Odds of being dealt 2 pair

[MarkD]

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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.

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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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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.

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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):

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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.


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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?

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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]

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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.


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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.