2 pair from a poker hand combinations, confused

[mr_coffee]

Hello everyone,

Another example from the book I'm going over and I'm not exactly sure how they got their answer:

The game of poker is played with an ordinary deck of cards. Various five-card holdings are given special names.

a. how many 5 card poker hands contain two pair?
Well I'm looking at the list, and I see two pair defined as this:
Two cards of one denomination, two cards of a second denomination, and a 5th card of a thrid denomination.

They said:
consider forming a hand with two paris as a four-step process:
Step 1: Choose the two denominations for the pairs.
Step 2: Choose 2 cards from the small denomination
Step 3: choose 2 card from the larger denomination
Step 4: choose one card from the remianing.


When they say denomination do they mean like:
A,2,3,4,5,6,7,8,9,10,J,Q,K

also why do they say choose 2 cards rfom the small and large denomination? couldn't u do it by getting like 2 kings, and 2 10's, and 1 other card like a 2?
These are all large denominations.

When i did this problem i thought it would be solved like this:
You have 52 cards to pick from, but you only want 2 denominations, so
Step 1: 52 choose 2
Step 2: You already used 2 cards so you only have 50 to choose from now, so 50 choose 2

Okay now you should have 4 cards, now you only need 1 more card, but you can't choose a card you picked in step 1 or 2, so u must take 52-4 = 48, 48 choose 2

(52 choose 2) * (50 choose 2) * (48 choose 1)

But they did it the following way:

total number of hands with two pairs:
Step 1: 13 choose 2
Step 2: 4 choose 2
step 3: 4 choose 2
step 4: 44 choose 1

i'm not sure how they got this...
Also the 44 choose 1, I am not sure how they got 44. Any explanation would be great!

 

[mattmns]

This is a great problem, but a little tricky

I believe denomination is as you said A,K,Q,J, ..., 2. Also, they said larger denomination. Large denomination has no meaning (I know you are probably getting at 10, J, Q, K, A being large, but this is not what the book is getting at).

They get 13 choose 2 from the following. They are first choosing the denominations, and there are 13 denominations (A,K,Q,J, ..., 2), and they are choosing 2 of them. Then they choose 2 cards from 4 (there are four suits) for the first denomination and they do the same for the second denomination, hence 4 choose 2, twice.

Now for the 44 part, the reasoning is as follows: You are counting 2 pair, not 3 of a kind or full house, or anything else. So if you choose a pair of K's then you cannot choose the other 2 kings for your final card. Hence for each pair you pick, you must remove 4 cards. Thus we have 52 - 4 - 4 = 44.

 

[mr_coffee]

Ahh i think i understand, i was getting confused becusae i was physically thinking, if you choose a denomination you have already chosen a suit as well, because a card containas both a number and a suit.

As an example, it could have been like:
They choose 2 numbers or denominations (A,K,Q,J...2);
Say they chose, a 3 and a 10 (we don't know the suit yet)
Now they need to select which suit they are going to match the 2 denominations up with, say now they choose 3D and 10H (3 of diamonds and 10 of hearts)

Now they must pick another suit, but same cards, so (3C and 10D), which is the other 4 choose 2.

Now you have 52-4, i would think...you've still only chosen 4 cards out of the 52 in all havnt u?

oh wait u explained that too, There is a total of 4 kings in a deck right?
What if you choose the following cards (KH, KS, 3H,3D) you have 2 pair, 2 kings, and 2, 3's. So why can't you choose another King as your 5th card? becuase then it would transform it into 3 of a kind instead of 2?

I think just answered my question but i want to make sure, i should have probably learned to play poker then these problems might be easier :P

 

[mattmns]

Wait one second. They are choosing the suits for each denomination one at a time. That is, they first choose the suits for the 3's then they choose the suits for the 10's, separately. You have 4 possible suits for the first denomination (say 3D and 3H) so you have 4 choose 2, then you have four possible suits for the second denomination (say 10C and 10D) which is another 4 choose 2. For your other questions you are basically right. You want exactly two pair, so you cannot have a third king because then you would have something better than 2 pair (in your case you would have a "full house" which is 3 of a kind and a pair). And Yes there are 4 kings in a deck (there are four of each denomination in a deck).

 

[mr_coffee]

otal number of hands with two pairs:
Step 1: 13 choose 2
Step 2: 4 choose 2
step 3: 4 choose 2
step 4: 44 choose 1

Well from this 4 step process, they are choosing 13 choose 2, this is where they are picking the numbers, such as a 3 and a 10 right?

Then they are choosing 2 suits for the first number =>3 (4 choose 2), then another 2 suits for the 2nd number => 10, (4 choose 2) I thought...I think that's what you just said above, sorry i confuse myself somtimes hah

 

[Max Eilerson]

There are 13 choose 2 for the ranks of the pairs. There are 6 combinations of each pair (4C2), and 44 choices for the remaining card.
[tex](13C2)(4C2)^2(44C1)[/tex]

 

[drpizza]

I've seen a students suddenly get it if I put it this way:
(13C1)(4C2) * (12C1)(4C2) * (44C1)
Choose a denomination, choose two suits. Choose another denomination (don't repeat the first denomination), ...

 

[mattmns]

I've seen a students suddenly get it if I put it this way:
(13C1)(4C2) * (12C1)(4C2) * (44C1)
Choose a denomination, choose two suits. Choose another denomination (don't repeat the first denomination), ...

That is a natural way to do it, but incorrect. You need to divide that answer by 2 since you are over counting, for example AAKK and KKAA.

Note that, [tex]\binom{13}{2} = \frac{\binom{13}{1} \binom{12}{1}}{2}[/tex]