Let's see, considering you hold a pair (let us use 77 as an example) and your opponent has another pair (let us assume 44), we still have 48 cards left. Considering that we have 17296 possible flops. To a set- over-set to occur we need one of the remaining 7s and one of the remaning 4s, so we have about 176 ways of a flop like this ocurring (2*2*44), considering that our probability of success is ~1.01^(-2) or

**~1.02%**, let's call it event A.

After that the probability of having a quad on the turn is around

**~2.22%**, lets call It event B.

Considering that the probability of event A and B happening is around

**0.0226%** (probability of A*probability of B)

That was a quick calculation I made here by hand and calculator, I might be wrong.

Now about the probability of that happening twice in 80 hands, I won't make the calculation, but the probability of that happening is approximately zero.

By the way, that's the calculation on a heads-up situation, if we consider that more pairs see the flop the odds can change.

EDIT

When I say "after that the probability of having a quad on the turn is around

**2.22%**" I'm talking about the probability of

**YOUR OPPONENT** turning a quad, the probability of

**YOU or YOUR OPPONENT** turning quads is around

**4.44%** and, considering that, the probability of

**EITHER YOU OR YOUR OPPONENT** having set-over-set and then quads is about

**0.0452%**.

Cheers