Probability of getting two consecutive 7s without getting a 6 when two dice are rolled Two dice are rolled at a time, for many time until either A or B wins. A wins if we get two consecutive 7s and B wins if we get one 6 at any time.
what is the probability of A winning the game??
 A: Let $p$ be the probability that $A$ wins.  Let $h$ be the probability that $A$ wins given that neither $A$ nor $B$ has yet won, and a $7$ has just been thrown, so that $A$ is "hot."
We break the calculation into cases, depending on what happens on the first throw. There are two ways that $A$ can (ultimately) win. Either (i) the first throw is a $7$ (probability $1/6$), making $A$ hot, in which case the probability of a win is $h$, or (ii) the first throw is something other than $7$ or $6$ (probability $25/36)$), in which case for all practical purposes the game starts again. In that case, the probability $A$ ultimately wins is $p$. We thus obtain the equation
$$p=\frac{1}{6}h +\frac{25}{36}p.\tag{$1$}$$
If $A$ is hot, then $A$ wins if (i) the next throw is $7$, or (ii) the next throw is something other than a $6$ or $7$, but $A$ ultimately wins. In case (ii), the probability of an ultimate win for $A$ is $p$. Thus
$$h=\frac{1}{6}+\frac{25}{36}p.\tag{$2$}$$
We have two linear equations in two unknowns, and it is not hard to solve for $p$.
From $(1)$ we obtain $11p/6$. Substituting in $(2)$, we find that $p=6/41$.
A: Let $p_A$ and $p_B$ the winning probabilities for $A$ and $B$, and $p_6$ and $p_7$ the probabilities to roll 6 and 7.
Now, by regarding the different possibilities for the first roll (and if one starts with 7, also the second roll), we find:
$$p_A=p_6 \cdot 0 + p_7 (p_6 \cdot 0 +p_7+(1-p_6-p_7)p_A) + (1-p_6-p_7)p_A.$$
This is a linear equation for $p_A$.
We have $p_6=\frac5{36}$, $p_7=\frac{6}{36}$, and therefore:
$$p_A=\frac{p_7^2}{p_7p_6+p_7^2+p_6} =\frac{6}{41}.$$
