Maximum Fair Penalty in a Dice Payout Game
You roll two fair $n$-sided dice (faces $1$ through $n$). Let $A$ and $B$ be disjoint subsets of the $n^2$ possible outcomes.
- If the roll lands in $A$, you receive $\$p$.
- If the roll lands in $B$, you pay $\$q$.
- If the roll lands in neither $A$ nor $B$, you re-roll at no cost, and keep re-rolling until the outcome falls in $A$ or $B$.
Given that $p$ is fixed, find the maximum value of $q$ such that your expected payout is non-negative.
Evaluate for the specific case $p = 100$, $n = 10$, $|A| = 25$, and $|B| = 50$.
Hints
- The re-roll mechanism means outcomes outside $A \cup B$ never matter. What does the game look like if you condition on landing in $A \cup B$?
- Since the dice are fair, the conditional probability of landing in $A$ given $A \cup B$ is just $\frac{|A|}{|A| + |B|}$. Write the expected payout using these conditional probabilities.
- Set the expected payout $\frac{|A|}{|A|+|B|} \cdot p - \frac{|B|}{|A|+|B|} \cdot q \geq 0$ and solve for $q$.
Worked Solution
How to Think About It: The re-roll mechanism is the key simplification here. Because you keep rolling until you land in $A \cup B$, everything outside that set is irrelevant -- it is as if those outcomes do not exist. The game effectively reduces to: draw uniformly from $A \cup B$, win $p$ if you land in $A$, lose $q$ if you land in $B$. The probability of each event is just the fraction of $A \cup B$ that it occupies. This is a bread-and-butter conditional probability observation.
Quick Estimate: With $|A| = 25$ and $|B| = 50$, you land in $A$ one-third of the time and in $B$ two-thirds of the time (conditional on being in $A \cup B$). So the expected payout is roughly $(1/3)(100) - (2/3)(q)$. For breakeven, $(1/3)(100) = (2/3)(q)$, giving $q = 50$. Done -- that is the answer, and you can get it in 15 seconds.
Approach: Condition on the outcome being in $A \cup B$ and set the expected payout to zero to find the breakeven $q$.
Formal Solution:
Since $A$ and $B$ are disjoint, $|A \cup B| = |A| + |B|$. The re-roll rule means we only care about outcomes in $A \cup B$. Given that the dice roll lands in $A \cup B$, each outcome in that set is equally likely (the dice are fair), so:
$$P(\text{outcome} \in A \mid \text{outcome} \in A \cup B) = \frac{|A|}{|A| + |B|}$$
$$P(\text{outcome} \in B \mid \text{outcome} \in A \cup B) = \frac{|B|}{|A| + |B|}$$
The expected payout is:
$$E[\text{payout}] = \frac{|A|}{|A| + |B|} \cdot p - \frac{|B|}{|A| + |B|} \cdot q$$
Setting $E[\text{payout}] \geq 0$:
$$\frac{|A| \cdot p}{|A| + |B|} \geq \frac{|B| \cdot q}{|A| + |B|}$$
$$q \leq p \cdot \frac{|A|}{|B|}$$
Plugging in $p = 100$, $|A| = 25$, $|B| = 50$:
$$q \leq 100 \cdot \frac{25}{50} = 50$$
Answer: The maximum value of $q$ for non-negative expected payout is $q = p \cdot \frac{|A|}{|B|}$. For the given parameters, $q = 50$.
Intuition
The re-roll trick collapses a complicated-looking game into a simple one. Whenever a game says "keep going until X happens," you should immediately condition on X and forget everything else. Here, conditioning on landing in $A \cup B$ turns two fair dice with 100 outcomes into a weighted coin flip between winning $p$ and losing $q$. The weights are just the relative sizes of $A$ and $B$.
This pattern shows up constantly in trading and market-making. If a contract pays out only when a specific event occurs (like an option that only matters at expiry, conditional on being in-the-money), the irrelevant scenarios wash out and the fair price depends only on the conditional distribution over the events that matter. The breakeven condition $q \leq p \cdot |A|/|B|$ is just the statement that the odds you are getting ($p : q$) must be at least as favorable as the odds against you ($|B| : |A|$).