Probability of getting 4 aces while drawing 6 cards.

Question

Drawing 6 cards from a standard deck of 52 cards, what is the probability of getting 4 aces?

This answer explains how to calculate the odds of a getting any 4-of-a-kind in a 5-card draw. I was hoping it would be obvious how to adjust that solution to solve my problem, but I'm not confident that I understand it well enough.

Given that there are 13 possible 4-of-a-kinds, and I am concerned with only 1 of them, I would assume the odds are 13 times less likely than simply drawing a 4-of-a-kind, but that's where I run out of steam.

Answer 1

$$\frac{{4\choose4}{48\choose 2}}{52\choose6}$$

Note this is $\frac1{13}$ of the (fairly easy) adaptation from your link to $6$ choices.

Answer 2

There are 52 cards, 4 of them aces. The probability the first card drawn is an ace is 4/52= 1/13. There are then 51 cards, 3 of them aces. The probability the second card drawn is an ace is 3/51= 1/17. There are then 50 cards, 2 of them aces. The probability the third card drawn is an ace is 2/50= 1/25. There are then 49 cards, 1 of them an ace. The probability the fourth card drawn is 1/49. After that of course the last two cards must be non-aces. So the probability of "four aces followed by 2 non-aces, in that order" is (1/13)(1/17)(1/25)(1/49). There are a total of 6!/(4!(2!)= 15 different orders: AAAANN, AAANAN, AANAAN, ANAAAN, etc. I will leave to you to show that the probability of "four aces, two non-aces" in any of those orders is the same, (1/13)(1/17)(1/25)(1/49), so the probability of "four aces, two non-aces" is 15(1/13)(1/17)(1/25)(1/49).