Here’s another way of looking at it:

Suppose you roll the two dice, A and B, N times.

The probability of getting a 1 on A and not on B is (1/6)(5/6) → 5/36

The probability of getting a 1 on B and not on A is (5/6)(1/6) → 5/36

So the number of *single *1s you expect to get is N*10/36

The probability of getting a 1 on both A and B is (1/6)(1/6) → 1/36

So the number of *pairs *of 1s you expect to get is just N*1/36. But for each pair you see *two *1s.

This means that the number of 1s you would expect to see in N throws of two dice is:

(10/36 + 2*1/36)*N→ 12N/36 or N/3

The expectation on a single throw of the pair is therefore just 1/3.