There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces (AS, AC), (AH, AD) out of 4 cards. (AS, AC), (AH, AD), (AS, AH) (AS, AD), (AC,AD) (AC, AH) So the answer is 6/1326 = 1/221. The probability of this happening twice in a row is (1/221)2 = 1 in 48,841.
How ever this would depend on where you are seated at the table and how many players are playing.
If you are in the 2nd seat there are 51*50/2 = 1/1275...but that is assuming the 1st seat did not get an ace.
So it is a complicated problem.
I do not think you would paly the A-A any differently...same odds on the 2nd hand
