View Full Version : Odds of being dealt any pair or specific pair in first 3 cards

09-26-2003, 12:26 PM
Can someone show me how to calculate this? I searched the archives and came up empty.

09-26-2003, 02:04 PM
This is the same as the probability of a paired flop in hold 'em. The probability of a pair is 1 minus the probability of no pair, and this is:

1 - 52*48*44/(52*51*50) = 17.2% or 4.8-1.

This includes sets which are 0.35%. Divide by 13 to get the probability of a specific pair or set = 1.3% or

Earlier I posted this for the whole 5 card board as a proposition bet. The board is almost even money to be paired in 5 cards (49.2%).

09-26-2003, 05:36 PM
Wow, that's an amazing statistic. So is that partly why if you flop a set, you should generally be jamming it on the flop and turn?? Even if you are up against a made flush or straight, or flush or straight draws, or even two pair, you are very likely to fill up?

09-27-2003, 12:39 AM
No, if you have a set on the flop, you will fill up just 28% of the time. The 49.2% is for the whole board, including times the flop pairs. You often jam the flop and turn with a set to make draws pay or get them to fold.

09-27-2003, 01:14 PM
if you have a set on the flop, you will fill up just 28% of the time.

Actually, you will fill up 29.1% of the time. You will fill up or make quads 33.4% of the time.

You don't often see the calculation of the odds for filling up or making quads starting from a set. Earlier I gave 3 methods (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=326 125&Forum=probability&Words=draw&Match=Entire%20Ph rase&Searchpage=2&Limit=25&Old=6months&Main=325790 &Search=true#Post326125) of computing the odds for completing a draw. This problem uses one of the alternative methods (method 2). We are calculating the odds for the board pairing again, but if it doesn't pair on the turn (7 outs) we have 3 additional outs to pair on the river (10 outs) because then we can pair the turn card.

7/47 + 40/47 *10/46 = 33.4%