plo combinatorics

Really quick combs question:

Odds of flopping a flushdraw with a doublesuited hand? no card cancellation, so only info is our hand.

I have this:
(((11 choose 2) * (37 choose 1)) + ((11 choose 2) * (37 choose 1))) / (48 choose 3) = 0.235314524
I'm adding the combs for the 2 flushdraws and dividing by total flop combs.

cheers,
semp

looks legit

that question has nothing to do with combs

=22/48 * (10/47 + 26/47 * 10/46 + 11/47 * 20/46) + 26/48 * (22/47 * 10/46) = 0.25439

edit: probably wrong but not sure why

I agree, this isn't a combinatorics question.

Using the PQL runner from ProPokerTools.com I receive a count of approximately 23.5% for a flush draw or better. If you want to know the odds of flopping only a flush draw, then I believe this is the correct math:

(22/48)*(10/47)*(37/46)+(22/48)*(37/47)*(10/46)+(37/48)*(11/47)*(10/46) = 0.196

p sure this is the def of a combinatorics question insideman

On July 23 2012 06:57 byrnesam wrote: that question has nothing to do with combs

unless he's using the teeth on combs to count the combinations

if he was, he probably should have specified that in the OP though.

On July 23 2012 17:41 KoeBawlt wrote: p sure this is the def of a combinatorics question insideman

I agree with you, I didn't think that through in my original statement.

On July 23 2012 01:25 SemPeR wrote: Really quick combs question: Odds of flopping a flushdraw with a doublesuited hand? no card cancellation, so only info is our hand. I have this: (((11 choose 2) * (37 choose 1)) + ((11 choose 2) * (37 choose 1))) / (48 choose 3) = 0.235314524 I'm adding the combs for the 2 flushdraws and dividing by total flop combs. cheers, semp

This is absolutely correct.

select count(fourFlush(p1,flop)) as doubleSuitedFlushDraw
from game='omahahi', p1='QcJsTs9c'
Results:
Trials DOUBLESUITEDFLUSHDRAW
600000 141150 (23.53%)

select count(exactHandType(p1,flop,flush)) as flopFlush
from game='omahahi', p1='QcJsTs9c'
Results:
Trials FLOPFLUSH
600000 11197 (1.87%)


wouldnt it be 23.53 - 1.87?

On July 24 2012 11:54 Daut wrote: select count(fourFlush(p1,flop)) as doubleSuitedFlushDraw from game='omahahi', p1='QcJsTs9c' Results: Trials DOUBLESUITEDFLUSHDRAW 600000 141150 (23.53%) select count(exactHandType(p1,flop,flush)) as flopFlush from game='omahahi', p1='QcJsTs9c' Results: Trials FLOPFLUSH 600000 11197 (1.87%) wouldnt it be 23.53 - 1.87?

That seems right yes.

i'll probably just use pql from now on for questions like this.
thanks everyone.

if anyone can see where my math is missing the -1.87%, please let me know.
I can't see it.
Maybe I'm discounting cards wrong for the second part (37 choose 1).
I suspect it has something to do with us knowing we have 2 of the 13 of one specific suit, twice, so somehow adding 11 choose 3 might not be the right way to run it.

What is combinatronics?

I guess i get where the 1.87% mistake come from:
lets say you hold AdKdQcJc
when you count ((11 choose 2) * (37 choose 1)) at the first time you don't remove the possibility of club card coming after two diamonds fall, and the second time you count (37 choose 1) you don't remove diamond, and when you sum both products you count some flop possibilities twice

edit: no actually you don't
and i don't see why would Daut and InsideMan want to remove flopped flushes out of our calculations when we already removed flopped flush possibilities here

i only did it cause he did it lol

just figured out how often you flop exactly a 4 flush and how often you flop exactly a flush. figured that would be helpful

On July 25 2012 07:22 okyougosu wrote: I guess i get where the 1.87% mistake come from: lets say you hold AdKdQcJc when you count ((11 choose 2) * (37 choose 1)) at the first time you don't remove the possibility of club card coming after two diamonds fall, and the second time you count (37 choose 1) you don't remove diamond, and when you sum both products you count some flop possibilities twice edit: no actually you don't and i don't see why would Daut and InsideMan want to remove flopped flushes out of our calculations when we already removed flopped flush possibilities here

The reason I was trying to remove flopped flushes was because I misunderstood the PQL syntax.

"fourFlush: Flop games only. Given a player and a street (one of flop or turn), evaluates to true if the player has at least one four flush. For example, fourFlush(p1, flop) would return true for a holdem hand of AhKh and a board of JhTh9s."

I thought this meant four or five flush, but as it can be used to evaluate the probability of holding a four flush on the turn, this simply means one or two four flushes.
My equation for flopping a four flush now reads as follows:

(22/48)*(10/47)*(37/46)+(22/48)*(37/47)*(10/46)+(37/48)*(11/47)*(10/46)+(37/48)*(11/47)*(10/46)=23.53%

I actually derived this solution earlier, but dismissed it due to my misunderstanding of the PQL syntax. Thanks for helping me find the error.