A Dice Duel

Two players roll a pair of fair dice repeatedly and record each sum. Player A wins as soon as a sum of $12$ appears. Player B wins as soon as two consecutive sums of $7$ appear. They roll until one of them wins. What is the probability that A wins?

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#Two live states

Only one thing about the past matters for the next roll, whether the previous sum was a $7$ . That gives two transient states, $S_0$ fresh (the last sum was not a $7$ ) and $S_1$ primed (the last sum was a $7$ ), plus an absorbing state $A$ reached by a $12$ and an absorbing state $B$ reached by a $7$ right after a $7$ . Per roll, $\PP(12) = \tfrac{1}{36}$ , $\PP(7) = \tfrac{6}{36}$ , and $\PP(\text{neither}) = \tfrac{29}{36}$ .

From either live state a 12 wins for A and a non-7 non-12 roll resets to fresh. A first 7 only primes the chain; a second 7 in a row is what wins for B.

#The transition matrix

Ordering the states $S_0, S_1, A, B$ , one roll moves the chain by

P = \begin{pmatrix} \tfrac{29}{36} & \tfrac{6}{36} & \tfrac{1}{36} & 0\\[2pt] \tfrac{29}{36} & 0 & \tfrac{1}{36} & \tfrac{6}{36}\\[2pt] 0 & 0 & 1 & 0\\[2pt] 0 & 0 & 0 & 1 \end{pmatrix}. \tag{1}

From $S_0$ a non- $7$ non- $12$ stays fresh, a $7$ primes the chain, and a $12$ wins for A. From $S_1$ the same $12$ wins for A, another $7$ wins for B, and anything else snaps the streak back to $S_0$ .

#Absorption probabilities

Let $a_i = \PP(\text{A wins} \mid \text{state } i)$ , with $a_A = 1$ and $a_B = 0$ . Reading the two transient rows,

a_0 = \tfrac{29}{36}a_0 + \tfrac{6}{36}a_1 + \tfrac{1}{36}, \qquad a_1 = \tfrac{29}{36}a_0 + \tfrac{1}{36}. \tag{2}

The first rearranges to $7a_0 = 6a_1 + 1$ . Substituting the second, $36 a_1 = 29 a_0 + 1$ , and clearing denominators gives $78 a_0 = 42$ , so

\PP(\text{A wins}) = a_0 = \frac{7}{13} \approx 0.538. \tag{3}

A 12 is rarer on any single roll than a 7, yet A is the favorite, because B needs two 7s in a row and the $\tfrac{29}{36}$ reset keeps snapping that streak before it completes.