[Luzion]

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Yes, you are correct about the outs but what I really wanted to know is if you calculate your probability and have 2 chances is there a method to calculate the increase in the probabiliy because you have two shots. I tried unsucessfully to make the example as vague as possible. Just assume you have a 25% chance after the flop, is there a way to estimate the increase in probability because you have two more chances?

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Yes, this is very simple to do. BruceZ, KJL, and I used this method in the post already.




Heres some real simple reasoning behind it.
1) You flip a coin. The probability of getting heads is 50%. Therefore the odds of NOT getting a heads is 50%.
2) You roll a dice. The probability of rolling a six would be 1/6. Therefore the odds of NOT rolling a six would therefore be 5/6.
3) A bag has 3green balls, and 1red ball. The probability of picking the red ball on your first try is 25% The probability of NOT picking the red ball is 75%.




See where I am getting at?

Now lets find the probability of NOT hitting your 9outs by the river. That would be (38/47) * (37/46) = 0.65.... Now you can reason that therefore the odds of HITTING one of the 9 outs at least once by the river is 0.35.

I hope I explained it as simply as possible... Basically find the probabillity of something happening, then subtract that result from 1, to find the probability of it NOT happening.



Heres a simple exercise so you can remember the logic. You hold a pocket pair before the flop. Can you tell me the probability you flop a set or better? (Hint: Find the probability of NOT hitting a set or better on the flop)