Card probability - AAKK out of a deck

Factual Questions
1

If dealt exactly 4 cards out of a deck of 52, what are the odds of winding up with exactly two Aces and exactly two Kings? (Suit is irrelevant.)

Here’s how I calculated it, just curious to see if others think these are correct or in error.

My formula is total probability P = P1*(P2aP3aP4a + P2bP3bP4b)

For the first card dealt, any Ace or King will suffice, which means eight out of fifty two.

P1 = 8 / 52

Now there are seven cards remaining that we can use. Three of them are the same as the first card, and four are different.

Case a: the second card pulled is the same as the first.

The odds of pulling the same card (Ace or King) as the first are 3 / 51.

P2a = 3 / 51

Once we have pulled two straight Kings, for example, we need to pull two of the remaining four Aces in the deck (vice versa if Aces were the first cards pulled.)

P3a = 4 / 50
P4a = 3 / 49

Case b: the second card pulled is not the same as the first (i.e., first two cards are AK)

P2b = 4 / 51

At this point there are three Aces and three Kings left in the deck, any of which can be used as the third card dealt.

P3b = 6 / 50

Now we either have AAK or KKA, and in either case, only three remaining cards in the deck can be used to form the target hand of AAKK.

P4b = 3 / 49

So by substitution we get:



      8         3      4      3          4      6       3     
P =  ---- * [( ---- * ---- * ---- )  + (---- * ----  * ---- )]
      52        51     50     49         51     50      49

Reducing that, we get

P = (8 / 52) * (3 / 49) * [ ((3 / 51) * (4 / 50)) + ((4 / 51) * (6 / 50)) ]

P = (8 / 52) * (3 / 49) * [ 12 / (5150) + 24 / (5150)]

P = (8 * 3 * 36) / ( 52 * 51 * 50 * 49) ~= 0.000133

or about 1 in 7520.

Did I do this right, math dopers? :slight_smile:

P.S. This is not a HW question, and if any of you play poker, you will recognize AAKK as the best starting hand in Omaha High.

2

I used ((4 choose 2)^2) / (52 choose 4). Google calculator returns 0.000132976267

3

Bit wise is correct in his calculation. but why is AAKK a better starting hand than AAAA? I could see that two AK’s of the same suit might be better than a random AAKK as it leaves open the possibility of a royal flush. But wouldn’t AAAA be a better start than AC AD KH KS?

BTW I don’t play Ohama High.

4

Now I’m confused. neuroman’s answer seems more technically correct than bitman’s, accounting for the reduction in size of the overall pack and the chance of different methods to arrive at the same hand. Yet **bitman’s ** numerical value is the same as neuroman’s. Have I been duped by years of statistics teachers into performing calculations in more complex ways then necessary? :eek:

5

I make it :

270,725 possible ways of getting 4 cards assuming order is irrelevent.

144 of them are AAKK
144/ 270,725 = 0.000531905

6

In Omaha High (and High/Low) you use exactly two of your starting four cards to combine with exactly three of the “board” cards to make a hand. Thus with AAAA, you can’t pair anything on the board, reducing your odds dramatically. In fact, in Omaha it’s generally considered a good idea to fold all “quad” hands immediately.

Pullet, I think maybe I just described the looong way what I could have done much more easily if I was more familiar with probability. But at least it looks like I was correct. :smiley:

7

neuroman’s answer: (8 * 3 * 36) / ( 52 * 51 * 50 * 49)
my answer: ((4 choose 2)^2) / (52 choose 4)

Work out the arithmetic, and you will see that these are the same.

How I arrived at my answer:

  1. There are (52 choose 4) four-card hands.
  2. There are (4 choose 2) pairs of aces and (4 choose 2) pairs of kings.
  3. There are (4 choose 2)^2 hands with two aces and two kings.
  4. The probability of such a hand is ((4 choose 2)^2) / (52 choose 4).
8

In Omaha, you play exactly two of your four hole cards. If you were dealt AAAA, you would have a pair of Aces. AAKK is a stronger starting hand because you can improve with any of the remaining Aces and Kings, as well as having the possibility of making the best possible straight (AKQJ10).

9

You are overcounting. (ace of spades, ace of hearts) and (ace of hearts, ace of spades) should be counted as the same.

10

The first number is right and it is 52 choose 4 as BitWise suggests. The second number is not correct. There are only 36 ways of getting AAKK. There are six AA pairs: (Clubs, Diamonds), CH, CS, DH, DS, HS, coupled with 6 ways to get two kings for 36 total. And 36 is (4 choose 2) squared again agreeing with BitWise. This is also the same answer that the OP gives since 52 choose 4 is (52515049)/(43*2) (see his last line). I did not mean to imply the OP was wrong, just that Bitwise’s method was simpler.

Where did 144 come from?

11

Thanks. So I guess two AK’s of the same suits would be the very best starter since you’ve have two flush possibilities as well.

12

434*3

13

I don’t doubt that both neuroman and bitwise are correct, I’m just amazed that one answer can be condensed to the other. Years of carful training by the public school system have left me incapable of doing math shortcuts in my head. I simply must do math the long way like a good girl.

I’m still not clear on what the phrase “chooses” means. What exactly are you picturing when you say

That there is some number x that represents the number of combinations you get when you choose 4 random cards out of a deck of 52? How is it that we can bypass consideration of the deck growing smaller with each selected card?

Forgive my simplicity, but my weakness in statistics is part of the reason I don’t table-top game. Even though neuroman’s answer is longer, I have an easer time following it.

Chicken kisses.

14

“n choose m” is the number of ways to select m objects from n objects where the order of the m objects is not relevant. You can find the formula here in the section on combinations. It involves factorials. The way that bitwise did it was the most efficient way, but it requires some basic knowledge of combinatorics. Basically, the trick boils down to knowing “how to count”.

15

Ahh OK but this counts Ace of spades first Ace od hearts second as different from Ace of hearts first and ace of spades second. So you want 43/2 * 43/2

16

I appreciate the link, because it actually does answer my question. My problem is that I was never taught factorials. I changed schools a lot and the last one always seemed to assume that I had already been taught various givens. For example, I never learned how to diagram a sentence.

Contrary to your post, though, I do know how to count. Managed to take that skill all the way through high school and college calculus. Your asshattery was not needed or appreciated.

17

Actually, I’m not so sure that nivlac intended the remark on knowing “how to count” to be insulting in any way. Combinatorics is the mathematics of counting, and there is nothing simple about some of the advanced methods of, “simply”, counting.

18

Then nivlac phrased it poorly.

That’s all. I don’t mean to pick a fight with anyone and I probably just over-reacted in my previous post.

Don’t want to hyjack the thread.

19

Ask any mathematician how hard counting is, and he’ll tell you it’s frightfully difficult. That’s not widely known outside of our circle, so I can see where somebody might take offense, but none was meant.

Anyway, bitwise is using a generalized form of the hypergeometric distribution, which might be of interest to you. It makes problems like this quite simple.

20

Yep. You start with the best possible pair and the best possible draws (nut straight and two nut flushes).