Beginners question

How can you calculate the chance of flopping a flushdraw using combinations?


Thank you.

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

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.

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

Hey Irchans,


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


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

I knew I would make a mistake!


Thanks for the catch Pseudo.


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