Two players with aces

Any two players: C(8,2)*C(4,2)² / C(52,4) ≈ 1/269

Any villain given that Hero has them: 7*C(4,2) / C(48,2) = 7/188

(My combinations aren't counting choices of cards for players, but choices of places for the aces.)

I can’t believe the formula is so simple. Thanks.

Sure thing, and now I'll show a way to do it when you don't see my shortcut.

To preface, consider the probability that Hero gets AA in Hold'em: (4/52)(3/51)
That's this general law in action: P(A and B) = P(A)*P(B|A)
We took P(1st card is ace) and multiplied by the conditional probability of an ace given that one was removed.

We can apply that law here. Take two arbitrary players P1 and P2.
P(P1 & P2 have AA) = P(P1 has exactly two aces) * P(P2 also has AA given that P1 does)

Next, for P(any two players have AA), we get to just multiply by 8C2. Those are all the possible combos of two players, each combo being equally probable and mutually exclusive. (By contrast, the probability of one player having AA isn't 8x that of an individual, because two individuals aren't mutually exclusive to have AA. But if Hero has already looked down at AA, the villains are mutually exclusive because only two aces remain.)

All told: (4C2)(48C2)/(52C4) * (46C2)/(48C4) * 8C2 ≈ .0037233355 (same answer as before)

This method is more complicated but also more intuitive.