Titaniumman: Request for permission from John Auston

[Titaniumman]

John, a little more than a year ago, you answered a question on Don's Domain by explaining how to true count KO. I told Wildcard about this, but the post appears to no longer be available.

Wildcard is a KO player, and would like to have this info. I only have a hard copy that I printed out last year. With your permission, I would like to re-type and post your explanation.

Don may want to consider archiving the post, and may feel free to edit out my name from the post (since I have no claim to its brilliance) should he so desire to archive.

[John Auston]

[Titaniumman]

Originally Posted by: John Auston
Original Date: Sunday, 15 July 2001, at 9:15 p.m.

In response to: A note to John Auston (Nick)

A while back you had indicated to me that using Vancura & Fuchs Knockout system, that in a 6 deck game which starts with an IRC of
-20 that at -4 it was nothing more then a coin flip. John as far as you are concerned at what count are the cards decisively in my favor using this system [sic]

You can figure this out for yourself once you realize that unbalanced counts can be true counted, just as balanced counts can, and even though you still play them by running count alone.

Just do the following, using 6 decks as an example.

Use IRC = # of decks * 'the unbalance per deck'

This is 6 (decks) times -4, for an IRC of -24.

Balanced counts use the same formula, only the second factor is always 0, so the answer (IRC) is always 0.

Now, at the start of the shoe, the KO true count is -4 (-24 divided by 6 decks left). This roughly equates to a hi-lo TC of 0. A KO TC of -3 would then be about a hi-lo of +1, a KO -2 would be about a hi-lo of +2, a KO -1 is about a hi-lo of +3, and a KO TC of 0 would be about a hi-lo TC of +4.

If we use a hi-lo of +2 to indicate "cards decisively in my favor" (a hi-lo +1 being the first small edge), then this is a KO True Count of -2.

Now lets see what the KO RUNNING COUNT would have to be, at say, 4 decks left, to be a KO TC of -2. Well, an RC of -8, right?
Because TC is RC/decks left, so RC of -8 / 4 (decks left) = -2 TC.

Now, with an IRC of -24, to get to -8 requires a net +16. If you use -20 as IRC, instead of -24, then -20 + 16 = -4. So with 2 decks gone (4 to go), an RC of -4 would be where the "cards are decisively in [your] favor".

A different RC would be the answer at 3 decks gone, 4 decks gone, etc. But in order to not actually True Count, a single RC has to be used, so the system authors pick one and "recommend" it.

But now that you know the mathe behind it, you can pick/decide on your own how to play it.

-John Auston

[Wildcard]