How many flops are possible in holdem?

This is an ambiguous question, so clarifications follow:

- I don't care about the order of the three cards.

- I want to account for suit symmetries; if you have one flop and obtain another by changing all the s to s, all the s to s, they are "suit-symmetric."

For example, 7 8 9 and 9 7 8 are the same flop. (However, neither of these is the same as 8 7 9. JTs for an open-ended straight flush draw is impossible on the last.)


I got , but I wanted to check because I'm not sure.

unpaired, rainbow: 13c3 = 286
unpaired, 2-flush: 13c3 * 3 = 858
unpaired, 3-flush: 13c3 = 286
paired, rainbow: 13*12 = 156
paired, 2-flush: 13*12 = 156
trips: 13

total: 1755

not sure how useful this number is, as the different types of flops have different probabilities.

Is this calculation correct?
Number of cards 50
50x49x48
divided by 1x2x3
+ 19600

No, read my original post more carefully.

How did you get 1599? Taking all the possible flops into consideration ther are 19600. Lets see your calculation.

Actually there are 22,100 total flops possible. But that number counts As 2s 3s as different from Ad 2d 3d, which isn't what he was trying to calculate. I thought his OP was pretty clear.

I get also 1755 – but a little bit more detailed:

3-Straight-Flush = 12
Trips = 13
3-Straight 2-flush = 12 * 3 = 36
3-Straight rainbow = 12
3-Flush = c(13,3)-12 = 274
Paired rainbow = 13 *12 = 156
Paired 2-flush = 13 * 12 = 156
High-Card-Flops 2-flush: [c(13,3)-12] * 3 = 822
High-Card-Flops Rainbow: [c(13,3)-12] = 274

As DarkMagus already said, this consideration makes not much sense. A split-up of the c(52,3) = 22,100 different flop combinations looks as follows:

3-Straight-Flush = c(12,1) * c(4,1) = 12 * 4 = 48

Trips = c(13,1) * c(4,3) = 13 * 4 = 52

3-Straight 2-flush = c(12,1) * c(4,1)*c(3,1)^3 = 12 * 36 = 432

3-Straight rainbow = c(12,1) * c(4,1)*c(3,1)*c(2,1) = 12 * 24 = 288

3-Flush = [c(13,3)-12] * c(4,1) = 274 * 4 = 1,096
Axx = (c(12,2) - 2 ) = 64 *4 = 256
Kxx = (c(11,2) – 1) = 54 * 4 = 216
Qxx = (c(10,2) – 1) = 44 * 4 = 176
Jxx = (c(9,2) – 1) = 35 * 4 = 140
Txx.= (c(8,2) – 1) = 27 * 4 = 108
9xx = (c(7,2) – 1) = 20 * 4 = 80
8xx = (c(6,2) – 1) = 14 * 4 = 56
7xx = (c(5,2) – 1) = 8 * 4 = 36
6xx = (c(4,2) – 1) = 5 * 4 = 20
5xx = (c(3,2) – 1) = 2 * 4 = 8

Paired rainbow = c(13,1) * c(4,2) * c(12,1) * c(2,1) = 78 * 24 = 1,872

Paired 2-flush = c(13,1) * c(4,2) * c(12,1) * c(2,1) = 78 * 24 = 1,872

High-Card-Flops = [c(13,3)-12] * [c(4,1)^3 - 4] = 274 * 60 = 16,440

High-Card-Flops 2-flush: 274 * 36 = 9,864
Axx = (c(12,2) - 2 ) * c(4,1)*c(3,1)^3 = 64 *36 = 2,304
Kxx = (c(11,2) – 1) = 54 * 36 = 1,944
Qxx = (c(10,2) – 1) = 44 * 36 = 1,584
Jxx = (c(9,2) – 1) = 35 * 36 = 1,260
Txx.= (c(8,2) – 1) = 27 * 36 = 972
9xx = (c(7,2) – 1) = 20 * 36 = 720
8xx = (c(6,2) – 1) = 14 * 36 = 504
7xx = (c(5,2) – 1) = 8 * 36 = 288
6xx = (c(4,2) – 1) = 5 * 36 = 180
5xx = (c(3,2) – 1) = 2 * 36 = 72

High-Card-Flops rainbow: 274 * 24 = 6,576
Axx = (c(12,2) - 2 ) * c(4,1)*c(3,1)*c(2,1) = 64 *24 = 1,536
Kxx = (c(11,2) – 1) = 54 * 24 = 1,296
Qxx = (c(10,2) – 1) = 44 * 24 = 1,056
Jxx = (c(9,2) – 1) = 35 * 24 = 840
Txx.= (c(8,2) – 1) = 27 * 24 = 648
9xx = (c(7,2) – 1) = 20 * 24 = 480
8xx = (c(6,2) – 1) = 14 * 24 = 336
7xx = (c(5,2) – 1) = 8 * 24 = 192
6xx = (c(4,2) – 1) = 5 * 24 = 120
5xx = (c(3,2) – 1) = 2 * 24 = 48

After we know our 2 hole-cards only C(50,3) = 19,600 flop combos are possible. 2,500 less than shown above.

Sorry in this formula the term c(4,1)*c(3,1)^3 is wrong!
Correct is c(4,1) * c(3,2) * c(3,1) = 36

Axx = (c(12,2) - 2 ) * c(4,1)* c(3,2) * c(3,1) = 64 * 36 = 2,304
The same for all other High-card-Flops 2-flush. However, all abovementioned results are correct.

