Roll Again on 4, 5, 6: Expected Total Payoff
You play the following game with a fair six-sided die. You roll the die and collect the face value in dollars. If the roll is a $4$, $5$ or $6$, you roll again (collecting the new face value as well) and the same rule applies to every subsequent roll. If the roll is a $1$, $2$ or $3$, you collect that face value and the game ends.
What is the expected total payoff of the game?
Hints
- Let $E$ be the expected total payoff. Condition on the first roll: with probability 1/2 the game ends, with probability 1/2 you are back at the start with some money already banked.
- Alternatively, the total is $S_N = X_1 + \cdots + X_N$, where $N$ is the (random) number of rolls. $N$ is a stopping time; use Wald's identity $E[S_N] = E[X]\,E[N]$.
- $N$ is geometric with success probability $1/2$, so $E[N] = 2$, and $E[X] = 3.5$. Hence $E[S_N] = 7$.
Worked Solution
How to Think About It: The payoff is a sum of a random number of i.i.d. rolls. Two clean routes: a one-line renewal equation (condition on the first roll), or Wald's identity, which says the expected sum equals the expected roll times the expected number of rolls, valid because whether you continue is decided by the rolls already seen.
Quick Estimate: You always get at least one roll (worth $3.5$). Half the time you get another, and so on, so you expect $2$ rolls in total, worth about $2 \times 3.5 = 7$.
Formal Solution:
*Step 1 -- Renewal equation.* Let $E$ be the expected total payoff from the start. Conditioning on the first roll $X_1$:
$$E = E[X_1] + P(X_1 \ge 4)\,E = 3.5 + \tfrac{1}{2}E \quad\Longrightarrow\quad E = 7.$$
The continuation is a fresh copy of the game because the rolls are independent and the rule does not depend on money already collected.
*Step 2 -- Wald's identity.* Let $N$ be the number of rolls. $N$ is a stopping time for the roll sequence (the decision after roll $k$ depends only on $X_k$), the rolls are i.i.d. with $E[X] = 7/2$, and $E[N] < \infty$. Wald's identity gives
$$E[S_N] = E[X]\,E[N].$$
Each roll ends the game with probability $1/2$ independently, so $N \sim \text{Geometric}(1/2)$ on $\{1, 2, \ldots\}$ and $E[N] = 2$. Hence $E[S_N] = \tfrac{7}{2}\cdot 2 = 7$.
*Step 3 -- Check by decomposition.* The last roll is uniform on $\{1,2,3\}$ (mean $2$), and each of the $N - 1$ earlier rolls is uniform on $\{4,5,6\}$ (mean $5$). $E[N-1] = 1$, so $E[S_N] = 2 + 5 \cdot 1 = 7$.
Answer: The expected payoff is $\$7$.
Intuition
Every roll is worth $3.5$ on average and you get two rolls on average (the number of rolls is geometric with parameter $1/2$), so the game is worth $7$. That is Wald's identity: for i.i.d. terms and a stopping time with finite mean, the expected sum is the expected term times the expected count, even though the decision to keep rolling depends on the values rolled. Wald-type reasoning is how you value any "keep collecting until a random stop" payoff, from expected fills over a random number of trades to the expected loss over a random number of defaults.