Yes - all unbalanced counts can...
By the Unbalanced True Count Theorem:
Net unbalance per deck for K-O: U = +4
Therefore choose Initial Running Count (IRC) = - N_decks * U
Eg for six decks, IRC = -6 * 4 = -24
Unbalanced True Count (UTC) = IRC/(# decks remaining)
Now the Equivalent Balanced Count (EBC) = (K-0) - (U/4))/13 =
(-14,12,12,12,12,12,12,-1,-1,-14*4)/13
= approx (-1, 1, 1, 1, 1, 1, 0, 0, 0, -1*4) [Hi-Lo]
Now UTC = EBTC - U, K-O(UTC) ~= Hi-Lo(TC) - 4
So if you have access to K-O true indices fine, otherwise since the EBC is approximately Hi-Lo (except for the 7's), take Hi-Lo indices and subtract 4 to give True count K-O indices.
For example, when the true count is zero for Hi-Lo, this corresponds to a UTC of -4. That is a neutral deck has UTC = -4, which is obviously true since at the beginning of the shoe, IRC = -24, UTC = -24/6 = -4.
Therefore all the Hi-Lo true count indices we know and love such as Insure at TC=+3, has a TKO UTC of -1. At the Pivot (UTC = 0), this corresponds to a Hi-Lo TC=+4.
Insure UTC >= -1 (TC=3)
Stand 16-10 UTC >= -4 (TC=0)
etc.. you get the picture.
All the same goes to betting.. if you only want to play in ADV>0 hands, this corresponds to a Hi-Lo TC of +2, so UTC >= -2. If you leave at TC=-1, leave at UTC=-5.
Cheers,
Brett.
Cheers,
Brett.