# Pair in a poker, again

1. Oct 6, 2012

### Astr

Im New on this forum, so i hope this is the right place to ask this question. I've tried to solve the problem of finding how many hands are in a poker game with exactly one pair, this way:

I can choose the first card from 52 ways. for the second card I can choose it from 3 ways (to match the pair). For the third card I can choose it from 48 possibilities (for ensure I´ve got no trios in the hand). For the fourth and fifth i can choose them in 44 and 40 different ways, respectively. And because order is not important I must divide for 5! (ways of permuting 5 cards). My (wrong) answer is:

(52x3x48x44x40)/5!=109 824

This is a tenth of the correct answer. Please can you tell whats wrong in the procedure. (English isn't my first language, sorry if Ive got some mistakes)

2. Oct 7, 2012

### CWatters

That gives you too many because a QS + QD is the same pair as QD+SQ.

There are 6 ways to get a pair of (for example) Queens..

QS + QH
QS + QD
QS + QC
QH + QC
QH + QD
QD + QC

but there is a choice of 13 denomination (values) so for pair it's

6 * 13

then the other cards must not have the same value but can be of any suit.

3. Oct 7, 2012

### Astr

Thank you Cwatters. But I think i avoid the problem QS+QD=QD+QS, dividing by 5!, becasuse that ensures that the order of the 5 cards does not matter.

4. Oct 10, 2012

### CWatters

I can't immediatly see why your method is wrond but continuing from above..

Then the next three cards can have one of 12 other values (3 from 12 = 220) and be from any of 4 suits (4*4*4 = 64)

220 * 16 = 14080

Now that is also 48*44*40/3! so your last part seems ok?

5. Oct 10, 2012

### Astr

Thank you again, CWatters. Well I know how to solve the problem, and this is out of discussion:
For the pair:
52x3/2!= 78 ways
For the three different cards:
48x44x40/3!=14 080 ways
then: 78x14 080= 1098240 ways to get exactly one pair.
but my problem is: Whats wrong in the procedure i post first?

6. Oct 10, 2012

### Ray Vickson

You should not divide by 5!; instead, you should divide by 2! × 3!, because the order of the cards within the pair does not matter and the order within the three non-pairs does not matter. That would give the number of hands as 1098240, as you want.

There are some subtle points here, so let me expand. When counting the hands containing exactly one pair, the person posing the question is distinguishing between hands such as PPNNN, PNNNP, PNNPN, NNNPP, etc., even though the actual cards involved are exactly the same (and so all would be regarded as a *single* hand by a poker player). Basically the question is asking for the number of different ways you could be *dealt* a hand containing a single pair, even though after being dealt the cards you would re-arrange them to your own, personal liking.

Another way to see this is to look at P{pair}, the probability of getting dealt a pair. Suppose, first that we have 2 aces and one each of 2 3 4. The probability, p, of getting that is p = C(4,2)*C(4,1)^3/C(52,5), where C(a,b) = "a choose b" = a!/[b!(a-b)!]. This holds because there are C(4,2) ways to pick 2 aces from 4 and for each of the 2 3 and 4 there are C(4,1) ways to choose the particular card, and because there are C(52,5) says of being dealt a 5-card hand althogether. Now, of course, the pair may not be aces and the non-pairs may not be 2 3 and 4. However, for any choice of ranks, the probability of the pair and the three non-pairs are all the same as the p above. Therefore, P{pair} is p times the number of ways to choose the rank of the pair (13) times the number of ways of choosing the three different ranks from the remaining 12 (which is C(12,3)). Thus, we have
$$P\{ \text{pair}\} = \frac{13 \, C(12,3) \, C(4,2) \, C(4,1)^3}{C(52,5)} = \frac{N_p}{C(52,5)},$$
where Np = number of hands containing one pair. Thus, we have
$$N_p = 13 \, C(12,3) \, C(4,2) \, C(4,1)^3 = 1098240.$$

RGV

Last edited: Oct 10, 2012
7. Oct 10, 2012

### Astr

Ray, Thank you for your answer. Can you explain me why i should not divide by 5!, because this is my whole problem. Thanks.