# Calculating a pair in Poker

1. Oct 5, 2006

### VonWeber

The correct way is $$\left(\begin{array}{cc}13\\1\end{array}\right \left(\begin{array}{cc}4\\2\end{array}\right$$(12;3)(4;1)(4;1)(4;1)

But why does (13;1)(4;2)(48;1)(44;1)(40;1) fail? I would have thought that after you get the two that match there are 48 left to choose from that don't macht, and 44 that don't macht, and so on.

Last edited: Oct 6, 2006
2. Oct 5, 2006

### Hurkyl

Staff Emeritus
Your approach treats QQJT9 and QQ9TJ as two different hands.

That's fine if you're counting permutations instead of combinations... but if you meant to count permutations, then you didn't count hands like JT9QQ... so you're wrong either way.

3. Oct 5, 2006

### VonWeber

And so then calculating two pair I did (13;2)(4;2)(4;2)(48;1) and subtracted Full Houses. I suppose this failed for the same reason?