Quck counting question

[rex4334]

I've read that given 2 suited cards hole cards your chances of improving to a 4 flush by the flop is 10.9 percent. I'm trying to verify this for myself mathematically. Given that C(n,r) is the number of r-combinations of a set with n-elements does the following make sense?

P(x) = ( C(11,2) * (50 - 11) ) / C(50,3)

Where P(x) is the prob. of drawing 3 more cards from the remaining 50 cards and improving to a 4-flush, given that you are delt 2 hole cards of similar suit

Thx

[ QUOTE ]
I've read that given 2 suited cards your chances of improving to a 4 flush by the flop is 10.9 percent. I'm trying to verify this for myself mathematically. Given that C(n,r) is the number of r-combinations of a set with n-elements does the following make sense?

P(x) = ( C(11,2) * (50 - 11) ) / C(50,3)

Where P(x) is the prob. if drawing 3 cards having 2 of the same suit from the 50 remaining cards.

Thx

[/ QUOTE ]

I don't think it's a combination problem as much as it's a pure probability problem:

To get a four flush after the flop, having started with two suited cards would go as follows.

There are three ways the flop can produce two same suited cards: 1st & 2nd, 1st & 3rd, 2nd & 3rd. Add those probabilities and subtract the probability of all three being the same suit, and you should have your answer.

[Tom1975]

You got it, that's the formula. If you crunch the numbers that works out to 10.9%

[alThor]

[ QUOTE ]
P(x) = ( C(11,2) * (50 - 11) ) / C(50,3)

[/ QUOTE ]

Yes, look up the hypergeometric distribution for further info.

Your answer is equivalently ( C(11,2) * C(39,1) ) / C(50,3)

alThor

[rex4334]

Thx for the responses folks.

To take it one step further if i wanted to improve to a 4 flush or better by the flop could we then express it as follows?

(( C(11,2) * C(39,1) ) + C(11,3)) / C(50,3)

Assuming C(11,3) adds all the combinations where we get all 3 cards flush on the flop.

Thx again