0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ The answer given by user247327 doesn't seem to match this one. Also, this isn't 1/13th of the answer given in the example I linked. By my calculation, this is 3/13 of that answer. Can you explain how you arrive at this answer? $\endgroup$ – automagically Oct 1 '18 at 14:29
  • $\begingroup$ The two answers match. I checked it. I adjusted the solution for the fact that $6$ choices are made rather than $5$. $\endgroup$ – Chris Custer Oct 1 '18 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.