Dice Race: 12 vs Two Consecutive 7s
Two fair six-sided dice are rolled repeatedly. Player A wins when a sum of 12 appears before Player B gets two consecutive sums of 7.
1. Reduce each roll to 3 outcomes
6 combinations out of 36
Only (6,6)
Breaks a possible streak of 7s
Final answer
2. The key idea: only remember whether the previous roll was 7
We do not need the full history. The only relevant information is whether the immediately previous roll was a 7.
S₀
S₁
Roll 12
Roll 7 from S₁
- From S₀, a 7 moves us to S₁.
- From S₀, anything except 7 or 12 keeps us in S₀.
- From S₁, another 7 makes B win.
- From S₁, 12 makes A win.
- From S₁, any other sum resets us to S₀.
3. Set up equations
Let a be the probability that A eventually wins starting from S₀.
Let b be the probability that A eventually wins starting from S₁.
Why? From S₀: roll 12 → A wins; roll something else → stay in S₀; roll 7 → move to S₁.
From S₁, rolling a 7 makes B win, so that branch contributes 0 to A's winning probability.
4. Solve
5. Interview shortcut
When you see a repeated stochastic process, ask: “What is the minimum information from the past that affects the future?”
Here the answer is only one bit of memory: Was the previous roll a 7? That gives exactly two transient states, so the infinite game becomes two simple equations.
6. Monte Carlo check
Run a simulation and compare it with the exact probability 7/13 ≈ 53.846%.