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09-30-2011, 08:12 PM   #26
EvilSteve

Join Date: Mar 2006
Location: 2p2 afterlife
Posts: 13,199
Re: Number of possible flops in Holdem

Quote:
 Originally Posted by phaddad DarkMagus, I am very interested in your post. I would like to be able to get to the bare number of flops. For example, Ad 7s 2h would be the same as 7s 2h Ad. But it would also be the same as 7d Ah 2s. How many distinct flop possibilities are there?
Not sure what you're asking. His 1755 number counts those three as the same flop, A72 rainbow. And actually the first two are the exact same cards, just listed in a different order. If you want to count each specific three card combination separately, that's 52c3 = 22,100. And if for some unknown reason you care about the order the cards come down, that's 52*51*50 = 132,600 ordered three card sequences.

02-02-2015, 04:42 PM   #27
DGAF
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Join Date: Jan 2009
Location: West Coast
Posts: 5,077
Re: Number of possible flops in Holdem

Quote:
 Originally Posted by TheCesspit Well, that's merely 50 Choose 3: 50*49*48 / 3*2*1 = 19,600 Total possibly flops are 52 Choose 3 = 22,100
Then what is the median flop (if you have no hole cards)? Looking to do a little over/under gambling and kind of burnt out on red/black. Thx.

05-05-2017, 07:01 AM   #28
CaptnCanary
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Join Date: Jan 2007
Location: England
Posts: 17
Re: Number of possible flops in Holdem

Quote:
 Originally Posted by N.Brosseau Basically, what I'm asking is, in keeping things in the perspective of what I was asking previously about pair, 2 pairs and so on showing up within the common cards (at flop, turn, river) should I take into consideration the cards that I have in evaluating the probability of these event or not when making a general evaluation. So wouldn't the total number of possible flop be 22,100 instead of 19,600 if I'm just checking the common cards in general and not in relation with the specific pocket hand I have ? The reason why I'm asking this is because I've been collecting data on the common cards that are popping up (my sample is at 2000 hands right now) and I'm seeing discrepancies of about 2% to 3% on the number of time a pair is showing in the common cards. To give you a specific example, it was previous established that at least two card of the same value would show up in the common card by the river 49.29% of the time but I'm getting a value of about 46.45%. The previously established value was based on the idea that we were considering 52 possible cards on the board, but should I only be considering 50 and take each time into consideration the hand that I have and the number of player present to determine the odds of a pair or more cards of the same value, in general, showing up on the board ?
I know this is very very old but I just wanted to say that if you were observing flops in holdem games, the stats may be biased by the hands that most players see a flop with. Players will usually see a flop with pocket pairs and will always fold many unpaired hands. So the flops you see will have a slightly reduced probability to show up with a pair on the board. This may account for a slight difference from theoretical and observed flop textures.

05-05-2017, 01:52 PM   #29
whosnext
Pooh-Bah

Join Date: Mar 2009
Location: California
Posts: 3,895
Re: Number of possible flops in Holdem

Quote:
 Originally Posted by CaptnCanary I know this is very very old but I just wanted to say that if you were observing flops in holdem games, the stats may be biased by the hands that most players see a flop with. Players will usually see a flop with pocket pairs and will always fold many unpaired hands. So the flops you see will have a slightly reduced probability to show up with a pair on the board. This may account for a slight difference from theoretical and observed flop textures.
This same issue came up in a fairly recent thread. Pairs are tricky in poker. If I have a pocket pair, that increases the chances that another player has a pocket pair, not decreases. If you think about it for a little bit, the reason becomes obvious.

Same logic applies to flops as well. So if I have a pocket pair, the chance that a pair appears on the flop actually is increased compared to if I do not have a pocket pair.

It is not hard to work through the actual tallies via combinatorics:

Number of flops having at least a pair given that I hold a pocket pair = 12*C(4,3)+12*C(4,2)C(46,1)+1*C(2,2)*C(48,1)
= 3,408

Number of flops having at least a pair given that I do not hold a pocket pair = 2*C(3,3)+11*C(4,3)+2*C(3,2)C(47,1)+11*C(4,2)C(46,1 )
= 3,364

Of course, to derive the respective probs the common denominator is C(50,3) = 19,600

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