Edit: I'm unclear what you mean. Maybe you're asking "what are the chances of rolling a 1 and a 16 (as components of your advantaged 16) twice in a row?" Well, the first one doesn't count, for reasons I describe further in. However, the second one to match it, your two dice have 400 possibilities, and two meet your criteria: 1/16, or 16/1. So the simple answer is 2 in 400, or 1 in 200.
That applies in your case, but doesn't apply globally, as user2357112 discuses in an answer.
What are the chances of rolling 16 twice? 1 in 166.49.
160,000 total possibilities; 961 possibilities result in advantage rolls being both 16.
But hold on. Why is 16 special? 16 is only special because the first roll was 16. Hence we have a fallacy. You only consider the second one special because it matched the first. But this condition could have happened with any first number!!!
Get it? The chance of the first number being a number is 100%. So the 16 is not special at all; it had to be some number. The only chance here is the chance of the second number matching the first. That depends on the second number; for instance in advantage, you only have a 1 in 400 chance of rolling an advantaged 1, but a 39 in 400 chance of rolling an advantaged 20. In this case, you have a 31 in 400 chance (1 in 12.9) of rolling an advantaged 16.
Not terribly remarkable, and even less remarkable for an advantaged 20 (almost 1 in 10). Higher numbers are more probable; that's what advantage is all about.
However if we let go of the 16, since it isn't special, let's look at your core question. What are the chances of rolling two consecutive same-numbers? Without advantage, that's easy: 1 in 20. (as said, the first number is anything; the second simply matches it.) With advantage, it gets more complicated, but also more likely, since high numbers are more probable than low ones.
The answer is 1 in 15.00938.
160,000 total possibilities; 10660 result in advantaged rolls being equal.