let's do the math !
number of 2 cards games : C(52,2)=1326
number of 2 aces games : C(4,2)=6
probability of having 2 aces :6/1326=0.45% that is 221:1
basically if youwant to go quickly you can multiply the above result by 6, you'll have a good approximation of what you want
but you count things twice !
multipling by 6 gives the same result than [C(50,2) x C(4,2) ] / C(52,4) = 2.71% (36.8333 : 1)
and explained in a non-mathematical manner it's the number of 2 aces among 2 cards (C(4,2)=6) multiplied by the number of any 2 cards among the 50 that are not your 2 aces (C(50,2)=1225) divided by the number of 4 cards games (C(52,4)=270 725)
ok, you understand easily that this number is too high because when you count C(4,2) x C(50,2) there are times when your aces are counted twice....but it's a good approximation.
Now let's try to find the correct answer :
to have AAxy where xy can be anything (including a third and fourth ace) you need to calculate :
the number of games with 2 aces among 4 cards (C(4,2)=6) and multiply it with the number of 2 cards games among 48 cards (that is a deck minus the 4 aces) C(48,2)=1128.
Ok now you need to add the number of hands with 3 aces (C(4,3)) and multiply it by 48 or if you prefer C(48,1)
And finally you add the number of 4 aces games, C(4,4) x C(48,0)
You divide that number by the number of 4 cards games that is still C(52,4)
And you find :
(C(4,2)xC(48,2) + C(4,3) x C(48,1) + C(4,4) x C(48,0))/C(52,4)
(6x1128 + 4x48 + 1x1) / 270 725 = 2.57% or 38.9 : 1