Rolling a die until two rolls sum to seven Here's the question:

You have a standard six-sided die and you roll it repeatedly, writing
  down the numbers that come up, and you win when two of your rolled
  numbers add up to $7$. (You will almost surely win.) Necessarily, one of the
  winning summands is the number rolled on the winning turn. A typical
  game could go like this: $1, 1, 4, 5, 3$; you win on the 5th turn
  because $3 + 4 = 7$. How many turns do you expect to play?

Here's what I've tried: 
We seek $E(N)$ where $N$ is a random variable counting the number of turns it takes to win. Then $N \ge 2$, and $$E(N) = \sum_{n=2}^\infty n P(N=n) = \sum_{n=1}^\infty P(N > n).$$
I want to find either $P(N=n)$, the probability that I win on the $n$th turn, or $P(N > n)$, the probability that after $n$ turns I still haven't won. Note that $P(N = 1) = 0$. Let $X_k$ be the number rolled on the $k$th turn. Then $$P(N = 2) = P(X_1 + X_2 = 7) = \sum_{x=1}^6 P(X_1 = x)P(X_2 = 7-x) = 6\cdot \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{6}.$$
So far so good. To compute $P(N > 3)$ I let $A_{i, j} = \{\omega \in \{1, \dotsc, 6\}^3 : w_i + w_j = 7\}$ and used the inclusion-exclusion principle and symmetry to find $$|A_{1,2} \cup A_{2,3} \cup A_{1,3}| = 3|A_{1,2}| - 3|A_{1,2}\cap A_{1,3}| = 90$$
so $P(N > 3) = \frac{126}{216} = \frac{7}{12}$. This is the probability that no two of three dice sum to seven. Similarly, I found $P(N > 4)$ to be $\frac{77}{216}$. 
I don't see how to generalize the above. I also thought that $$P(N > n) = P(X_i + X_j \ne 7 \text{ for all }1 \le i\ne j \le n) = (1 - P(X_i + X_j = 7))^{\binom{n}{2}} = \left(\frac{5}{6}\right)^{n(n-1)/2}$$ 
but that's false because the events are not independent. 
I also tried $$P(N = n) = P(X_n = 7 - X_k \text{ for some } 1 \le k < n \text{ and }N \ne n - 1)$$
where that last clause is shorthand for "and the previous rolls did not secure your victory". 
This yields the recursion $p_n = (1-(5/6)^{n-1})(1-p_{n-1})$, $p_1 = 0$, which didn't agree with my previously computed probabilities. (Perhaps I made an error.)

The answer is: 4.
Reasoning: A six sided die presumably has numbers in $\{1,...,6\}$.  Every one of those numbers can be used to add to $7$.  So take your first roll.  It doesn't matter what the number is.  Now look at your die. There is one number that can add to seven with your first number, and you can roll it with probability $1/6$.  Thus, the probability of winning on each successive turn is $1/6$.  The probability of winning on a successive turn is given by the binomial distribution where $p = 1/6$.
The interesting part about this problem is how it is phrased: it is asking for how many turns you expect to need until you win.  To me, that means the number of rolls required to find a probability of you winning $\ge 1/2$.
The expected value of the binomial distribution is $np$.  So, we want $np \ge 1/2$.  Since $p = 1/6$, $n$ must be at least $3$.  Therefore, you need your first turn ($1$) plus your $3$ "expected" turns to win: $1 + 3 = 4$.
