View Full Version : question on figuring odds and percantages.

01-07-2004, 06:30 PM
hi all I'm new to poker and odds and such. I just picked up sklanskys hold em poker and also lee jones winning no limit poker. I'm trying to figure out how they came up with the odds using material from both books. the puzzling thing to me is how sklansky came up with the %'s he did with the odds he gives in his example on page 106.
With 9 outs, he gives 38:9 (which i understand) but he gives this 35% chance to win. Should this not be 24%? or perhaps if I'm wrong again then maybe 19%?

The reason why I ask is that he goes down to show how to change the % to odds. If you use his forumla given on the bottom of page 106 and going back to our previous example. if you subtrace 35 from 100 and dived that sum by 35 I get 1.9 : 1 odds. Which to me does not equal 38:9 or 4.2:1. Am I smoking crack or just stupid as hell or both by thinking there is something missing or not cottect in Sklanskys calculations on page 106 and 107?

Thanks for any insight and help in sorting my confusion. I'm to the point now where I want to start using pot odds along with my play but I'm afraid to go into it unsure on how to really use them.

01-07-2004, 07:20 PM
The case of 9 outs being 35% happens ONLY when you consider the chance of hitting on the turn OR the river (or both). You are right that 9 vs. 38 (4.2:1) -> 9/(38+9)=19%.


01-07-2004, 08:24 PM
yeah I re read that and saw the magic 2 cards thing. My question I guess now is how do you calculate for 2 cards and such? Its been 8 years or so since taking stats class. I'm so rusty on this. I guess I'm wanting to figure out how they derive these odds and such. I'm wanting to make a table of odds to print out to memerize and compare to each hand and such. Thanks again for the help.

01-08-2004, 01:31 AM
To calculate the probability for 2 cards to come, invert the probabilities for hitting on each of the remaining cards, multiply them together, and then invert the result. The concept is that you are combining the chances of NOT hitting your outs on either card, then subtracting that from 100%.

So, for a typical nut flush draw on the flop in Hold'Em:

P(flush,2 cards) = 1 - (38/47 * 37/46)