Expected Payoff of a Recursive Die Game
You roll a fair six-sided die repeatedly. On each roll:
- If the outcome is 4, 5, or 6, you get paid that amount (in dollars) and continue rolling.
- If the outcome is 1, 2, or 3, you get paid that amount and the game stops.
What is the expected total payoff?
Hints
- Think about how the game resets after a high roll. Can you write the total expected payoff in terms of itself?
- Split each roll into two cases: with probability $1/2$ you roll high (expected payment 5, game continues) and with probability $1/2$ you roll low (expected payment 2, game stops). Write the recursion $E = \frac{1}{2}(5 + E) + \frac{1}{2}(2)$.
- Subtract $E/2$ from both sides to get $E/2 = 7/2$, giving $E = 7$. Verify by noting the expected payment per roll is 3.5 and the geometric number of rolls is 2.
Worked Solution
How to Think About It: Each roll pays you its face value regardless of whether the game continues. The game continues (with probability 1/2) or stops (with probability 1/2) depending on whether you roll high or low. This is a geometric-type recursion: the expected payoff equals one roll's payment plus (with probability 1/2) the expected payoff of the entire game again.
Quick Estimate: The expected number of rolls is geometric with parameter $1/2$, giving $E[\text{rolls}] = 2$. The average payment per roll is $(1+2+3+4+5+6)/6 = 3.5$. So the expected total is roughly $2 \times 3.5 = 7$. We will confirm this is exact.
Formal Solution:
Let $E$ denote the expected total payoff. On each roll, two things can happen:
- High roll (4, 5, or 6) -- probability $1/2$, expected payment $(4+5+6)/3 = 5$, game continues.
- Low roll (1, 2, or 3) -- probability $1/2$, expected payment $(1+2+3)/3 = 2$, game stops.
By the law of total expectation:
$$E = \frac{1}{2}(5 + E) + \frac{1}{2}(2)$$
$$E = \frac{5}{2} + \frac{E}{2} + 1$$
$$E - \frac{E}{2} = \frac{7}{2}$$
$$\frac{E}{2} = \frac{7}{2}$$
$$E = 7$$
Verification via Series Decomposition:
The game reaches roll $n$ if the first $n-1$ rolls were all high, which occurs with probability $(1/2)^{n-1}$. Conditional on roll $n$ occurring, that roll is high with probability $1/2$ (paying 5 in expectation) and low with probability $1/2$ (paying 2 in expectation), so the expected payment on roll $n$ given it happens is $(1/2)(5) + (1/2)(2) = 3.5$. Summing over all possible rolls:
$$E = \sum_{n=1}^{\infty} 3.5 \cdot \left(\frac{1}{2}\right)^{n-1} = 3.5 \cdot \frac{1}{1 - 1/2} = 7 \quad \checkmark$$
Note that the payment amount and the continuation decision are correlated -- high rolls both pay more and continue the game. This means Wald's identity does not apply in its naive form here, but the recursive approach and the conditional series both handle this correctly.
Answer: The expected total payoff is $\boxed{7}$.
Intuition
This problem is a clean example of the "recursive game" pattern that appears constantly in interview questions. The key structural observation is that after a high roll, you are in exactly the same position as when you started -- the game "resets." This self-similarity is what makes the recursion $E = \frac{1}{2}(5 + E) + \frac{1}{2}(2)$ valid.
A subtle point worth noting: the payment per roll and the stopping decision are correlated (high rolls pay more and continue), yet the expected total payoff equals $E[\text{payment per roll}] \times E[\text{number of rolls}] = 3.5 \times 2 = 7$. This works because each roll contributes its unconditional expected value of 3.5 regardless of whether it turns out to be the stopping roll. This is a nice example where the "product of expectations" gives the right answer despite dependence, because of the memoryless structure of the game.