The probability of this occurring in which your other two cards are any two 10-point cards is 4*COMBIN(6,4)*COMBIN(6*16,2)*(4/6)*(3/5)*(1/2)/combin(312,6) 1 in 22,307,231. However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. I would have to do a computer simulation to consider all the other combinations. However to make a rough guess I’d say the 7 million looks about right.