Odds of being dealt 2 pair

[Matt R.]

I'm trying to figure out the exact way to calculate the probability of being dealt 2 pair in a 5 card hand, and for some reason I'm getting 2x the correct answer. Could someone please expalain what is incorrect about this?

I would think the answer should be:

[13*12*11*(4 2)(4 2)(4 1)] / (52 5)

Where the (4 2) things are combinations... i.e. n=4 and r=2. Reasoning is that there are (4 2) ways of choosing 1 pair, with 13 ranks to choose from; (4 2) ways of choosing the second, with 12 ranks; and (4 1) ways of choosing the 5th, non-paired card with 11 ranks left. This divided by (52 5), the total # of possible hands.

This comes out to precisely 2 times what the correct answer is, and I can't figure out where or why you divide the numerator by a 2! (or just 2, same thing). Any help would be appreciated.

[BruceZ]

[ QUOTE ]
I'm trying to figure out the exact way to calculate the probability of being dealt 2 pair in a 5 card hand, and for some reason I'm getting 2x the correct answer. Could someone please expalain what is incorrect about this?

I would think the answer should be:

[13*12*11*(4 2)(4 2)(4 1)] / (52 5)

Where the (4 2) things are combinations... i.e. n=4 and r=2. Reasoning is that there are (4 2) ways of choosing 1 pair, with 13 ranks to choose from; (4 2) ways of choosing the second, with 12 ranks; and (4 1) ways of choosing the 5th, non-paired card with 11 ranks left. This divided by (52 5), the total # of possible hands.

This comes out to precisely 2 times what the correct answer is, and I can't figure out where or why you divide the numerator by a 2! (or just 2, same thing). Any help would be appreciated.

[/ QUOTE ]

Your denominator (52,5) counts the total number of hands ignoring the order of the cards, so the numerator must do the same. Your numerator counts each 2-pair twice when you do 13*12, so this should be (13,2) = 13*12/2.

[Matt R.]

Thank you. I thought it was something along those lines -- but here is what I don't quite get yet. Since the 13*12 is incorrect and is counting each 2 pair twice, then why, once you include the fifth non-paired card, don't you do (13,3) since the order of the last card should be ignored as well (Edited to say: the order of the last card with respect to the other 4)?

[MarkD]

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[ QUOTE ]
I'm trying to figure out the exact way to calculate the probability of being dealt 2 pair in a 5 card hand, and for some reason I'm getting 2x the correct answer. Could someone please expalain what is incorrect about this?

I would think the answer should be:

[13*12*11*(4 2)(4 2)(4 1)] / (52 5)

Where the (4 2) things are combinations... i.e. n=4 and r=2. Reasoning is that there are (4 2) ways of choosing 1 pair, with 13 ranks to choose from; (4 2) ways of choosing the second, with 12 ranks; and (4 1) ways of choosing the 5th, non-paired card with 11 ranks left. This divided by (52 5), the total # of possible hands.

This comes out to precisely 2 times what the correct answer is, and I can't figure out where or why you divide the numerator by a 2! (or just 2, same thing). Any help would be appreciated.

[/ QUOTE ]

Your denominator (52,5) counts the total number of hands ignoring the order of the cards, so the numerator must do the same. Your numerator counts each 2-pair twice when you do 13*12, so this should be (13,2) = 13*12/2.

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Someone walk me through this a bit as I think I could learn something about counting and combinations from this exercise.

(13,2) = number of ways to be dealt two cards of different rank, correct?
11 = number of different ranks available for the third card, correct?
(4,2) = there are 4 suits to choose the first rank from, and we need two of them of the same value to make a pair, correct?
(4,2) = there are 4 suits to choose the second rank from, and we need two of them of the same value to make a pair, correct?
(4,1) = there are 4 suits to choose the third rank from, correct?

And the denominator is obvious to me. Wow, I think this actually makes sense now that I broke it down to myself. Hopefully I will be able to apply this in the future when I want to count stuff.

Ok, simple follow up counting question.

Let’s say I want to count the number of different ways to be dealt AK23 double suited in Omaha. How would I do this?

[Matt R.]

