Bennie E.

asked • 01/26/15

..Can someone show me how to set this up? Should be an experiment with Equally Likely outcomes

A poker hand consists of five cards drawn from a deck of 52 cards.Each card has one of 13 denominations (2,3,4...10,Jack,Queen,King,Ace) and one of four suits (Spades,Hearts,Diamonds,Clubs).Determine the probability of drawing a poker hand consisting of two pairs ( two cards of one denomination, two cards of a different denomination, and one card of a denomination other than those two denominations).

Eric M.

Okay Bennie,
 
Sorry about doing this incorrectly the first time. I made two big mistakes, so let's start over by erasing everything I've told so far!
 
The number of possible hands of five cards is correct: 2,598,960.
 
To count the ways of getting two pairs, let's look at the number of ways to get just one pair (meaning what are the possible combinations of suits for a single pair): you can get spade-club, spade-diamond, spade-heart, club-diamond, club-heart, and finally diamond-heart. That's six ways to get a single pair of cards.
 
Now all we need do is count the total number of ways to get two pair, disregarding suits. Let's say a pair of aces and 2's can be written "Ax2".
 
With aces as one of the pairs, then, the possible combinations of pairs with one pair being aces are
Ax2, Ax3, Ax4, Ax5, Ax6, Ax7, Ax8, Ax9, Ax10, AxJ, AxQ, and AxK. That's 12 ways. (We eliminated two pairs of aces because that's four of a kind.)
 
With deuces being one of the pairs, we first eliminate 2xA, since we already counted Ax2 in the "one pair is aces" list. We also eliminate 2x2. We are left with
2x3, 2x4, 2x5, 2x6, 2x7, 2x8, 2x9, 2x10, 2xJ, 2xQ, and 2xK. That's 11 more ways.
 
There's a pattern. For the 3's we eliminate 3xA, 3x2, and 3x3. The other 10 ways are 3x4, ..., 3xK.
 
Thus for the 4s we have 9 combinations, for the 5's we have 8 combinations, etc. on up to the Queens we have 1 combination. (We have zero combinations for the Kings because two pairs of kings are 4 of a kind, and all the other combinations were already counted, if you look at the last combination of each list.)
 
The total number of ways of getting 2 pairs, then (not including suits) is 12 + 11 + 10 + ... + 1 = 78.
 
But remember that, for each of these number-combinations, we have 6 ways of getting each pair, if you take suits into account. Thus the total ways of getting two pairs is 78*6*6 = 2808.
 
We must include our fifth card possibilities now. After getting our two pair, there are 48 cards left in the deck...but we must exclude the two cards of each number that would make our hand a full house. That is, if we had a pair of Aces and 9s, we'd need to exclude the remaining two Aces and the remaining two 9s so as not to get a full house.
 
This leaves 44 cards in the deck that will leave us with only two pair.
 
Now our grand total possible number of ways to draw two pair in 5 cards is 2802 * 44 = 123.552. The probability, then, of getting two pair is 123,552/2,598,960 = 0.04754, which is about 1 in 21.
 
By now, of course, you've finished your PhD and this information is just a tinge late, but hey, to restore my credibility -- what little I had, at least -- I must be the first to point out my error and correct it forthwith.
 
Q.E.D.
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03/10/15

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Eric M. answered • 03/05/15

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Hello Bennie.
 
I'm not sure what you mean about "Equally Likely Outcomes." We assume, of course, a fair deck of cards. A probability is a ratio between zero and unity defined by the number of ways get your desired outcome divided by the number of ways to get any outcome.
 
To calculate this probability, the numerator will be the number of ways you can get two pairs, and the denominator will be the number of ways to get five cards of any kind.
 
Let's do the denominator first. Hopefully, you know about combinatorial notation. You want to know what is 52 things taken 5 at a time, where order doesn't count. This is written 52C5 = 52!/(5!47!) = 48*49*50*51*52/120 = 2,598,960.
 
How many ways are there of getting two pair in a five-card draw? We'll break this down into 3 parts:
 
1) How many combinations of suits are there for a pair
2) How many number-combinations are there of pairs
3) How many fifth-card choices are left
 
1) In a pair, the suit combinations are 4 things taken 2 at a time, independent of order. This is 4C2 = 4!/(2!2!) = 3*4/2 = 6. Note this factor of 6 applies to each pair, so we're going to multiply by 62 down below.
 
2) Let's start with two pair where one pair is aces. The possible combinations, then, are aces and 2's (abbreviate as "A x 2"), A x 3, A x 4, ... A x K. This is 12 possible combinations where aces are one pair. (Why didn't I include A x A?)
 
To enumerate pairs where one pair is 2's, we omit 2 x 2 (since that's 4 of a kind) and 2 X A (since that was covered in the aces list). We thus have 2 x 3, 2 x 4, ... 2 x K. That's 11 combinations.
 
We continue this pattern, eliminating 4 of a kind hands and hands already counted in previous lists. For example, the two-pair combinations with 5's as one pair are 5 x 6, 5 x 7, 5 x 8, ... 5 x K, or 8 combinations.
 
We thus have a total of 12 + 11 + 10 + ... + 1 = 78 ways to get two pairs, counting numbers only (independent of suit combinations).
 
3) Once we have all these combinations of pairs, there are 48 cards left in the deck. But 4 of those cards would leave us with a full house! So we have a choice of 44 cards to pick from as our fifth card.
 
The total number of ways to get two pair in a five-card draw, then, are 6*6*78*44 = 123,552.
 
Our probability of two pairs, then, is 123,552/2,598,960 = 0.04754, or one in 21.
 
This number agrees with published values for two pairs.
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Eric M.

The published odds for two pair in a 5-card poker hand is 1 in 21. This from two different online sources. I am working on where I messed this one up!
 
 
 
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03/10/15

Eric M.

To correct my initial attempt: there are 78 number combinations that give you two pair. Start by excluding aces and aces, deuces and deuces, etc., because these hands are four of a kind, not two pair. Also exclude repeats like aces and 4s, and 4s and aces. Enumerating correctly you get 78 total combinations.
 
For a single pair, there are six suit combinations possible: spade-club, spade-heart, spade-diamond, club-heart, club-diamond, and heart-diamond.
 
Thus we have 78*6*6 = 2808 ways of drawing two pair...but that's in four cards!
 
To include the 5th card possibilities, first exclude the 4 cards remaining in the deck that would make our two pair a full house. Remembering to exclude also the 4 cards in our two pair, the remaining 44 cards are possible for our fifth card, so we still have two pair.
 
Thus the total is 2898 * 44 = 123,552 combinations. With 2,598,960 possible 5-card hands, the probability of two pair is 123,552/2,598,960 = 0.04754, which is about 1 out of 21 hands.
 
Sorry I messed this up last time!
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03/10/15

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