This would apply in a loose game, such as 1/2NL.

I have KK vs. 5 opponents

Flop: AK6

What would be the odds of one or more of my opponents holding a flush draw?

Notice that Ax Qx and down the line is available to opponents. Thanks!

Your flopped set is irrelevant to the chance someone has a flush draw here. There are two factors: 1) what is the chance one of five opponents was dealt two diamonds, given that we have seen five cards and two are diamonds. 2) What is the chance he stayed in for the flop.

1. The chance any specific player here was dealt DD is C(11/2) / C(47,2) = 55/1081 = 5.09%. We can approximate the chance at least one among five players has DD by 1-(1-.0509)^5 = about 23%.

2. What is the chance a player in a loose game sees the flop with any suited hand? It would be helpful if we know the preflop betting, and player tendencies, but without any of that we'll just say 50% chance.

23% * 50% = about 11.5% chance one of your opponents flopped a flush draw.

Here is a way to get the exact probability under the assumption of random card distribution. You can break this down by the total number of diamonds dealt to your five opponents, and then tally how many of these ways exhibit one or more flush draws.

Clearly the number of ways for your five opponents to have exactly D diamonds between them is C(36,10-D)*C(11,D) since there are 11 diamonds not accounted for, 36 non-diamonds not accounted for, 10 total cards dealt out to them, where C(x,y) is the combination function.

There are 945 ways of dealing of 10 cards into 5 2-card hands where the cards are ordered so that the first card of each hand is greater than the second card and the five first cards are ordered from high to low.

That is, C(10,2)*C(8,2)*C(6,2)*C(4,2)*C(2,2)/5! = 945.

I cannot remember what the technical term is for the following, but it is straightforward to calculate how many of these deals with D diamonds exhibit a flush in one or more hands for each D.

They are 0, 0, 105, 315, 585, 825, 945, 945, 945, 945, 945 for D=0 to 10.

Combining these two sets of data, we find that, under the assumption of total randomness, the probability of one or more of your opponents having a flush draw is 1,156,379,935,095 / 4,893,273,079,695 = 23.632%.

I approached the problem a bit differently using a simulation program. If one sees the flop, it is reasonable to think their hand has some value. I assumed of the 5 opponents, one plays atc, two play top 50% and two play top 25%. Then the probability at least one has a flush draw is 25.7%. Naturally, the results will vary based on what you think villain ranges are.

By inclusion-exclusion:

5 * C(11,2) / C(47,2) -
C(5,2) * C(11,4) / C(47,4) +
C(5,3) * C(11,6) / C(47,6) -
5 * C(11,8) / C(47,8) +
11 / C(47,10)

= 23.632%

(Note: with PIE, neither partitions nor compositions are necessary for this problem because it doesn't matter which diamonds go to which villain.)

Edit: Just noticed your flashy new color, whosnext! Glad to see

Nice answer heehaww. One day, perhaps, my brain will flash to PIE methodology for these types of problems, but I am not yet there.

Yes, I am now green!

to be clear I'm speculating here, so I'd greatly appreciate comments why this is wrong/improper.


using this methodology
and figuring for this


55/1081 =.0509 = ATC will have flopped a FD
35/473 =.074 = a top 50% hand will have flopped a FD
20/227 =.0881 = a top 25% hand will have flopped a FD

1-0509 =.9491
1-.074 =.926
1-.0881 =.9119

1-(.9491*.926^2 *.9119^2)
1-(.67675)

=32.325%

in contrast to this

intuitively, it seemed to me that the chance should get greater as people get more selective about their starting hands

and this

this was greater but surprised me that it wasn't even more, ran a sim myself and arrived at 34.27%

Thanks guys!
I was surprised by the low percentage answers. I thought it would be much closer to 50% vs. 5 opponents.

Since the pot was straddled to $6, 1 caller & then I raise to $30, get the 5 callers. One of which was the straddler. They are probably holding hands in the top 25-30% or so.

A quarter of all random hands are suited (23.5%). That doesn't increase much taking top 50% or top 25% hands, only very slightly.

Ducking back in just to point out the double factorial method that is useful in many poker applications.

The number of disjoint ways to form N 2-card hands from 2*N cards is N!!.

Which is most easily calculated as 1*3*5*7*...*(2*N-1). That is, the product of the first N odd numbers.

So the number of disjoint ways to form 5 2-card hands from 10 cards is 5!! = 1*3*5*7*9 = 945.

This is useful in many poker problems.

That's 9!! and the formula should be (2n-1)!!

5!! is 5*3

See if you can apply the concept to Omaha (ie generalize the double-factorials): how many ways are there to form N unlabeled (not belonging to specific players) 4-card hands from 4*N cards?

Oh jeez, I cannot even get the fricking formula correct.

Thanks heehaww for the correction.