Probability of rolling a sum of 4 and a sum of 5 before 2 sums of 7's. I can't seem to grasp my mind around this question.
My attempt: P(roll 7) = 6/36
            P(roll 4) = 3/36
            P(roll 5) = 4/36
There are three combinations:
1) 4,5,7
2) 4,5
3) 7,4,5
I am stuck here.
 A: The first time you get a number that's either a $4$, a $5$, or a $7$, what's the probability that it's a $4$ or a $5$?  In other words, you want the conditional probability that what you get is either a $4$ or a $5$, given that it's either a $4$, a $5$, or a $7$.
A: Amended to inset "$|\text{no } 7 \text{s}$" and recalculate 
There is probably a simpler way than this, but you might look at the probability of $k$ failures to roll $7$ before the second success which is $(k+1)(6/36)^2(30/36)^k$ 
$\Pr( \text{roll } 4 \text{ and } 5 \text{ in } k \text{ rolls}|\text{no } 7 \text{s})  =  1- \Pr( \text{don't roll } 4 \text{ or don't roll  } 5 \text{ in } k \text{ rolls}|\text{no } 7 \text{s}) $  
$= 1 -  \Pr( \text{don't roll } 4  \text{ in } k \text{ rolls}|\text{no } 7 \text{s}) -  \Pr( \text{don't roll } 5  \text{ in } k \text{ rolls}|\text{no } 7 \text{s})  $ 
$+ \Pr( \text{don't roll } 4 \text{ and don't roll  } 5 \text{ in } k \text{ rolls}|\text{no } 7 \text{s}) $ 
$= 1 - (27/30)^k - (26/30)^k + (23/30)^k.$
So your answer is 
$$\sum_{k=0}^\infty (k+1)(6/36)^2(30/36)^k(1 - (27/30)^k - (26/30)^k + (23/30)^k)$$
You could work this out.  I make it $15536/38025 \approx 0.4085733$.
If you do it following Michael Hardy's suggestion of ignoring all rolls except 4,5 and 7 this becomes 
$$\sum_{k=0}^\infty (k+1)(6/13)^2(7/13)^k(1 - (3/7)^k - (4/7)^k + (0/7)^k)$$ where $(0/7)^k =0$ unless $k=0$ in which case $(0/7)^0 =1$.  You then get the same result.   
A: It seems that you would need to use the negative binomial distribution here.
