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HELP--Pot ODDS: Sklansky v. Negreanu, epper, 17. Oct 2003 08:01
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It seems as though pro poker players calculate pot odds differently. Negreanu stated that he takes the number of outs and puts it in a ratio with outstanding cards. Ex: 14 outs means 46-14, which is 3.3-1.

Sklansky does it differently. He takes the number of winning cards and places them in a ration with losing cards : 14 wins out of 46 so that means 32 losses, which means 32-14 or 2.3-1.

I think that Sklansky is correct because if you had 23 outs with 46 unseen cards you would be even money, not 2-1 underdog. Am I right or wrong? Thanks.

I am confused. Thanks for your help, and I enjoy your columns.
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Re: HELP--Pot ODDS: Sklansky v. Negreanu, shorn, 17. Oct 2003 08:18
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I vote for Sklansky on this one. The sample set of total outcomes cannot equal more than the total number of cards and therefore Negreanu's way doesn't work.
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Re: HELP--Pot ODDS: Sklansky v. Negreanu, pt_Gatsby, 17. Oct 2003 08:49
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I may be wrong in this, as I'm not sure of the question...

But Odds and Probabilities are expressed differently.

Probability is expressed as (Number of wanted outcomes) / (Number of total outcomes), or 14/46. It basicly states that the reduced version (ie: ~1/3.3) gives you the probability of an event occuring.

Odds are expressed as (Number of wanted outcomes) : (Number of unwanted outcomes), and thus gives a different number.


It expresses the same information but can be used in different contexts. The basic way to derive each one is to add/subtract the wanted outcomes from the other (ie: to get odds from probability, take the bottom, subtract the top and stick a semicolon inbetween).

I think the DR Math site (www.drmath.com) has a better explanation, if that doesn't help.
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Re: HELP--Pot ODDS: Sklansky v. Negreanu, ICIP, 17. Oct 2003 08:59
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Actually each one is correct, but the way it is phrased isn't. While I'm not familiar with Negreanu's style of calculating odds, what he is doing is calculating the percent chance of hitting the winner which would be 14/46 or 30.4%.

The way that Sklansky phrases these is a versus situation. It is like putting lottery balls into a hopper. Who ever the odds are against gets 1 ball and the other person gets as many as their advantage is, so in this case 2.3 balls.

Also note that the odds are 2.3 to 1 and 3.3 to 1 which has a difference of 1. Whenever you see this, the percentage chance is the larger of the two or about 1 out of 3 times, in this instance. It is simply different ways of looking at the same thing. Hope this clears it up for you.

ICIP
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Re: HELP--Pot ODDS: Sklansky v. Negreanu, Phish, 17. Oct 2003 09:13
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As Mason Malmuth has said, you don't really need to calculate odds all that precisely. Just need a general feel to do the right thing. Easiest way to develop that is to develop a basic strategy that will apply to almost all situations. For example: if you have flush or open ended straight draw, generally go to the river. If only a gut-shot, peel for one small bet but dump it on the turn unless pot is really big. Raise and reraise on the flop if there are more than 2 opponents and you have a strong flush draw. etc.
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