# Thread: Wolfram Explanation of Probability of Two Pair

1. ## Wolfram Explanation of Probability of Two Pair

So I'm looking at Wolfram's Probability of Poker Hands page and I don't completely understand the one for two pairs. I understand and agree with everything except division by 2!.

I figure: $(13)$ $\binom{4}{2}$ $(12)$ $\binom{4}{2}$ $44$ tells us the number of ways to get two of a kind for thirteen face-card values, then two of a kind for the remaining twelve face-card values, and then 44 cards which are none of these two face cards. But this would represent getting one pair first, then the other pair next, then the onesy card last. To correct for this sort of double-counting, we divide. But I don't get why we divide by 2!. I mean, there's the first pair, the second pair, and onesy. Three positions, so to speak. So shouldn't we divide by 3!?

(Note I: I asked about calculating the probability of two pair before, but my question was different--more about the general issue of ordering rather than the particular issue in this example.)

(Note II: What is wrong with this LaTeX math code? It keeps giving me a compile error. 13 \binom{4}{2} \cdot 12 \binom{4}{2} \cdot 44)

2. Originally Posted by ragnar
I figure: $(13)$ $\binom{4}{2}$ $(12)$ $\binom{4}{2}$ $44$ tells us the number of ways to get two of a kind for thirteen face-card values, then two of a kind for the remaining twelve face-card values, and then 44 cards which are none of these two face cards. But this would represent getting one pair first, then the other pair next, then the onesy card last. To correct for this sort of double-counting, we divide. But I don't get why we divide by 2!.
Say you pick 'kings' first then 'twos' and a 'nine'.
Is the same as first picking 'twos' then 'kings' and a 'nine'.
Look at $\binom{13}{2}$ $\binom{4}{2}$ $\binom{4}{2}$ $44$

3. I thought it would count 3! times, though. Kings then twos then nine is the same as twos then kings then nine, is the same as nine then kings then twos, etc.

4. Originally Posted by ragnar
I thought it would count 3! times, though. Kings then twos then nine is the same as twos then kings then nine, is the same as nine then kings then twos, etc.
Sorry to say, but you cannot count.

5. Cute, but on the other hand, you can't explain math.

For anyone willing to help, if you call the position of the first pair A, position of the second pair B, position of the onesy C, you would have to divide out by the number of permutations of these three, i.e. 3!. So why is this not an accurate representation of how you divide out by the number of ways that a hand can be rearranged, if we've already divided out by the rearrangements of each pair since we used $\binom{4}{2}$.

Perhaps another way of asking: Why is it that counting the number of two-pair hands as $(13)$ $\binom{4}{2}$ $12$ $\binom{4}{2}$ $44$ $\frac{1}{2}$ has already eliminated counting (9H)(9D)(3S)(3H)(AH) (where 9H is the 9 of hearts, etc.) as distinct from (AH)(9H)(9D)(3S)(3H)?

6. Just figured this out--my reasoning was basically right, the way of counting (up to dividing by two) does give an ordered listing, even counting order of the onesy--but only counting as the onesy coming last. If I wanted the full listing of ordered hands I'd have to multiply by 3 for the three positions of the onesy, but in the next step I'd be dividing out so as to disregard order and so just be dividing by the three, making it all a superfluous step in calculation (though, for the logic, a subtle but necessary step).

7. Originally Posted by ragnar
Just figured this out--my reasoning was basically right,
Well that is very magnanimous of you. I have written and published on this for over thirty years. And now you agree I may be correct?
Originally Posted by ragnar
an ordered listing,
Ordered listings have absolutely nothing to do with ‘hands’ in a game of cards. In a game of cards, only content of a ‘hand’ makes any difference.
Order has absolutely nothing to do with the ‘hand’.