Figuring outs percentages |
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Posted Tue Oct 14, 2003 1:07 pm GMT by trenhes
As I understand it you take your outs and divide that by cards left, ie
9/46=19.6% or 9/47=19.1%. In Ken Warren Teaches Texas Hold-em
he shows the odds of 9 outs w/2 cards to come as 35%. I don't know how to figure this, but when you think about it aren't your odds greater with 2 cards to come because you have 2 chances of hitting your out, even though you have 1 more card in the deck you have 2 draws to catch your card. Does this make sense and how would the math look. Hope I have explained this clear enough. Otherwise there's very little difference in odds between the turn and the river. Thanks for any replies.
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Posted Tue Oct 21, 2003 8:18 pm GMT by Adam Marshall
There used to be an example of this on this site under "Calculating Odds" somewhere. Musta been lost in the changeover.
Whenever you have to figure out the chance of whether an event will occur with multiple chances, you have to calculate the chance that won't happen by subtracting that probability (represented as a decimal!) from 1. That way you don't run into problems hitting it both times. In the flush example, that'd mean finishing your flush on the turn, and getting that same suit on the river.
So the example we're thinking about is a flush draw. To visualize, let's say we've got A5 of hearts. The flop comes 2h, 7h, Qd. What are your chances of hitting a flush on the turn AND/OR river?
We already know the chance on the turn and river individually so we can use that info too. Just to restate, the chance of hitting it on the turn would be 19.1% (.191) and the river would be 19.6% (.196).
The chance of your NOT hitting on the turn or river individually would be 1 minus the probability of those...
(1 - .191) = .809
(1 - .196) = .804
Multiply them together to figure out your cumulative chance of NOT hitting them (I say NOT because the numbers are 1 minus)...
(.809 X .804) = 0.650436
1 minus it back to figure out your chance of actually hitting it...
(1 - .650436) = 0.349564
Round it up to an easy .35 and convert back to a % to come out with 35%.
I should mention that these aren't TOTALLY mathematically accurate because the 19.1 and 19.6 weren't the exact, precise figures, but you get the idea.
There's an example like this about a straight draw (which is pretty similiar) on this site at http://www.texasholdem-poker.com/odds2.php .
P.S. I'm gonna find out what the hell the math term is for what a number is after it gets 1-.x 'ed. I'll edit this once I find out.
Posted Wed Oct 22, 2003 1:04 pm GMT by Geno
| Adam Marshall wrote: | | P.S. I'm gonna find out what the hell the math term is for what a number is after it gets 1-.x 'ed. I'll edit this once I find out. |
I don't think there is one dude.....I did lots of Math to a pretty high level and never came across it. Even mathematicians are pretty lame, when we did probability using binomials, we used P(x) for the probability of x and (1-P)(x) was given the notation Q(x) so really, I don't think there is a term for that Q as such.
I'd love you to prove me wrong by the way 
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