Expected Payout Excluding Six
Eric rolls a fair six-sided die repeatedly until he gets any result other than $6$. He is then paid that final face value in dollars. What is Eric's expected payout?
Hints
- The number of rolls before the final one is irrelevant to the payout -- focus on what distribution the final roll has.
- The final roll is a die roll conditioned on not being $6$; by symmetry, the five remaining outcomes $\{1, 2, 3, 4, 5\}$ are equally likely.
- Apply the conditional expectation formula: $E[X \mid X \neq 6] = \frac{\sum_{k=1}^{5} k \cdot P(X = k)}{P(X \neq 6)}$, which simplifies to the average of $\{1, 2, 3, 4, 5\}$.
Worked Solution
How to Think About It: The number of rolls does not matter for the payout -- Eric rolls until he gets a non-six, and then he keeps whatever he rolled. The six is just a "roll again" instruction. So the question is really: given that we know the outcome is not a $6$, what is the expected value of a fair die roll?
Quick Estimate: The non-six outcomes are $\{1, 2, 3, 4, 5\}$, symmetric around $3$. Expected value is exactly $3$.
Approach: Conditional expectation. Condition on the event $\{X \neq 6\}$ where $X$ is a single die roll.
Formal Solution:
Let $X$ be the outcome of a single fair die roll, uniform on $\{1, 2, 3, 4, 5, 6\}$. Eric's payout is the first roll that satisfies $X \neq 6$. Because each roll is independent and identically distributed, the payout has the same distribution as $X$ conditioned on $X \neq 6$.
Given $X \neq 6$, the five remaining outcomes $\{1, 2, 3, 4, 5\}$ are equally likely (by symmetry of the uniform distribution over the remaining mass). Therefore:
$E[\text{payout}] = E[X \mid X \neq 6] = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$
Answer: $E[\text{payout}] = \$3$.
Intuition
This is a clean illustration of conditional expectation: once you condition out the impossible outcome, the remaining distribution is just the original one restricted to the feasible set. The re-normalization step (dividing by $P(X \neq 6) = 5/6$) ensures the conditional probabilities sum to
The deeper lesson is about recognizing when repeated sampling does not add complexity. The geometric waiting time (how many sixes you roll before a non-six) affects the total number of rolls, but it does not affect the distribution of the payout, because each roll is independent. This independence structure -- where the stopping rule is decoupled from the payout distribution -- appears frequently in market microstructure models, where you wait for a tradeable quote but the trade price has a distribution independent of the waiting time.