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View Full Version : What are the odds I have the best hand?

Mikey
09-29-2005, 04:33 AM
I've been doing a little thinking about something but I'm not sure if I can figure out this.....

Odds of having the best hand.

Now lets assume I have in my hand 9d9c.

The board reads 6h 3h 9s 2h Js.

Now I'm trying to figure out what are the odds I have the best hand here.

I see that there are 59 possible combinations that beat me.

9 flushes with the Ace
8 flushes with the King
7 flushes with the Queen
6 flushes with the Jack
5 flushes with the T
4 flushes with the 9
3 flushes with the 8
2 flushes with the 7

Never counting a specific flush twice....

There are also

12 possible straights with the 45.

There are also

3 possible holdings of JJ which is higher than my set.

That means there 59 possible hands that are higher than mine.

So how do I know what type of a favorite I am here to hold the best hand?

I always here of talking of how I should call a river bet here because I'm getting x to 1 from the pot if this is a Limit HE game.

Is this the way to figure out your odds you have the best hand compared to the pot odds you are getting?

Luzion
09-29-2005, 05:07 AM
I think how everyone does it is pretty much the standard; which is including any reads you have on your opponent. Its sort of ridiculous to assume your opponent could be holding all possible suited heart combinations. Put your opponent on a range of hands that he could possibly have. You cant really assume your opponent could always be holding hands like T3s or 82s can you?

You use logical guesswork to eliminate some possibilities. You can then sum up all the possibilities by using C(N,r) or just manually counting. Then you can simply figure out the hands that beat you and divide by the range that he could possibly have.

LetYouDown
09-29-2005, 09:09 AM
I come up with 63: 44 flushes, 1 straight flush, 15 straights, 3 trip Jacks.

So if you're up against a random hand, the probability you have the best hand is:

1 - 63/C(45,2) = 93.6363...%

As Luzion noted, logic factors in here, but this should be right for one random hand.