Can anybody show me the calculation to determine what the odds are that three people will all flop a flush? the situation came up the other day while i was playing and I was curious what the chances are of that happening.

I know that there are 6 of the suit in three hands and there are only 7 left so the flop had to contain exactly three of those 7. but, I didn't know how to do the calculation.


Thanks

First, we need to count the total number of possible flops. We know the identity of 6 cards, so that leaves us 46 cards left choose from. Therefore there are C(46,3)=(46*45*44)/(3*2*1)=15180 possible flops.

Now we need to count the number of those flops which will contain 3 flush cards. There are 7 flush cards left, therefore there are C(7,3)=(7*6*5)/(3*2*1)=35 flops.

So the final answer is 35/15180=.0023

Thank you sir. I was on the wrong end of this hand but I figured the odds of it happening were pretty darn slim.

If hero has suited cards and we assume random suit distribution for villains then we have hero hitting a flush on the flop C(11,3) = 165.

So, you and two villains all have the same suit of cards, the chance that you flop a flush is about 1/5 the chance of you getting the desired monochrome flop when villains have random suit distribution.

alternatively, you can count it by considering all the possible villain holdings (i.e. 100% range) vs their flush holdings. You can see 5 cards so villan's combined hole cards are C(47,4) = 47*46*45*44 / 4 * 3 * 2 = 178365

and the combinations where they hold the same suit as the board are C(8,4)=70. So, when you see that your suited cards have flopped a flush (about 1% of the time), two villains who came along with 100% PF ranges will also have a flush 70/178,365 = 0.039% of the time (3.9e-4). This is more of a first person perspective (i.e. how likely is it that they both have the same suit given that I just hit my flush).

Of course villains generally don't play a 100% preflop range. When we get to the weakest part of a players range, most prefer suited over unsuited. E.g. villain might have 45s in his range but not 45o.

FYI: the N chose K formula is

N! / (N-K)! * K!

Where "!" is the factorial function.

So if you have N cards that satisfy a condition and you want to know how many different flops can be made with these cards that is N chose 3.

So with this proposition, it am assuming that the odds of 3 people starting suited is irrelevant, and your question is based on the assumption that you and 2 other people are already suited. So this is what I figure:
The odds of YOU flopping a flush with any 2 suited cards is at best 117.79:1, or %.84(homepokergames.com/odds). Since there are 13 of each suit, it means that the odds I just gave are assuming there are 11 cards of your suit left in the deck. Now with 2 other people holding suited cards, the assumption is that there are at most 7 cards left in the deck of your suit. So what are the odds of 3 of the 7 max remaining cards coming out on the flop, if they aren't dead already?

I think its this- %0.84(7/11), or %0.535, or 161.00:1
Some extra homework:
Odds of 4 people flopping a flush: %0.84(5/11), or %0.382
Odds of 5 people flopping a flush: %0.84(3/11), or %0.229
Odds of 6+ people flopping a flush: %0 (%0.84x 1/11), but you need 3+ outs to make a flush.

Hope my math for this situation was proper. Now remember, the odds given in these situation are BEST CASE scenarios, assuming that no other cards from your suit besides the ones dealt to you and the selected players are dead. If you want to incorporate what the odds are of you and 2+ other people starting with suited cards AND flopping the flush, that's a different story, and its too late for me to crunch that last easy part of this equation right now. I need to go hit the sack.

Two people flopped flushing on me last night when I held top two. It always seems like it is higher when it happens to you. But I see the odds are so low that I don't feel too bad about calling off on the flop.

N! / [(N-K)! * K!]

or

N! / (N-K)! / K!

Damnit why didn't I stay in school.

Confirmed via software simulation:

PQL Query:
select count(handshaving(exactHandType, flop, flush) = 3)
from game='holdem', p1='2s9s', p2='8sKs', p3='Qs3s'

Results:
Trials COUNT 1
610477 1380 (0.23%)

this is a similar, but a different question.
5 people sees the flop and the flop is monotones what is the probabilty of one of us having flopped the straight. and also what is the probabilty of someone cathcing a flush draw on a flop like ororor ?
I know in HU chances are pretty low, but also what is the probablity of the person who plays only suited cards matching the monotoned flop?

If the flop is only ever seen when villain holds a suited hand, then 1/4 of the monotone flops will match his suit.

The monotone flop will happen 5% of the time that a flop is seen. 5% * 1/4 = 1.25% or 1 in 80 seen flops he will have the flush.

However, he will flop a flush draw a lot more often than that, around 14% of seen flops.

I´m using this useful free online odds calculator for free:

http://www.pokerprediction.net

very helpful

agreed. the equations make my stomach upset

https://rumble.com/vqwo5k-three-flop...dem-poker.html