BruceZ,
I answered my own question with a little internet research. It actually is (13,3), but you multiply by a factor of 3 somewhere else. I couldn't think of an explanation on my own at all, so I thought this might be of some interest.

You're selecting 3 different ranks from 13 total ranks, (13,3). The two pair you select is (4,2)^2. Then, the 5th non-paired card can be selected 3 ways since you have 3 ranks to choose from (i.e., you can have JJTT2, 22JJT, 22TTJ -- the non-paired card has 3 possible values) -- 3*(4,1). This gives:

[3*(13,3)*(4,2)^2*(4,1)] / (52,5)

Thanks again for the help.

[Matt R.]

Very close. This is what I thought it was initially, but I was ignoring that you actually have cards of *3* different ranks (since the 5th card cannot be the same rank as the 2 pairs), so it's (13,3). And since there are 3 chosen ranks that you pick this non-pair card from, you get 3*(13,3)

As far as your Omaha hand goes, I don't play Omaha, but I'll try it for practice.

4*(4,1)^2 -- choose 1 of 4 suits twice, 4 different suits to choose from
(1,1)^2 -- choose 1 of 1 suit twice, since you've already picked your 2 suits


So, it would be 4*(4,1)^2*(1,1)^2) = 64 ways of being dealt AK23 double suited. Can someone verify this?

[BruceZ]

[ QUOTE ]
BruceZ,
I answered my own question with a little internet research. It actually is (13,3), but you multiply by a factor of 3 somewhere else. I couldn't think of an explanation on my own at all, so I thought this might be of some interest.

You're selecting 3 different ranks from 13 total ranks, (13,3). The two pair you select is (4,2)^2. Then, the 5th non-paired card can be selected 3 ways since you have 3 ranks to choose from (i.e., you can have JJTT2, 22JJT, 22TTJ -- the non-paired card has 3 possible values) -- 3*(4,1). This gives:

[3*(13,3)*(4,2)^2*(4,1)] / (52,5)

Thanks again for the help.

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You can look at it that way, but there is really no need to consider the factor of (13,3). All we need to do is take the number of 2-pair combinations (13 2)*(4 2)*(4 2), and then multiply that by the number of possible 5th cards, which is 44 if we are excluding full houses.

When we count the 2-pairs, 13*12 counts each pair of ranks twice, so we need to divide by 2. When we include the 5th card by multipying the number of 2-pair combinations by 44, this counts each combination only once, so there is no order issue involved with the 5th card. I hope you understand the difference between these two cases.

[Matt R.]

I'm seeing it now. You're starting from a different point than my last explanation. I started by choosing (13,3) different ranks, you're starting by choosing (13,2) 2 pairs then adding the 5th card. I'm trying to play around with your method (and the one I started with) to make sure I understand both.

[BruceZ]

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Let’s say I want to count the number of different ways to be dealt AK23 double suited in Omaha. How would I do this?

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I use C(x,y) for (x,y).

There are C(4,2) ways to choose the 2 suits, and for each pair of suits, there are C(4,2) ways to choose which 2 ranks go with each suit. So C(4,2)*C(4,2) = 6*6 = 36 ways.

[Matt R.]

Bleh, I'm still ass-backwards when thinking about combinatorics.

[MarkD]

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[ QUOTE ]
Let’s say I want to count the number of different ways to be dealt AK23 double suited in Omaha. How would I do this?

[/ QUOTE ]

I use C(x,y) for (x,y).

There are C(4,2) ways to choose the 2 suits, and for each pair of suits, there are C(4,2) ways to choose which 2 ranks go with each suit. So C(4,2)*C(4,2) = 6*6 = 36 ways.

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Woh, can you try that again. I'm missing something in the second C(4,2) term. I'm not sure how you come up with that (I do understand the first C(4,2) term.

