dice())

        if sequence[-1] == 5:
            break
        elif sequence[-1] == 6:
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))

Original Explanation

If Dan did not see a $5$ , that means the condition is $6$ came before $5$ . When rolling a fair dice, this will occur half the time, thus the probability of the condition is $\frac12$ .

The probability of rolling a $6$ given the condition is $\frac{\frac16}{\frac12} = \frac13$ . Because rolls are independent of each other, this distribution is geometric with $p = \frac13$ resulting in: $\mu = \frac1p = \boxed3$

(Note if he does not see $5$ the probability of rolling a $6$ is not $\frac15$ . That suggests Dan is equally likely to roll a $1,2,3,4,$ and $6$ which can not be true since you know $6$ comes before $5$ which is not necessarily the case with the other numbers)

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000
for i in range(num_iters):

    sequence = []
    while(True):

        sequence.append(roll_dice())

        if sequence[-1] == 5:
            break
        elif sequence[-1] == 6:
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))

第 2 小问

题目详情

Dan 掷骰子，直到获得 $6$ 。鉴于他看到了 $5$ ，Dan 掷骰子的预期次数是多少？

Dan rolls a dice until he gets a $6$ . Given that he saw a $5$ , what is the expected number of times Dan rolled his die?

解析

这与第 1 部分的情况相反，其中第一个 $5$ 在第一个 $6$ 之前出现。使用第 1 部分，我们知道找到预期的卷数，看到第一个 $5$ 是 $3$ 。

一旦我们看到了第一个 $5$ ，就没有更多的条件了。当没有条件时，滚动 $6$ 的概率就是 $\frac16$ ，因此我们期望在看到第一个 $6$ 之前滚动骰子 $6$ 次。要消除这种情况，平均需要掷 $3$ 次，一旦这种情况消失，我们预计会掷 $6$ 次，因此 Dan 预计会掷骰子 $\boxed9$ 次。

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000
for i in range(num_iters):

    sequence = []
    while(True):

        sequence.append(roll_dice())

        if sequence[-1] == 6:
            if 5 in sequence:
                sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))

Original Explanation

This is the opposite case as Part 1 where the first $5$ is seen before the first $6$ . Using Part 1, we know find the expected number of rolls to see the first $5$ is $3$ .

Once we have seen the first $5$ , there is no more condition. When there is no condition, the probability of rolling a $6$ is simply $\frac16$ so we expect to roll a die $6$ times before seeing our first $6$ . To get rid of the condition will take on average $3$ rolls and once the condition is gone we expect to roll $6$ times therefore Dan can expect to roll the die $\boxed9$ times.

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000
for i in range(num_iters):

    sequence = []
    while(True):

        sequence.append(roll_dice())

        if sequence[-1] == 6:
            if 5 in sequence:
                sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))

第 3 小问

题目详情

Dan 掷骰子，直到获得 $6$ 。鉴于他只看到偶数，丹掷骰子的预期次数是多少？

Dan rolls a dice until he gets a $6$ . Given that he saw only even numbers, what is the expected number of times Dan rolled his die?

解析

仅看到偶数意味着 $6$ 出现在 $1, 3,$ 和 $5$ 之前。 $4$ 总数中第一个出现的单个数字的出现概率为 $\frac14$ ，因此该条件的概率为 $\frac14$ 。

使用与第 1 部分类似的逻辑，我们得到在给定的滚动中将 $6$ 滚动为 $\frac{\frac16}{\frac14} = \frac23$ 的概率，因此： $\mu = \boxed{\frac32}$

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000

for i in range(num_iters):
    sequence = []

    while(True):
        sequence.append(roll_dice())

        if sequence[-1] == 1:
            break
        elif sequence[-1] == 3:
            break
        elif sequence[-1] == 5:
            break
        elif sequence[-1] == 6:
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))

Original Explanation

Seeing only even numbers means $6$ came before $1, 3,$ and $5$ . A single number coming first among $4$ total numbers has probability $\frac14$ of occuring, thus the probability of the condition is $\frac14$ .

Using similar logic to Part 1 we get the probability of rolling a $6$ in a given roll to be $\frac{\frac16}{\frac14} = \frac23$ , thus:

\mu = \boxed{\frac32}

import random

roll_dice = lambda: random.randint(1,6)

sequence_lengths = []
num_iters = 10_000

for i in range(num_iters):
    sequence = []

    while(True):
        sequence.append(roll_dice())

        if sequence[-1] == 1:
            break
        elif sequence[-1] == 3:
            break
        elif sequence[-1] == 5:
            break
        elif sequence[-1] == 6:
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths)/len(sequence_lengths))