[Buzz]

[ QUOTE ]
you are about 33 to 1 to get dealt 3 unpaired wheel cards.

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Mike - Not a big deal, but I think 34.5 to 1 against getting dealt three unpaired wheel cards is correct.

Here's my math:
52*51*50/(20*16*12) = 34.5.
(Both the numerator and denominator of 52*51*50/(20*16*12) are actually divided by six, but the sixes cancel).

Doesn't change your conclusion. Waiting for three wheel cards is too tight.

Here's waiting for three low cards, six or better:
52*51*50/(24*20*16) = 17.2.

Here's waiting for three low cards, seven or better:
52*51*50/(28*24*20) = 9.9.

I think you'll be playing about one hand out of ten if you wait for three low cards, seven or better. That seems reasonable.

Trouble is, the person with the lowest card showing has to make a forced "bring-in" bet. And the forced "being-in" bet complicates things.

In a ten handed game, you figure to have the lowest card showing one time out of ten.

Assuming you have the low card showing, and assuming it is a seven or lower, and not considering the cards your opponents have showing, you figure to only have two more low cards, seven or better, in the hole only about one hand in five.
Here's the math:
51*50/(24*20) = 5.3.

So you're bringing it in one time out of ten, but of these times you bring it in, assuming you want to stick to your standard of seven or better, you'll only want to continue one hand in five. The other four times out of five when you bring it in, you won't like your hand much.

Geez, thinking a bit ahead of what I'm typing, this game sounds like a bluffers paradise!

But enough! Omaha-8 is my game. I'll put this in my razz file and if I ever have to play razz, maybe I'll finish it. Better yet, I'll re-read Sklansky on razz (found in the book <font color="white">_</font>Sklansky on poker). That held me in very good stead the last time I had to play razz (as part of a HORSE tournament).

Buzz