Computing Two Pair Odds

[katakana]

Can anyone tell me what the formula would be to compute the chance of getting two pair in the flop when you have none in your hand?

Ex: 8 7 o

Flop: 897 or Flop: 899

Thanks

Robert

[kluCAR]

I didnt go calculating because i think it is rather complicated. So instead I ran a generator for texas holdem poker (250 000 games). I ve gathered that in 22% of time you get two pairs.

[mrfloppy]

my formula definitely wrong on this 1. Hopefully someone can point out where my logic is flawed. Hitting 2 pair 1 out of 5 hands on the flop can't be right. My formula may be correct now.


Let's assume you have that 87o...that leaves 6 cards left to give you 2 pair. That 1st card can be any 8 or 7 so 6. At that point 1 of your next 2 cards has to be the other, so that leaves 3. So let's assume we hit 1st pair on 1st card. Miss the 2 pair on the 2nd card, but hit it on that 3rd flop card.

6x1x3=18/117600=.000153

Any corrections appreciated, as this information is valuable as general knowledge to myself or anyone.

[kluCAR]

Again I have to corect myself. I thought here that you meant the odds of hitting two pair on the TABLE. I mixed FLOP and the TABLE again. As I said, I am not good poker player but I am still good matematician.
The odds of hitting two pair on the floop if you have two unequal cards is poor 4%. If you have a pocket pair the odds goes up to 16%. But if you have a pocket pair you are lookin for a three of a kind. The odds are (FLOP) 11% and 20% after a turn and river.[/b]

[Adamm]

Lord help me, here we go...
You got yer two random cards, be they 87 or A2, whatever...
Two possible situations...
1. You pair each of your cards once (you have 87 with a flop of 827)
2. You pair one of them and board pairs also (you have 87 with a flop of 899)

Situation 1 gets divided into three more sub-situations...
As the three cards get dealt you will pair two outta three...
1.1 First two each pair, last one doesn't make a boat...
( 6/50 )( 3/49 ) ( 44/48 )
1.2 First one pairs, second one junks, last one pairs the other...
( 6/50 )( 44/49 )( 3/48 )
1.3 First one misses, second hits, thirds hits the other...
( 44/50 )( 6/49 )( 3/48 )
Simplify all of those to ( 3*6*44 )/( 50*49*48 )!
or... 0.0067347
Got it so far?
Situation two with three more sub-situations!
2.1 First one matches one of yours, second two pair each other...
( 6/50 )( 44/49 )( 3/48 )
2.2 Second one matches one of yours, other two pair each other...
( 44/50 )( 6/49 )( 3/48 )
2.3 First two pair each other, third one matches one of yours...
( 44/50 )( 3/49 )( 6/48 )
But wait, we can simplify all of those to ( 3*6*44 )/( 50*49*48 )
or 0.0067347, which is the same number again.
6 situations, so multiply it by 6 and get 0.0404082 or 4.041%.
Half of the time (2.02%) it'll be with a pair on the board, the other half it'll be with each of your cards pairing just once.
This doesn't take into account straights or flushes because they would be impossible anyway. If anything pairs, there is no chance of a straight or flush.
I checked in a poker book, and that's the answer, but I'm not sure if it's the best (or correct!) way to do it.