Probability : Roll of Die $\textsf{A}$ and $\textsf{B}$ are playing a game with $2$ standard dice.

  
*
  
*Both the dice are rolled together and the total is counted.
  
*$\textsf{A}$ says that a total of $2$ will be rolled first.
  
*$\textsf{B}$, whereas, says that two Consecutive totals of $7$′s will be rolled first.
  
*They keep rolling the dice till one of them wins !.
  

What is the probability that $\textsf{A}$ wins the game ?.
For a total of $2$, $\{(1,1)\}$ and for a total of $7$, $\{(1,6),(6,1),(2,4),(4,2),(3,4),(4,3)\}$ are the required scenarios. I don't understand how we need to incorporate the probabilities of $\textsf{A}$ winning, i.e., $1/36$ and $\textsf{B}$ winning, i.e. $6/36$ into a game of infinite rounds, i.e. until $\textsf{A}$ wins. –
 A: Let $T_n2$ be the event consisting of throwing a total of $2$ on the $n$th throw of a pair of dice, $T_n7$ be throwing a total of $7$, and $T_n2,7$ be throwing a total of either $2$ or $7$.
By counting cases it is readily shown that: $P(T_n2)=1/36$, $P(T_n7)=1/6$, $P(T_n2,7)=7/36$.
If a pair of throws gives any total other than $2$ or $7$, the status of the game at the next pair of throws will be exactly as at the start of the game.  So we can focus on conditional probabilities, conditional on $T_n2,7$.  We have:
$P(T_n2 | T_n2,7) = (1/36) / (7/36) = 1/7$
$P(T_n7 | T_n2,7) = (1/6) / (7/36) = 6/7$
In the first case A has won. In the second we need to consider what happens at throw $n+1$.  We have:
$P(T_n7 \ \&\ T_{n+1}2 | T_n2,7) = (6/7)(1/36) = 1/42$
$P(T_n7 \ \&\ T_{n+1}7 | T_n2,7) = (6/7)(1/6) = 1/7$
If throw $n+1$ yields any total other than $2$ or $7$ the game reverts to its initial status, so it suffices to consider the relative probabilities of A and B winning during throws $n$ and $n+1$.  We have (conditional on $T_n2,7$, and noting that B cannot win at $n$):
P(A wins) = P(A wins at $t$) + P(A wins at $t+1$) = 1/7 + 1/42 = 1/6
P(B wins) = P(B wins at $t+1$) = 1/7
If we now remove the conditionality, the relative probabilities will not change. Therefore:
P(A wins)
 $= \frac{1/6}{1/6 + 1/7} = \frac{7}{7+6}=\frac{7}{13}$
A: To me this sounds like a good application of an absorbing Markov chain.  There are four states:


*

*“Neutral” where neither A nor B is ahead.

*“B up” where B can win on the next roll.

*“A wins”

*“B wins”


The first two states are transitional and the last two terminal.  The transition matrix between these four states is
$$
    P = \begin{bmatrix} 29/36 & 1/6 & 1/36 & 0 \\
                        29/36 & 0   & 1/36 & 1/6 \\
                            0 & 0   &    1 & 0 \\
                            0 & 0   &    0 & 1 \end{bmatrix}
    = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix}
$$
Here $p_{ij}$ is the probability of moving from state $i$ to state $j$. The $2\times 2$ matrices $Q$ and $R$ are the top left and top right blocks of $P$.  The fundamental matrix of this Markov chain is
$$
    N = (I-Q)^{-1} = \begin{bmatrix} 216/13 & 36/13 \\ 174/13 & 42/13 \end{bmatrix}
$$
The probability of absorption matrix is
$$
    NR = \begin{bmatrix} 7/13 & 6/13 \\ 6/13 & 7/13 \end{bmatrix}
$$
This means that from neutral (state 1), A has a $7/13$ chance of winning (state 3).  However, once a 7 is rolled (state 2), B has a $7/13$ chance of winning (state 4).   
