Probability of rolling two dice multiple times

Question

For this question, I know that the sample space is $6^3$, and that (1,6), (2,5), (3,4) are the possibilities for rolling a sum of seven. I'm not sure how to continue after that. Here is what I have so far. Can anyone please help me out?

Suppose we repeatedly roll two fair six-sided dice, considering the sum of the two values showing each time. What is the probability that the first time the sum is exactly 7 is on the third roll?

$S = 6^3 = 216$

$(1, 6) (2, 5) (3,4)$ are the possibilities for the sum eqaul to 7

Answer 1

Best you consider the two dice indepenedly.

The ways to roll a 7 the become $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

There are 6 ways to roll a 7.

There are 36 combinations of outcomes from 2 dice.

The chance of getting a 7 on any roll then is $\frac {6}{36} = \frac {1}{6}.$ The chance of not getting a 7 on any roll then is $1- \frac {1}{6} = \frac {5}{6}.$

Now you need to calculate the likelihood of not getting a 6 on the first roll, not getting a 6 on the second roll and getting a 6 on the 3rd roll.

Answer 2

There are $6^2=36$ possible outcomes for each roll of both dice. There are $6$ combinations that result in a sum of $7$, those you mentioned, and those also with digits reversed.

So the success probability is $\frac{6}{36}=\frac16$ and the failure probability is $\frac{30}{36}=\frac56$. You are saying there are two failures and one success, so that probability is: $$\bigg(\frac56\bigg)^2\cdot\frac16=\frac{25}{216}$$