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Re: 9 vs 2 in K.O. Why not a Category B play?(long)

Posted by Pete Moss on 13 May 1998, at 3:16 a.m., in response to 9 vs 2 in K.O. Why not a Category B play?(long), posted by NickTheGreek on 1 May 1998, at 3:49 a.m.

would it be acceptable to take the insurance at a count of 2 on a 6 deck since the corresponding true count would be greater than 3 if no more than 4 decks are dealt?

It's not "acceptable" to calculate that way, but no harm would come from this particular example.

I haven't read Ken's response, so I may be duplicating what he said. In any case, that's not the way you calculate true count. If you start your six deck running count at -20, to calculate true count you must first subtract 4 from the running count, then divide by the number of deck remaining, then if you want the number on the original scale, add 4 back in. It's easier just to start the running count at -24. (See the article above titled "Unbalanced counts to not exist".) That makes your A index zero. If you do that, you can get the true count by just dividing by the number of decks unseen.

Let's solve the particular example: IRC = -20. Running count +2, two decks remaining unseen. Subtract the pivot 4 from the running count +2, giving -2. Divide by the two decks unseen, giving -1. If you want it with the "pivot = 4" convention, now add the +4 pivot back in, giving +3. What you get is a true count of pivot minus one, which makes insurance a roughly break-even proposition with the K-O tags.

Pete


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