cool, thanks for an actual answer. what about 3 ppl flopping a flush?
~36294 to 1
Here's the calculation:
Chance you get suited cards (12/51) x chance the flop come 3 of your suit (11c3)/(50c3) x chance at least 2 other people have been dealt 2 cards of the same suit (let's call that p)
p is the tricky part:
the probability that 2 other players were dealt those suited cards p2 = (8c4)/(47c4)
the probability that 3 other players were dealt those suited cards p3 = (8c6)/(47c6)
the probability that 4 other players were dealt those suited cards p4 = (8c8)/(47c8)
2 players could be dealt those hands (9c2) ways
3 players could be dealt those hands (9c3) ways
4 players could be dealt those hands (9c4) ways
Using inclusion/exclusion we have this event NOT happening = 1 - (9c2)(p2) + (9c3)(p3) - (9c4)(p4) = 0.986090311
so p = 0.013909689
so this event happens (12/51) x (11c3)/(50c3) x 0.013909689 = 0.000027552205 = ~36294 to 1
Pretty sick when it happens. Last 2 players in a tourney, both flop a flush. What's worse, we had the 2 nut flushes (A & K). Both slowplay it until the river: bet, raise, reraise all-in, call. I was going broke no matter what.
This morning playing heads up I have K rag suited.... villian has A rag suited and flop was suit suit suit. Needless to say it was a cap fest. Unfortunately there was no way for my 2nd nut flush to suck out.
Comments
I have no idea. But, I'll guess: slim.
10 handed... the odds of atleast 2 people (including you) flopping a flush is ~2302 to 1
God doesn't hate you... He just has a twisted sense of humour.
~36294 to 1
Here's the calculation:
Chance you get suited cards (12/51) x chance the flop come 3 of your suit (11c3)/(50c3) x chance at least 2 other people have been dealt 2 cards of the same suit (let's call that p)
p is the tricky part:
the probability that 2 other players were dealt those suited cards p2 = (8c4)/(47c4)
the probability that 3 other players were dealt those suited cards p3 = (8c6)/(47c6)
the probability that 4 other players were dealt those suited cards p4 = (8c8)/(47c8)
2 players could be dealt those hands (9c2) ways
3 players could be dealt those hands (9c3) ways
4 players could be dealt those hands (9c4) ways
Using inclusion/exclusion we have this event NOT happening = 1 - (9c2)(p2) + (9c3)(p3) - (9c4)(p4) = 0.986090311
so p = 0.013909689
so this event happens (12/51) x (11c3)/(50c3) x 0.013909689 = 0.000027552205 = ~36294 to 1
Somehow I think 50/50 would have been easier.