最高点数期望

因为 Jim 会在第一次掷出 $4$ 时停止，所以他在整个过程中掷到的最大值 $H$ 只可能是 $4,5,6$。

$\textbf{1) }H=4$：等价于在第一次出现 $4$ 之前从未掷出 $5$ 或 $6$。 只看掷骰序列中落在集合 $\{4,5,6\}$ 的那些次序（忽略 $1,2,3$），第一次出现的值在 $4,5,6$ 之间对称且等可能，因此

$\textbf{2) }H=6$：等价于在第一次出现 $4$ 之前先出现了 $6$。 只看集合 $\{4,6\}$ 的首次出现，$4$ 与 $6$ 对称且等可能，所以

$\textbf{3) }H=5$：只能是剩余概率

因此

import random

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

results = []
num_iters = 10_000

for _ in range(num_iters):
    highest = 0
    while True:
        roll = roll_dice()
        highest = max(highest, roll)
        if roll == 4:
            break
    results.append(highest)

print(sum(results) / num_iters)

Original Explanation

Because Jim stops when he first rolls a $4$, the possible values of the highest roll are $4$, $5$, and $6$. Let $H$ be the highest value rolled.

$\textbf{Case 1: }H=4.$ This happens exactly when Jim never rolls a $5$ or $6$ before the first $4$. Consider only the subsequence of rolls that are in $\{4,5,6\}$. The first such value is equally likely to be $4$, $5$, or $6$, so

$\textbf{Case 2: }H=6.$ This happens exactly when Jim rolls a $6$ before the first $4$. Looking only at the values $\{4,6\}$, the first one to appear is equally likely to be $4$ or $6$, hence

$\textbf{Case 3: }H=5.$ This is the remaining probability:

Therefore,

import random

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

results = []
num_iters = 10_000

for _ in range(num_iters):
    highest = 0
    while True:
        roll = roll_dice()
        highest = max(highest, roll)
        if roll == 4:
            break
    results.append(highest)

print(sum(results) / num_iters)