Expected Winnings of a Repeat-Stop Dice Game
You have a 6-sided die with faces $(1, 2, 3, 1, 2, 3)$ -- so each of the values $\{1, 2, 3\}$ appears with probability $\frac{1}{3}$.
You roll repeatedly. Each roll pays you coins equal to the face value shown. The game ends the moment you roll the same value as the previous roll.
What is the expected total payout?
Hints
- Think of this as a Markov chain: the only state information that matters is the value of the last roll. Define $E_v$ as the expected future earnings given the last roll was $v$.
- On each step, you stop with probability $\frac{1}{3}$ and continue with probability $\frac{2}{3}$. Since all three values sum to $6$, the equation for $E_v$ simplifies to $E_v = 2 + \frac{E_a + E_b}{3}$ where $a, b$ are the other two values.
- Sum all three equations to get $S = E_1 + E_2 + E_3 = 6 + \frac{2S}{3}$, which gives $S = 18$. Then use the individual equation to show $E_v = 6$ for all $v$, and add the expected first-roll value of $2$.
Worked Solution
How to Think About It: This is a Markov chain stopping problem. After each roll you earn that value and check: did I repeat? The key observation is that the die has only three distinct values, each equally likely at probability $\frac{1}{3}$. So from any given last roll $v$, you have a $\frac{1}{3}$ chance of stopping immediately (rolling $v$ again) and a $\frac{2}{3}$ chance of continuing with one of the other two values. That symmetry is what makes this tractable -- define conditional expectations $E_v$ and you will find they are all equal.
Quick Estimate: Think about it informally. The first roll is worth $2$ in expectation (the average of $1, 2, 3$). After that, conditional on not having stopped, you expect about one more roll before the game ends (stopping probability $\frac{1}{3}$ each step, so geometric with mean $3$ additional rolls -- but those include the stopping roll). Actually each "continuation phase" has a geometric number of rolls averaging $3$ rolls total before stopping: that is about $3 \times 2 = 6$ in additional expected coins after the first roll. So total is around $2 + 6 = 8$. That should be exactly right by symmetry.
Approach: Set up conditional expected future earnings $E_v$ given the last roll was $v$, then exploit symmetry to solve the linear system.
Formal Solution:
Let $E_v$ denote the expected additional coins earned given that the last roll was $v \in \{1, 2, 3\}$. On the next roll:
- With probability $\frac{1}{3}$, you roll $v$ again, earning $v$ and the game ends. Contribution: $\frac{1}{3} \cdot v$.
- With probability $\frac{1}{3}$ each, you roll one of the two other values $a$ or $b$, earning that value and continuing. Contribution: $\frac{1}{3}(a + E_a) + \frac{1}{3}(b + E_b)$.
So the equation for $E_v$ is:
$$E_v = \frac{v}{3} + \frac{1}{3}(a + E_a) + \frac{1}{3}(b + E_b)$$
Since $\{v, a, b\} = \{1, 2, 3\}$, we have $v + a + b = 6$, so:
$$E_v = \frac{v + a + b}{3} + \frac{E_a + E_b}{3} = 2 + \frac{E_a + E_b}{3}$$
This holds for all three values. Let $S = E_1 + E_2 + E_3$. Summing all three equations:
$$S = 3 \cdot 2 + \frac{1}{3}\bigl((E_2 + E_3) + (E_1 + E_3) + (E_1 + E_2)\bigr) = 6 + \frac{2S}{3}$$
$$S - \frac{2S}{3} = 6 \implies \frac{S}{3} = 6 \implies S = 18$$
So $E_1 + E_2 + E_3 = 18$. By symmetry of the equation $E_v = 2 + \frac{S - E_v}{3}$, we get:
$$E_v = 2 + \frac{18 - E_v}{3} \implies E_v = 2 + 6 - \frac{E_v}{3} \implies \frac{4E_v}{3} = 8 \implies E_v = 6$$
All three conditional continuation values are equal to $6$.
The first roll is unconditional. The expected value of the first roll is:
$$\frac{1+2+3+1+2+3}{6} = \frac{12}{6} = 2$$
The total expected payout is the first roll plus the expected future earnings given that roll:
$$E[\text{total}] = 2 + 6 = 8$$
Answer: $\boxed{8}$
Intuition
The symmetry here is doing a lot of heavy lifting. Even though the values $1$, $2$, and $3$ are different, the future continuation value $E_v = 6$ is identical for all three. That happens because the structure of the game -- stop with prob $\frac{1}{3}$, continue to a uniformly random other value -- is the same regardless of which value you landed on. Whenever you have this kind of symmetric Markov structure, summing the equations and solving for the total is the cleanest path. It avoids having to solve a $3 \times 3$ system directly.
In quant interviews, this problem is testing whether you recognize the Markov property and can set up conditional expectations properly. The common mistake is to try to compute the distribution of the stopping time $T$ directly, which gets complicated. The slicker move is to condition on the last roll and write recursive equations -- a technique that generalizes to much harder stopping time problems, including optimal stopping (when should you stop?) and ruin problems. The insight that continuation values can be equal despite different per-step payoffs also shows up in stochastic games and pricing models where symmetry arguments let you collapse a high-dimensional system.