Thread: Hold'em Odds
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Old 10-28-2004, 04:51 PM
BarkingMad BarkingMad is offline
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Join Date: Aug 2004
Location: Seattle
Posts: 33
Default Re: Hold\'em Odds

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I saw one site that calculated the odds of you NOT getting the cards you need, since that can be an "AND" situation and you can multiply

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Thats it. You can also use combinations to solve this problem (see pg 103 of Mike Petriv's Hold Em's Odds Book).

Here's why you have to use the odds of not completing the draw when using probabilities.

Addition Rule: To find the chances that at least one of two things will happen, check to see if they are mutually exclusive. If they are, add the chances.

Multiplication Rule: The chance that two things will both happen equals the chance that the first will happen, multiplied by the chance the second will happen given that the first has happened.

If you're looking to fill a draw on the turn or river, you can tell by it's definition that the addition rule doesn't apply because the draw could be filled on each street (the events are not mutually exclusive). You can also tell by it's definition that multiplication rule doesn't apply if you're using the odds of filling the draw in each case. By using the odds of not filling the draw, the requirements of the rule are satisfied. Then you can subtract the result from 1 since the chance of something (filling the draw) equals 100% minus the chance of the opposite thing (not filling the draw).

The method of using the odds of something not happening is the correct way to solve alot of probability problems.

Here's the same thing you did, but for flush draws.

Correct way:
1-(38/47)*(36/46) = 35% or 1.86 to 1

Note: The above result includes the chances of getting runner-runner flush cards. The chances of flopping a four flush and then getting only one of your needed suit on the turn or river is 31.6% or 2.16 to 1. I just learned this from Petriv's book, he uses combinations to calculate it.

Incorrect way:
(9/47)+(9/46) = 38.7% or 1.58 to 1

-Lance
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