PDA

View Full Version : Beginners question

08-28-2002, 02:05 PM
How can you calculate the chance of flopping a flushdraw using combinations?

Thank you.

08-28-2002, 02:19 PM
I'd say the easiest way is to type something like:

Take p = C(n,k) / n! ^ exp(k), where n is the number of suits, k is the expected value, and divide p by C(x, f), where x is the number of cards in a standard deck and f is the number of cards in a standard flop.

The sheer idiocy of what you type will then induce Mike Haven, Lorinda, BruceZ, or Irchans to calculate the correct chance for you.

PP

08-28-2002, 02:30 PM
Carlos,

I always find it difficult to explain probability, but I will give it a try.

C(m,n) is the number of ways that you can pick n items from a list of length m if the order that you pick them is unimportant.

To figure out the probability of flopping a flush draw, you need to calculate the number of possible flops and the number of flops with flush draws. Then

(prob flushdraw) = (number flushdraws) / (number flops)

The number of possible flops is C(52-2, 3) if you hold 2 cards.

If you hold two suited cards, there are C(13-2,2) ways to choose 2 cards in your suit from the remaining 11 cards. There will also be one other, non-suited card in the flop. There are 13*3 possible non-suited cards.

(number flushdraw) = C(11,2)* 13*3

(number flops) = C(50,3)

(prob flushdraw) = C(11,3) * 13*3/ C(50,3) = 0.328316

I will attach a url with some basic poker probability.

Lastly, it is very easy to make mistakes in reasoning when doing probability, so it is best to check answers before posting them. I will foolishly skip the verification step in this post do to lack of time. Hopefully, at least the reasoning is readable even if there is an error.

08-28-2002, 02:31 PM

08-28-2002, 02:40 PM
Okay, sorry about that...here's the real answer.

I'm assuming that your question is:

If I hold 2 suited cards, what is the chance that exactly 2 cards of the same suit will come on the flop?

So, pre-flop 50 cards are unknown to you, 11 of which are in the desired suit (leaving 39 in the other 3 suits).

The number of possible flops is: C(50,3)

The number of flops with 2 of your suit and one other is: C(11,2) * C(39,1)

So, your chance of flopping a flush draw is

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

or roughly 11%.

If this seems fishy, you can always double check it with the good ol' probability tree...

Chance of your suit then your suit then other:

(11/50)*(10/49)*(39/48)

Chance of your suit then other suit then yours:

(11/50)*(39/49)*(10/48)

Chance of other suit then two of yours:

(39/50)*(11/49)*(10/48)

Add them all together and you get:

(3*11*10*39) / (50*49*48)

This is what C(11,2)*C(39,1) / C(50,3) reduces to when you...er...reduce it.

Hope that helps,

PP

08-28-2002, 02:44 PM
Hey Irchans,

I was too slow adding my real answer -- you beat me to it /images/wink.gif.

I like your explanation, but I think you might have made a small typo in the line:

(prob flushdraw) = C(11,3) * 13*3/ C(50,3) = 0.328316

It should be C(11,2), not C(11,3). Make that change, and our two posts will agree.

PP

08-28-2002, 03:34 PM
I knew I would make a mistake!

Thanks for the catch Pseudo.

C(11,2) * 13*3/ C(50,3) = 0.109439

08-30-2002, 09:57 AM