Question for Rob McGarvey

Rob McGarvey

Well-Known Member
#2
It's not shaping up, it's complete. The 5 is +3 and the 9 is -1. The A is still +2 when rich. Gets you into the game quicker and out quicker when you are wonging in or out.
 

Rob McGarvey

Well-Known Member
#4
I find the RoR (SBA sim) goes up with the +3 for the 5 and the -1 for the 9, but so does the RoR...risk of reward. ;> If you ever lose 10% of your BR, you size down to protect yourself. Then you control your RoRuin and keep it to 0% Win 10% size up.
 
#5
The level-3 beheemoth...

...that the Robster just discribed is neither practical or nor more powerful than the standard HO2 - Uston said of his own Ace-neutral/level-3 'UAPC' "its obsolete" (Uston on BJ'86) - of course I may be mistaken because I haven't compared exhaustive sims of these McGravey hybrids - no pro in the world would use them, and no novice should consider them. zg
 

Rob McGarvey

Well-Known Member
#6
Learning from the level-3 beheemoth

I myself backed off of the Uston APC with it's 3 level count and stuck with Hi Opt II and have refined it ever since. All of these counts work, but what is most important to me is what we can learn from the effect each count has on the bottom line. Example: Hi Opt III, Uston APC, no matter, gets you into the game and out of it if you are wonging out. You can do the same thing by dropping out at -1TC with Hi Opt II instead of -2TC, and bet slightly more during + counts to get the exact same effect as these higher level counts can.
 

Rob McGarvey

Well-Known Member
#8
Re: Half a Bee-Moth

The Mayor himself uses (do you still use this count Eliot?) a level three count, Wong Halves which uses tags of .5 , 1.0 , and 1.5. It is called halves because these level three numbers have been cut in half from 1 2 and 3.
 

T-Hopper

Well-Known Member
#11
Re: Halves

Halves is also nearly as strong as AOII in double deck games with bad penetration and a large spread. You just don't get to make that many big strategy variations in the first 52 cards.
 
#12
Ace +2

Rob was describing his hi opt III.
When he said Ace is still +2 when rich, did
he mean that you add +2 to the true count for
every ace that is left over per one deck?
 

Sonny

Well-Known Member
#13
Not quite

> Rob was describing his hi opt III.
> When he said Ace is still +2 when rich, did
> he mean that you add +2 to the true count for
> every ace that is left over per one deck?

Almost. You add +2 to the RUNNING count for every extra ace. If you have a two-deck segment with five aces in it you would add +2 to the running count before you convert to true count. Likewise if there were only three aces you would subtract 2.

-Sonny-
 
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