Also, doing it the long way I come up with 36 ways, so I agree with your answer but am still foggy about the second term (I really want to understand your method):
<font class="small">Code:</font><hr /><pre>
As2s Ac2c Ad2d Ah2h
Kc3c Ks3s Kc3c Kc3c
Kd3d Kd3d Ks3s Kd3d
Kh3h Kh3h Kh3h Ks3s
- - - -
As3s Ac3c Ad3d Ah3h
Kc2c Ks2s Kc2c Kc2c
Kd2d Kd2d Ks2s Kd2d
Kh2h Kh2h Kh2h Ks2s
- - - -
AsKs AcKc AdKd AhKh
2c3c 2s3s 2c3c 2c3c
2d3d 2d3d 2s3s 2d3d
2h3h 2h3h 2h3h 2s3s
</pre><hr />

Also, if we were only interested in A2sK3s or A3sK2s from the long way I know this would be 24 hands, but how would you do it in a shorter faster methhod? With these question I'm really trying to understand the short hand methods so that I can perform some analysis of these situations fast on my own.

[BruceZ]

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[ QUOTE ]
[ QUOTE ]
Let’s say I want to count the number of different ways to be dealt AK23 double suited in Omaha. How would I do this?

[/ QUOTE ]

I use C(x,y) for (x,y).

There are C(4,2) ways to choose the 2 suits, and for each pair of suits, there are C(4,2) ways to choose which 2 ranks go with each suit. So C(4,2)*C(4,2) = 6*6 = 36 ways.

[/ QUOTE ]

Woh, can you try that again. I'm missing something in the second C(4,2) term. I'm not sure how you come up with that (I do understand the first C(4,2) term.

Also, doing it the long way I come up with 36 ways, so I agree with your answer but am still foggy about the second term (I really want to understand your method):

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From the first term, there are C(4,2) = 6 pairs of suits. For each pair of suits, pick one of the suits. For example, for (hearts, spades) pick hearts arbitrarily. Now take the 4 ranks A,K,3,2, and choose one of the C(4,2) pairs of ranks, for example, (K,3). Assign that pair to hearts, and then spades will get the remaining 2 ranks (A,2). You can do this for each of the C(4,2) pairs of ranks, so that is the second term. Then you do this for each of the C(4,2) pairs of suits, so there are C(4,2)*C(4,2) = 6*6 = 36 ways to do this all together.


[ QUOTE ]
Also, if we were only interested in A2sK3s or A3sK2s from the long way I know this would be 24 hands, but how would you do it in a shorter faster methhod?

[/ QUOTE ]

I assume you mean A2 is one suit, and K3 is another suit? Then there are 4 ways to choose the suit for A2, and for each of these, there are 3 ways to choose the suit for K3 since the suits cannot be the same. So there are 4*3 = 12 ways to make A2sK3s, and the same for A3sK2s, so 24 ways all together.

[Matt R.]

lol... okay, I'm going to take another stab at this. BruceZ, be ready to correct me if I'm wrong. Since you have a total of 4 ranks to choose from, but only two different suits, you have to choose 2 ranks to apply a suit to, or C(4,2). i.e. you can apply your suit of clubs to AK, A2, A3, K2, K3, or 23 -- 4*3/2 = 6.

[Matt R.]

As a side note (not that anyone cares), I think I'm learning -- I see why my method didn't work. When I said choose one suit twice (C(4,1))^2, I don't think this accounts for the fact that you only have 3 suits left to choose from when you pick your second suit. The second term is obviously just one, but I thought it accounted for assigning suits to the ranks. I managed to be wrong all the way around.

[MarkD]

[ QUOTE ]
From the first term, there are C(4,2) = 6 pairs of suits. For each pair of suits, pick one of the suits. For example, for (hearts, spades) pick hearts arbitrarily. Now take the 4 ranks A,K,3,2, and choose one of the C(4,2) pairs of ranks, for example, (K,3). Assign that pair to hearts, and then spades will get the remaining 2 ranks (A,2). You can do this for each of the C(4,2) pairs of ranks, so that is the second term. Then you do this for each of the C(4,2) pairs of suits, so there are C(4,2)*C(4,2) = 6*6 = 36 ways to do this all together.


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Bruce,

Thanks for your patience. I understand this now and hope that I will become better at counting using combinations. I have seen you solve so many problems so elegantly with them.

I have an engineering degree and math was never a problem... except combinations - I never did understand them very well, but am getting better. It's one of those things that my brain just doesn't work with very well so I really appreciate you taking the time here as I will definitely be using this stuff in the future.