# Probability: 2 dice rolled, two players

Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?

I am unable to formulate a repeating sequence or an appropriate formula. How can I solve this question?

• do you know what a markov chain is? Commented Jun 17, 2017 at 5:26

The probability of throwing a 12 equals $\frac{1}{36}$, the probability of throwing a 7 equals $\frac{6}{36}$ and the probability of throwing anything else equals $\frac{29}{36}$. In every turn, we must take into account these three possibilities. Let us denote $A$ the event in which a 12 is thrown first and $B$ the event in which two consecutive 7s are thrown first. We then get:
$$P[A] = \frac{1}{36} 1 + \frac{6}{36} \bigg(\frac{1}{36} 1 + \frac{6}{36} 0 + \frac{29}{36} P[A]\bigg) + \frac{29}{36} P[A]$$
$$\iff P[A] = \frac{6}{216} + \frac{1}{216} + \frac{29}{216} P[A] + \frac{174}{216} P[A] \iff \frac{13}{216} P[A] = \frac{7}{216} \iff P[A] = \frac{7}{13}$$
We can now either use $P[B] = 1 - P[A] = \frac{6}{13}$, or use a similar approach for the event B:
$$P[B] = \frac{1}{36} 0 + \frac{6}{36} \bigg(\frac{1}{36} 0 + \frac{6}{36} 1 + \frac{29}{36} P[B]\bigg) + \frac{29}{36} P[B]$$
$$\iff P[B] = \frac{6}{216} + \frac{29}{216} P[B] + \frac{174}{216} P[B] \iff \frac{13}{216} P[B] = \frac{6}{216} \iff P[B] = \frac{6}{13}$$