So the other day I was thinking that over half of the time I flop a set, there are either two suits or a single suit on the flop giving someone a potential flush draw. Now if the board is paired, no problem (generally), I've flopped a boat.

So I'm just wondering what the odds of there being both a non paired board and two suits or a single suit?

I know roughly 60% of flops will have two or one suit (40% rainbow). And I know about 17% of flops will be paired or three of a kind (83% no pair or trips)

I don't think I can just multiply this out because pairs on the board prevent single suits and probably change things. But maybe that doesn't matter or it's insignificant.
So if I could multiply it out, it would just be 0.6*0.83 which is about half. Maybe that's close enough.

Now if I really want to flop a clean set, I also have to worry about straight draws, but I think that's a lot more complicated and I'll save that for another day - I'm guessing it would be easier with computer simulations.

Thanks!

-d

non paired board with exactly 2 suits: 13C3*4C2*3C1 / 52C3 = 23.29%
non paired board with a single suit: 13C3*4C1 / 52C3 = 5.18%

this is in general, not when holding a set.

I think the first result is off by a factor of 2.

This may be a little more transparent for some for the no-pair 2 suiter.

There are C(4,2)=6 ways to select the two suits. For one of the suits there are C(13,2) =78 ways to select 2 cards leaving (13-2)= 11 ways to select a non-pairing card from the other suit. Thus the total number of successful ways = 6*78*11 for the two suits, no pair.

But given 2 suits, there are 2 ways for the 2-1 card split so multiply by 2 and divide that by the number of possible flops for the answer (0.466), which is twice that given above.

yes, thanks for the correction. I was doing all other possibilities yesterday and it didn't add up to 1, but I thought the mistake was somewhere else.

non paired with 2 suits: 4C2*2C1*13C2*11 / 52C3 = 46.59%
non paired rainbow: 4C3*13C3*3! / 52C3 = 31.06%
non paired with 1 suit: 4C1*13C3 / 52C3 = 5.18%
paired with 2 suits: 13C2*2C1*4C2*2C1 / 52C3 = 8.47%
paired rainbow: 13C2*2C1*4C2*2C1 / 52C3 = 8.47%
all cards with same rank: 13C1*4C3 / 52C3 = 0.24%

this is in line what OP's numbers say. paired flop 17.18%, rainbow 39.77%.

Perfect, thank you!

-d

It's simpler than this. It's already given that we've flopped a set and the suit of the key card is irrelevant so there are only two unknown cards.

There are 4*(12C2) ways to select suited cards different from the rank of our flopped set, and 12*11*3 ways to select two cards such that exactly one is suited with our flopped key-card, and the board not paired.
Total ways to select two cards: 49C2 = 1176

(4*(12C2)+12*11*3)/1176 = 55/98 = 56% of our set being vulnerable, 44% chance of a flopped set being "clean."