PDA

View Full Version : Holdem Math

SteveS
01-07-2003, 04:17 PM
Is the expectation of catching an A or K on the flop in holdem if you have A/K:

6/50 + 6/49 + 6/48 = 18/147 = 12.25% ?

If not please help me understand the math.

Same question as it applies to Omaha if you hold A2XX, what is the chance you would be counterfeited on flop? ...on the next two cards?

Is it:

6/48 + 6/47 + 6/46 = 18/141 = 12.75% ?

And then to end: 30/230 = 13.0% ?

If not, please show me how to calculate this. I would really like to learn the math.

pudley4
01-07-2003, 05:03 PM
First of all, when adding fractions, you do not add the top numbers and bottom numbers together.

Second, the easiest way to figure out most of these questions is to figure the probability of not catching on the flop, then subtract that number from 1.

The probability of not catching an A or K on the flop if you have AK is:

44/50 * 43/49 * 42/48 = .676

Subtract this from 1 and you get .324 = 32.4% chance at least one A or K is on the flop.

pudley4
01-07-2003, 05:21 PM
Here's the reasoning behind why it's usually easier to figure out the probability of not hitting:

We want to know how often at least one A or K hits. There are a number of different combinations of Aces and Kings. There could be 0 Aces and 0 Kings, 1 Ace and 0 Kings, 0 Aces and 1 King, 2 Aces and 0 Kings, etc etc etc. In all there are 9 different combinations of Aces and Kings, but only 1 way to have 0 Aces AND 0 Kings.

In order to figure out the probability of the board showing at least one Ace or King, we could calculate all of the probabilities for all boards that have at least one A or K and add them all together. But this means 9 probabilities to calculate.

To make it easier, we just figure out the probability of 0 Aces and 0 Kings. If the board doesn't have 0 Aces AND 0 Kings, then it has at least one of them. So we subtract our number from 1 (because adding all probabilities together equals 1) to get our answer.

cero_z
01-09-2003, 05:02 AM
Steve,
Hold'em's Odds Book, by Mike Petriv, is a great reference that explains hold'em probability in simple terms.
He shows the odds of flopping exactly one pair with AK thusly: Total # of 3-card combos (flops to AK)=50*49*48/3*2*1=19600.
Take out 4 Aces &amp; 4 Kings, and the # of 2-card combos=44*43/2*1=946.
Each of those 2-card combos can have as its 3rd card one of the 6 Aces or Kings you don't hold, so total # of ONE PAIR flops (Kxx or Axx)is 6*946=5676.
Probability of one pair on the flop w/ any 2 cards of different ranks (like AK) is 19600/5676= 29%!
I hope you can follow this; I've shortened it to reduce my typing. His version is very easy to follow.

BruceZ
01-09-2003, 08:21 AM
Probability of one pair on the flop w/ any 2 cards of different ranks (like AK) is 19600/5676= 29%!

Make that 5676/19600. An even easier way to do this using just fractions is to figure the probability of exactly a pair with the first card on the flop being A/K is (6/50)(44/49)(43/48) and then multiply all this by 3 since the A/K can occur in any of the 3 positions, to give 29%.