How many different flops, asks DaVitsche

[herttaberlin]

How many different flops are there in Hold'em? And with different we mean that for example 234 spades suited are not different from 234 diamonds suited.

I based my calculations on the logic that 3 card flop is same as 2 cards + 1 card. As we know there are 169 possible 2 cards, 13 pairs and 78 suited and off-suit non paired cards. For all these 169 options I add the third card. Below the spreadsheet. Presentation screwed up but I don't care.

So for example a pair of 22, if you add one more 2, there is only one way to do it, it doesn't matter which one of the remaining 2's it is. But then if you add a card with a value of 3 to A, the card can be either off suit (flop becomes a rainbow) or suited with one of the 2's.

Same thing for 23 suited. If the 3rd card is 2 or 3, it has to be off suit. For 4 to A it can be suited or off suit.

Last 23 off suit. If the 3rd card is a 2, it can be suited or off suit with 3, same for 3. If it's a 4 to A, it can be off suit (rainbow flop), suited with 2 or suited with 3. Finally we sum and multiply and end up to 5083.

Where do I go wrong @DaVitsche?

 

	Third card	2	3	4	5	6	7	8	9	T	J	Q	K	A	Sum	No.	Tot.
	Pocket pairs 22	1	2	2	2	2	2	2	2	2	2	2	2	2	25	13	325
	Suited 23	1	1	2	2	2	2	2	2	2	2	2	2	2	24	78	1872
	Offsuit 23	2	2	3	3	3	3	3	3	3	3	3	3	3	37	78	2886
																			5083
	Sum is a sum from left to right (third card 2 – A)
	No. is a number of different first 2 cards ( 13 different pocket pairs)
	Tot. = Sum x No.

[ArtyMcFly]

There are 22100 possible flops in Holdem, 1755 of which are strategically different.

[DaVitsche]

Do you know how many flops we would need to get a good representation of possible flop to reduce that number?

[ArtyMcFly]

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@DaVitsche wrote:

Do you know how many flops we would need to get a good representation of possible flop to reduce that number?

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The guys behind PioSolver say that near-perfect GTO solutions can be found by using subsets as small as 184 flops (there's a list of them at https://cdn.shopify.com/s/files/1/0769/9693/files/kuba_184_flops_5_11_2015.txt?18111346264467327999), but for approximate solutions that are quick to compute with solving software you can get away with using as few as 25 (!). i.e. If you understand the strategy for your range on the 25 flops listed at https://cdn.shopify.com/s/files/1/0769/9693/files/kuba_25_flops_5_11_2015.txt?18111346264467327999 you'll be pretty solid on any board.

[DaVitsche]

You've done your homework :)

I've actually been working with the 74 flop subset. Just to be clear, I knew the answers to what I was asking :) but sometimes asking the right question is more of an eye-opener than giving the answer.

Happy studying!

[herttaberlin]

@ArtyMcFly, @DaVitsche Ok, I finally figured it out correct.

1. 3 of a kind, AAA, 222,... 13 pcs

2. A pair with 3rd card suited with eiter of the pair cards, AdAs2d, AdAs3d... 13 x 12 = 156 pcs

3. A pair + 3rd card, all offsuit, same as above, 156 pcs

4. All suited, 13 x 12 x 11/(3 x 2) = 286 pcs

5. All offsuit, items 1-3 excluded, same as item 4. 286 pcs

6. 2 cards suited, 3rd offsuit, three options, suited are two lowest, two highest or highest and lowest cards, see item 5., 3 x 286 = 858 pcs

Total 13+156+156+286+286+858=1755 pcs. Heureka!

Now, how do I apply this knowledge in practice, that'll take some more thinking.

 

 

 

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