遇 6 清零的骰子游戏

Dynamic Dice Game

专题: Finance / 金融
难度: L4
来源: QuantQuestion

题目详情

赌场提供一个掷骰游戏：你可以一直掷下去，除非掷到 6。

掷到 1–5：分别获得 1–5 美元并累加到总额；
掷到 6：立刻失去一切，游戏结束且收益为 0。

每次掷到 1–5 后你可以选择停下并拿走累计收益，或继续掷。作为风险中性玩家，你愿意为进入游戏支付多少钱？最优策略是什么？

A casino offers a dice game where you may keep rolling as long as you like, unless a 6 appears. After each roll:

If you roll 1, you gain $1.
If you roll 2, you gain $2.
...
If you roll 5, you gain $5.
If you roll 6, you lose everything and the game ends immediately.

After each roll of 1–5, you can decide to stop and keep your accumulated total, or to continue rolling. As a risk-neutral player, how much would you pay to play this game, and what is the optimal strategy?

解析

设当前累计为 $n$ 。

若选择继续掷：以 $1/6$ 概率掷到 6，收益变 0；以 $1/6$ 概率掷到 $k\in\{1,2,3,4,5\}$ ，累计变为 $n+k$ 。

因此继续的期望为

\frac{1}{6}\bigl[(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+0\bigr]=\frac{5n}{6}+2.5.

当继续期望大于当前 $n$ 时才继续：

\frac{5n}{6}+2.5>n\iff n<15.

所以最优策略是：

若 $n<15$ ，继续掷；
若 $n\ge 15$ ，停下。

用递推（注意 $n=14$ 时最多再加 5 到 19）可算出游戏初始价值约为 6.15 美元。

Original Explanation

Let $n$ be your current accumulated total. If you decide to roll again:

With probability $1/6$ , you roll 6 and end up with $0.
With probability $1/6$ for each $k=1,2,3,4,5$ , your total becomes $n+k$ .

Hence the expected value if you continue is $\frac{1}{6}\bigl[(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+0\bigr] = \frac{5n}{6} + 2.5.$ You should continue if this exceeds your current $n$ , i.e. $\frac{5n}{6} + 2.5 > n \quad\Longrightarrow\quad n < 15.$ So if $n<15$ , you roll again; if $n\ge 15$ , you stop.

By solving this via backward recursion (noting the maximum possible total is 19 if you are at 14 and roll a 5), one obtains an initial expected value of about $6.15. Thus, you would pay up to $6.15 to play, and the stopping threshold is $n=15$ .