# Probability of Two Pair

• Apr 9th 2011, 11:27 PM
ragnar
Probability of Two Pair
I see in this Wolfram page the computation for two pair in a five-card hand, and I understand everything except the division by 2!. I would think that this division represents making sure that the order in which each pair occurs does not distinguish two poker hands. But I would think we should divide by 3! because we want to make sure that order doesn't matter between the pairs as well as for the place of the single card which does not match the face value of either pair.
• Apr 10th 2011, 04:21 AM
Plato
Quote:

Originally Posted by ragnar
I see in this Wolfram page the computation for two pair in a five-card hand, and I understand everything except the division by 2!.

You did not give us a link. So there is no way to comment.
But there are $\desplaystyle\binom{13}{2}\binom{4}{2}^2(44)$ ways to have two pairs.
• Apr 10th 2011, 09:33 AM
ragnar
Wolfram Poker

Sorry about that; the result of no sleep.

So as I understand your presentation of the answer, choose two face-values regardless of order, then for each choose two suits, regardless of order. Then choose a last card. But in the last part, we just multiply--so doesn't this distinguish hands by their order? Wouldn't this count the hand (2H)(2D)(3H)(3D)(4H) as distinct from (4H)(2H)(2D)(3H)(3D)? So do we need to divide by something, or since we just did the choice of the pairs and then the choice of the onesie, is the later kind of hand not counted at all?
• Apr 10th 2011, 09:51 AM
Plato
Quote:

Originally Posted by ragnar
Wouldn't this count the hand (2H)(2D)(3H)(3D)(4H) as distinct from (4H)(2H)(2D)(3H)(3D)?

Absolutely not.
Order is not considered at all.
• Apr 10th 2011, 10:13 AM
ragnar
But my question is more like why that's the case. It seems like it should. If I were finding the number of ways of arranging 5 people in three seats, I'd just multiply the possibilities for the first seat (5) by the possibilities for the second (4) by the possibilities for the third (3). If I want to disregard order I need to also divide by the factorial of the number of seats.

So I'm trying to understand why this case is different. Here we just multiplied the number of ways of choosing two pair by the number of ways of choosing the remaining card. It would seem to me that this is like the first case, and to disregard order we would need to divide by something.

Is that not true of this case because we basically count only the hands where the onesie card comes last, and equate such hands with all other orderings of the same cards--whereas in the case of people in seats, we cannot distinguish one seat from another in the way that we distinguish the onsie from the two pairs? Something like that?
• Apr 10th 2011, 10:47 AM
Plato
Quote:

Originally Posted by ragnar
But my question is more like why that's the case. It seems like it should. If I were finding the number of ways of arranging 5 people in three seats, I'd just multiply the possibilities for the first seat (5) by the possibilities for the second (4) by the possibilities for the third (3). If I want to disregard order I need to also divide by the factorial of the number of seats.

Arrangements are order driven.

Selections are content driven. Order makes no difference.
Think about a team of six children. It makes a different who is on the team(content). It makes no difference the order in which they are picked. Just who is selected.