Wednesday, August 29, 2007

HUC5, PLO and Hand Combos

I lost in two straight games to lifesagrind in the Heads-Up Challenge on Full Tilt. Lifesagrind is one of the more aggressive heads-up players I've faced, and he kept me at a disadvantage most of the time. Then he finished me off when I held QQ and KK, so what can you do.

I really prefer PokerStars's heads-up games that have no blind increases. The rising blinds really alter the feel and style of heads-up play, and they make for shorter matches.


I've still been playing some Omaha, and it's a really fun game when I win. When I lose, I feel like a lost idiot. One cool part of it is that you can get people's money when they have the worst of it by far, something that seems harder to accomplish in hold 'em:
pokenum -o 7d 6s 4d 6c - ad qc qd js -- tc 5d 8d 9c
Omaha Hi: 40 enumerated boards containing Tc 9c 8d 5d
cards win %win lose %lose tie %tie EV
6s 6c 7d 4d 1 2.50 39 97.50 0 0.00 0.025
Js Qc Ad Qd 39 97.50 1 2.50 0 0.00 0.975
That was my biggest PLO hand so far, an $800 pot in my first time playing 2/4. And no, my opponent didn't suck out the straight flush.

I'm still pretty bad at this game, but I just need practice. I think it'll help develop postflop skills in general.


Roy Cooke wrote an article in this month's CardPlayer that discussed a topic I have wondered about in the past. He discusses considering the odds that your opponent holds one of a limited number of hands that your opponent may have based on a narrow read. It should be easy enough to memorize and apply:

Keep these numbers in mind for hand-reading, and use them when defining your opponents' hands and reckoning your price. There are six possible combinations of any pair; for example: J J, J J, J J, J J, J J, J J♣. If any one of them is accounted for in your hand or on the board, the number of possible combinations drops to three; for example, if you hold the J: J J, J J, J J. If two are accounted for, there is one possible combination. There are 16 possible combinations of two unpaired cards, including the suited ones; for example, A-K or J-10. If one is accounted for - for example, you hold a king and are analyzing the possibility of your opponent holding A-K - the number of possible combinations drops to 12.

So, the odds that your opponent has:

Any pair: 6 possible combinations
Any pair with one card accounted for: 3 remaining combinations
Any pair with two cards accounted for: 1 possible combination

Two unpaired cards (like AK): 16
Two unpaired cards with one accounted for (like AK with one King out): 12
Two unpaired cards with two accounted for (like AK with one King and one Ace out): 9
Two unpaired cards with two accounted for (like AK with two Kings out): 8
Two unpaired cards with three accounted for (like AK with three Kings out): 4
Two unpaired cards with four accounted for (like AK with three Kings and one Ace out): 3


HighOnPoker said...

Cooke has an interesting theoretical article, but I can't seem to figure out how to use it to practical effect. Any ideas?

Gnome said...

Here's an example from an old 2+2 thread: 3/6 Set Kings turn play (
I private messaged a longtime poster who commented in that thread about counting these probabilities. Here's part of his response:

"As you point out, when hand ranges become wider it makes counting combos much more difficult. That's why on an overwhelming % of hands counting combos just isn't practical. But when you can limit your opponent's possibilities to maybe just two hands (like in that hand), then counting combos becomes helpful.

The easiest way to do it for me is just to figure out how many of one card is left, how many of the other is left, then multiply. So if we are trying to figure out how many AK combos are left, we multiply the unknown Aces by unknown Kings.

Obviously, always remember to discount based on the action/situation.

This is something that practice really helps with. You'll find that it becomes easier and easier the more you play/read the forum."

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Welcome to the Loser's Lounge!