For example 542 2-flush:
c(3,2) ways to get a 2-flush combination {5,4} {5,2} {4,2}
times c(3,1) for one card of other suit

542 542 542
542 542 542
542 542 542

542 rainbow:
542 542
542 542
542 542

This is the same wrong term. Correct is:
3-Straight 2-flush = c(12,1) * c(4,1) * c(3,2) *c(3,1) = 12 * 36 = 432

In Omaha we have the same flop combos. As soon as we know our 4 hole cards only c(48,3) = 17,296 different flops are possible.

c(52,3) - c(48,3) = 4,804.

It is relatively easy to explain these differences and also the 2,500 differences in Texas Hold'em for any hole card combination. However, this insight has not much value. More important is the consideration what hole card combos are still possible in the hands of our opponents after the flop, turn and river.

In Texas Hold'em after the flop: c(47,2) = 1,081
In Omaha after the flop: c(45,4) = 148,995.

Quick question on the subject at hand.

I'm doing some mathematical evaluation on the different kind of common cards that shows up on the flop, turn and river (for a Poker Analysis program I'm working on).

Comparing theory to practice I've hit numbers that don't match which is making me doubt my formula, for pair and 2 pairs in the common cards.

(FYI, my sample is based on about 700 hands, which is not much but the rest of the other event I'm checking are close to theory so this is why I'm really doubting my math here).

Are these formula right ?

Seeing at least 2 card of the same value on the board (pair):
Flop: 13*C(4,2)*C(50,1) = 17.65% in practice (10.92% in theory)
Turn: 13*C(4,2)*C(50,2) = 35.29% (31.97% [not to far but still under])
River: 13*C(4,2)*C(50,3) = 58.82% (45.59%)

Seeing at least 2 time 2 card of the same value on the board (2 pair), including quads:
Flop: 0
Turn: C(13,1)*C(4,2) * (C(12,1)*(C4,2) + 1 (for quads)) = 2.10% (0.51%)
River (C(13,1)*C(4,2) * (C(12,1)*(C4,2) + 1 (for quads))*48 = 10.52% (4.86%)

So is my math good or does my sample contain less Pair and 2 pair than it should ?

at least one pair on board

Flop : 13* (C(4,2) * C(48,1) + C(4,3)) / C(52,3) = 17,18%

Turn : (13 * C(4,2) * C(12,2) * C(4,1)^2 + C(13,2) * C(4,2)^2 +
13 * C(4,3) * C(48,1) + 13 * C(4,4)) / C(52,4) = 32,39%

River: (13 * C(4,2) * C(12,3) * C(4,1)^3 + C(13,2) * C(4,2)^2 * C(44,1) +
13 * C(4,3) * C(48,2) + 13 * C(4,4) * (48,1)) / C(52,5) = 49,29%

at least two pairs on board

Turn : (C(13,2) * C(4,2)^2 + 13 * C(4,4)) / C(52,4) = 1.04%

River: (C(13,2) * C(4,2)^2 * C(44,1) + 13 * C(4,3) * C(12,1) * C(4,2) +
13 * C(4,4) * C(48,1)) / C(52,5) = 4,92%

Please note the problem of double counting. Please look at Wikipedia for further explanation of the aforementioned formulas for Turn and River.

Thanks for the information.

First, I just noticed that I switch theory and practice in my previous listing so I apologize for the possile confusion there (I'll go and edit it up after this post).


I've check the link on Wikipedia but if I'm understanding the equation correctly you and they are extracting the quads and trips and other better hands out of the question, which is not what I'm trying to do.

By at least a pair I meant at least 2 or more card of the same value, which means including trips and quads.

Are the previous equations accurate based on that ?

I can't edit my previous message but I think I figure out where my math was wrong and what you meant in your equation. The formula on Wikipedia for a single pair only and this is what got me confused.

Quick follow-up question on that subject:

Can the number of player on the table who receive cards affect the probability of one or two pair showing up on the board ?

(I can't seem to find why or how it should but just I'm just asking in case I'm missing something)

Only your own hole cards will affect the flop probabilities. You do not know which cards your opponents hold. Therefore it does not matter how many opponents you have.

If I'm not mistaken, my pocket hand would affect the probability for the event at a specific moment but it shouldn't affect the general probability of the flop holding a pair or 3 of a kind, correct ?

In Super System, Mike Caro states that there are 19,600 flops possible to any given Hold 'Em hand. That is NOT the number of total possible flops in the game, but it is an interesting statistic.

Well, that's merely 50 Choose 3:

50*49*48 / 3*2*1 = 19,600

Total possibly flops are 52 Choose 3 = 22,100

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 ?

First, 2000 hands isn't that big a sample and 3% error is definitely not out of the question.

Second, if you're just counting every flop seen,then no you don't really need to consider what's in your hand, but you should be using the 22,100 figure.

Thank you for your answer Magus.

I knew that my sample is indeed pretty small to draw any kind of formal conclusion at this point but since previously I've made a few mathematical error and I'm seeing some numbers being exactly like the theory, I just wanted to make sure that I hadn't forgotten anything.

Obviously I am holding 2 cards that can not be in the flop so 19600 is correct.
22,100 can only be the right number if I am not holding any cars......... in which case the flop wouldnt concern me............. ffs

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?