http://archives2.twoplustwo.com/show...ue#Post1329041
^ This is actually the chance of flopping top pair or higher though. The chance of only flopping top pair is pretty straightforward though.
Here's an example for AK:
P(A) = Probability of only flopping an ace, which is automatically top pair: 3!/(1!*2!) * (44!/(2!*42!)) / (50!/(3!*47!) =
14,48%
P(K) = Probability of flopping a King, and no ace appearing on flop: 3!/(1!*2!) * (44!/(2!*42!)) / (50!/(3!*47!) =
14,48%
P(A) + P(K) =
28,96%. AK is the easiest example because if you only flop one pair, you autmatically have top pair.
Now for J9:
P(J) = Probability of only flopping a Jack, and every card on the flop is below a Jack: 3!/(1!*2!) * (32!/(2!*30!)) / (50!(3!*47!)) =
7,59%
P(9) = Probability of only flopping a Nine, and every card on the flop is below a Nine: 3!/(1!*2!) * (28!/(2!*26!)) / (50!(3!*47!)) =
5,79%
P(J) + P(9) =
13,38%
If you look carefully, it is very easy to come up with the probability of flopping top pair with any card.
3!/(1!*2!)=3 and
(50!(3!*47!))=19600 (the cardinal) will appear in every equation. The only thing you need to figure out is the middle term. Count the number of cards in the deck higher in rank than the card you wish to flop top pair with, add your hole cards to that number and substract from 52. Example: for a a hand like QT, there are 8 cards higher in rank (4 Aces, 4 Kings.) Since we're trying to figure out the probability of only flopping one pair, you also substract the 3 remaining Queens and the 3 remaining Tens. That's 52 - 4 - 4 - 3 - 3 = 38 = X. So you plug in X!/(2!*(X-2)!), or 38!/(2!*36!). Add up the probability for both hole cards and you get what you're looking for.
Last edited by u cnat spel; 11-07-2011 at 02:42 PM.
