View Full Version : odds calculation question

12-01-2003, 01:57 PM
In Sklansky's advanced hold em book there is a probablity table in back I do not understand. He calculates the probablity of completing a four-flush post-flop in hold em at about 35%. In other words, with two cards to come...
my understanding of the math is 9/47 + 9/46 or about 38.7%. What is the math used to come up with Sklansky's figure? Why is it wrong to add the percentages together for the turn and river cards? /images/graemlins/confused.gif

12-01-2003, 02:16 PM
You cannot add probabilities for events which are not mutually exclusive. Two events are mutually exclusive if the probability of both events happening is zero.

The easiest way to calculate this probability is to calculate the probability of the event NOT happening, and substract this from the full sample space, ie:

12-01-2003, 04:43 PM
And the hard way?

12-01-2003, 05:21 PM
"Two events are mutually exclusive if the probability of both events happening is zero."
So in other words, hitting a heart on the turn does not exclude the possibility of another coming on the river???
Tnanks for the reply.

12-01-2003, 06:16 PM
This post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=395803&page=&view=&sb =5&o=&vc=1) will tell you all about mutually exclusive events and independent events, and it contains a link near the bottom to 3 correct methods for doing this problem. This question gets asked all the time BTW.

12-02-2003, 06:32 AM
An infinitely difficult way would be to do it experimentally by the definition of probability.

1. Shuffle a deck of cards with a perfect shuffle (meaning a shuffle that makes each of the 52! possible permutations equally probable).
2. Remove four cards of one suit, and one card of another suit.
3. Reshuffle the deck perfectly (it's now 46! permutations that need to be equally probable).
4. Deal two cards.

Repeat steps 3 and 4 n times, and let m be the number of times that the cards in step 3 contain at least one card of the same suit as the four previously removed. Now you just have to let n approach infinity and divide m by n.

Hard enough? /images/graemlins/grin.gif

12-02-2003, 01:55 PM
Yup, as a result you need to subtract the odds of hitting on both streets in order to get the correct answer.

12-05-2003, 05:32 AM
The probability of getting the flush on the turn would 9/47. To this you would add the probability of not getting the flush on the turn but then getting it on the river which would be 38/47 * 9/46 so you should make flush draw
(9/47) + (38/47)*(9/46) = .191 + (.809)*(.197) = about .35