Comparing the probabilities of rolling a $12$ or two consecutive $7$'s first with a pair of dice A pair of dice is rolled repeatedly which event is more likely to occur first? 
Event $A$ both dice shows 6's. 
Event $B$ two consecutive rolls give a sum of 7 each.
By not solving I think it's A since it's more likely to happen since your just going to roll it once rather then event B.
Event A So I know that probability of getting a 6 in a dice is $\frac{1}{6}$ since it's two so it will be $\frac{1}{36}$  
Event B sum of 7 each roll of two dice so $\frac{6}{36}$  or $\frac{1}{6}$ but it's twice thrown so $\frac{1}{36}$
WAIT it's EQUAL?! I think I'm wrong here somewhere. 
 A: There are $4$ states this process can be in: 
S: start, or you have just rolled the dice and gotten some sum other than $7$ or $12$;
V: You have just rolled a first sum of $7$;A: You have just rolled a sum of $12$, so event $A$ has occurred;
B: You have just rolled a second consecutive sum of $7$, so event $B$ has occurred.
In completing the experiment, consider the final time you are at S.  From here you could go S, A; or S, V, A; or S, V, B.
In general, if you are at S (not necessarily the final time), $P(S,A)=\frac{1}{36}$, $P(S,V,A)=\frac{1}{6}\cdot\frac{1}{36}$, $P(S,V,B)=\frac16 \cdot \frac16$.
Thus from the final time you are at $S$, the probability of getting to A is $\frac{P(S,A)+P(S,V,A)}{P(S,A)+P(S,V,A)+P(S,V,B)}$.  This works out to $\frac{7}{13}$.
So the probability of event $A$ occurring first is $\frac{7}{13}$; the probability of event $B$ occurring first is $\frac{6}{13}$.
A: Event A can happen on any single roll.  Event B needs to happen on two consecutive rolls.  If a sum of 7 occurs, event A can still happen on the subsequent roll.
$P$, the probability that event A occurs before event B is not determined by simply comparing probabilities that they happen on the next few rolls.
The method to use is to use the Law of Total Probability and partition on the events of the next roll, $X$: a double 6, $Y$: a sum of $7$, and $Z$: the remainder.  
Let $P_X, P_Y, P_Z$ be the conditional probabilities of event A occurring before event B given a result of $X,Y,Z$ respectively.
As $X$ means event A has occurred, $P_X =1$.  If $Z$ occurs, events A and B become neither more or less probable, so $P_Z=P$.  
$$\begin{align}
P & = \Pr(X)P_X+\Pr(Y)P_Y+\Pr(Z)P_Z
\\[1ex] & = \tfrac 1 {36}+\tfrac 1 6 P_Y + \tfrac{29}{36}P \tag{1}
\end{align}$$
If $Y$ happens, then similarly partitioning on the subsequent roll gives us:
$$\begin{align}
P_Y & = \Pr(X)P_{YX}+\Pr(Y)P_{YY}+\Pr(Z)P_{YZ}
\\[1ex] & =\tfrac 1 {36}+\tfrac 1{6}\cdot 0+\tfrac {29}{36}P \tag{2}
\end{align}$$
Thus we have two simultaneous equations.  Solve them.
A: The answer can be much simplified by instead computing the probability that two consecutive $7s$ are rolled, and (if needed) find the complement to pinpoint which event occurs first.
It is enough to consider two states, $A$ (starting) and $C$ (one $7$ obtained), since the route to a sum of $12$ is in effect barred for our computation.
With eventual probabilities $a$ and $c$ of getting two consecutive$7's$ from states $A$ and $C$, we have
$a =\frac16c +\frac{29}{36}a$
$c=\frac{29}{36}a + \frac16\times 1$ (reach event $B$ with certainty)
This yields $a= \frac6{13}$, thus two consecutive $7's$ (event $B$) will occur first,
