Flush odds doubt usign cond. probability

[aaaaa]

Hi,

The prob of seeing a flush pre-flop and no flush cards on turn and river is (11c3)*(39c2)/(50c5)

If I use conditional probability --
event A: flush pre-flop
event B: no flush cards on turn and river

then P(A_and_B) = P(A)*P(B\A)
=(11c3)/(50c3) * (39c2)/(47c2)

I'm falling short tho'. what is wrong with my conditional probability set-up?

thanks

[jay_shark]

event A . What do you mean by flush pre-flop ?

This is the second thread where you are really confusing yourself and others . You're given two cards pre-flop so it's impossible to have a flush . Do you mean two suited cards ?

[Tom1975]

I understand what you're getting at, but you could have stated your question a little more clearly. What you're really asking is "Given that you hold two cards of the same suit, what are the odds of A) flopping a flush; and B) have no additional cards of that suit come on the turn or river?"

The reason for the inconsistency lies in the denominator of your first formula. There are indeed C(50,5) possible boards; however C(50,5) ignores the order those 5 chosen cards are in. There are ten possible ways to arrange 3 flush cards and 2 non-flush cards. Only one of those (fffxx) satifies your conditions. This is why your two answers are off by a factor of ten.

[aaaaa]

No reason to get mad dude. Yes i mean post-flop.

So something like AsKs and flop is 6s7s3s amd the river and turn are 8d 6h

[Cobra]

The prob of seeing a flush pre-flop and no flush cards on turn and river is (11c3)*(39c2)/(50c5)

This is actually the probability to getting exactly three of your suit by the river, in any order.

then P(A_and_B) = P(A)*P(B\A)
=(11c3)/(50c3) * (39c2)/(47c2)

This is actually the probability of flopping a flush and then having a blank on the turn and the river, blank being a card that is not your suit.

Cobra