The Setup

Two players engage in a dice game:

  • Player A wins if a roll of 12 appears before Player B achieves two consecutive rolls summing to 7.
  • Player B wins if at any point there are two consecutive rolls each summing to 7, occurring before the first 12.

The game continues with repeated rolls of two fair six-sided dice until one of these conditions is met.

Modeling the Game

To analyze this game, we model it as a Markov chain with the following states:

  • State S₀: No 7 was rolled on the previous turn (initial state).
  • State S₁: A 7 was rolled on the previous turn.
  • State A: Player A wins (a 12 is rolled).
  • State B: Player B wins (two consecutive 7s are rolled).

Transition Probabilities

The probabilities of rolling certain sums with two dice are:

  • P(12): Only one combination (6+6) yields 12, so \( \frac{1}{36} \).
  • P(7): Six combinations yield 7, so \( \frac{6}{36} = \frac{1}{6} \).
  • P(other): The remaining 29 combinations, so \( \frac{29}{36} \).

Based on these, the transitions are:

  • From S₀:
    • To A: \( \frac{1}{36} \)
    • To S₁: \( \frac{6}{36} = \frac{1}{6} \)
    • Stay in S₀: \( \frac{29}{36} \)
  • From S₁:
    • To B: \( \frac{6}{36} = \frac{1}{6} \)
    • To A: \( \frac{1}{36} \)
    • To S₀: \( \frac{29}{36} \)

Solving the System

Let:

  • \( p_0 \): Probability that Player A wins starting from S₀.
  • \( p_1 \): Probability that Player A wins starting from S₁.

We can set up the following equations:

  1. \( p_0 = \frac{1}{36} + \frac{6}{36} p_1 + \frac{29}{36} p_0 \)
  2. \( p_1 = \frac{1}{36} + \frac{29}{36} p_0 \)

Solving Equation 1:

\[ p_0 - \frac{29}{36} p_0 = \frac{1}{36} + \frac{6}{36} p_1
\Rightarrow \frac{7}{36} p_0 = \frac{1}{36} + \frac{6}{36} p_1
\Rightarrow 7 p_0 = 1 + 6 p_1 \quad \text{(Equation 3)} \]

Substitute Equation 2 into Equation 3:

\[ 7 p_0 = 1 + 6 \left( \frac{1}{36} + \frac{29}{36} p_0 \right )
\Rightarrow 7 p_0 = 1 + \frac{6}{36} + \frac{174}{36} p_0
\Rightarrow 7 p_0 = \frac{42}{36} + \frac{174}{36} p_0
\Rightarrow 7 p_0 - \frac{174}{36} p_0 = \frac{42}{36}
\Rightarrow \left(7 - \frac{174}{36}\right) p_0 = \frac{42}{36}
\Rightarrow \left(\frac{252 - 174}{36}\right) p_0 = \frac{42}{36}
\Rightarrow \frac{78}{36} p_0 = \frac{42}{36}
\Rightarrow p_0 = \frac{42}{78} = \frac{7}{13} \]

Final Answer

\[ \boxed{\frac{7}{13}} \]

Therefore, the probability that Player A wins (i.e., a 12 is rolled before two consecutive 7s) is 7/13.

Reference