Ten-Sided Die Payoff Game
You roll a fair 10-sided die with faces labeled
- What is the maximum amount you would pay to play this game?
- Now suppose you are allowed one optional re-roll. After seeing the first roll, you can either keep it or re-roll once, keeping the second result. What is the new fair price?
Hints
- For Part 1, what is the expected value of a discrete uniform distribution on $\{1, \ldots, 10\}$?
- For Part 2, think about when you should re-roll: compare your current roll to the expected value of a fresh roll. You only re-roll when keeping is worse than starting over.
- The threshold for keeping is $E[X] = 5.5$, so you keep rolls of 6 through 10 and re-roll on 1 through 5. Compute the weighted average of the two branches.
Worked Solution
How to Think About It: This is a classic expected value problem with an optionality twist. Part 1 is straightforward -- the fair price is just the expected payout of a uniform die. Part 2 is where it gets interesting: you have the option to re-roll, and you should only exercise that option when your current roll is worse than what you expect from a fresh roll. This is the simplest possible example of optimal stopping -- keep if you are above expectation, re-roll if you are below. Before doing any math, your gut should say: the re-roll option is worth something positive (optionality is never negative), and it should not be worth more than a couple of dollars since you are only improving on the bad outcomes.
Quick Estimate: Without the re-roll, you average $\$5.50$. With the re-roll, you keep rolls of 6-10 (half the time, averaging 8) and re-roll on 1-5 (half the time, getting 5.5 in expectation). So roughly $0.5 \times 8 + 0.5 \times 5.5 = 6.75$. The re-roll option is worth about $\
Approach: Compute the expected value directly for Part 1. For Part 2, use backward induction: compare each first-roll outcome to the expected value of a fresh roll to determine the optimal keep/re-roll threshold.
Formal Solution:
Part 1 -- Single Roll:
The die is uniform on $\{1, 2, \ldots, 10\}$, so:
$E[X] = \frac{1 + 2 + \cdots + 10}{10} = \frac{55}{10} = 5.50$
The fair price is $\$5.50$.
Part 2 -- One Re-Roll Allowed:
After seeing the first roll $x$, you choose between keeping $x$ or re-rolling and getting $E[X] = 5.5$ in expectation. The optimal strategy is:
- Keep if $x \geq 6$ (i.e., $x > E[X] = 5.5$)
- Re-roll if $x \leq 5$ (i.e., $x < E[X] = 5.5$)
The expected payoff under this strategy is:
$E[\text{payoff}] = P(\text{keep}) \cdot E[X \mid X \geq 6] + P(\text{re-roll}) \cdot E[X]$
$= \frac{5}{10} \cdot \frac{6 + 7 + 8 + 9 + 10}{5} + \frac{5}{10} \cdot 5.5$
$= \frac{1}{2} \cdot 8 + \frac{1}{2} \cdot 5.5 = 4 + 2.75 = 6.75$
The value of the re-roll option is $6.75 - 5.50 = \
Answer: The fair price for a single roll is $\$5.50$. With one optional re-roll, the fair price is $\$6.75$. The re-roll option is worth $\
Intuition
This problem distills the core logic of optionality down to its simplest form. The re-roll is a free option: it lets you replace bad outcomes with a fresh draw while keeping good ones. The optimal strategy is trivially simple -- re-roll if and only if your current payoff is below what you expect from a new draw. This is exactly backward induction with one step remaining, and it generalizes to multi-step stopping problems where you compare the current state to the continuation value.
In practice, this kind of thinking shows up constantly. Every American option exercise decision follows the same logic: exercise now if the immediate payoff beats the expected continuation value, hold otherwise. Market makers quoting a spread on an illiquid name are implicitly doing the same calculation -- should I hit this bid now, or wait for a better one? The $\