True fudging, by a practitioner
Playing an unbalanced count against the AC monsters, I developed some approximations that may be what MathProf means by "true fudging."
I play the Red 7, which is unbalanced by 2 points per deck. At the pivot (twice the number of decks in play, i.e., a running count of 16 if you're playing an 8-decker), the player has gained about 1% over the initial situation. In AC, you start with a house advantage of just under half a percent, so at the pivot you're up by about half a percent. I would Wong in with $50 at this RC.
The problem I faced was what to do if the RC went down. The Wonging is heaty enough without sitting out every time a few tens appear, then jumping back in if the count rebounds. On the other hand, to keep playing when you're below pivot could be costly, because you could be at negative expectation, depending on where in the shoe you are. I wanted a rough guideline for the situations where the RC was below pivot.
The first issue was to identify negative expectation situations. If the house starts out half a percent ahead, then a Red 7 player is at roughly an even game when the running count equals the number of decks in play (8 in an 8-decker) plus the number of decks in the discard rack. (Explanation if you don't want to take my word for it: Let c be the Red 7 running count, n the number of decks in play, and d the number of decks in the discards. A High-Low counter, who wouldn't be counting the red 7's, would have a running count that's lower than yours by the number of red 7's seen, which will be, on average, 2 per deck. So you estimate the High-Low running count as c-2d. The High-Low true count would be (c-2d)/(n-d). Each High-Low true count point is worth about half a percent to the player, so your current percentage advantage, taking account of the initial 0.5% house edge, is ((c-2d)/(n-d))*0.5 - 0.5. Set that expression as equal to zero, to find the break-even point, and it simplifies to c=n+d.)
You don't have to go through the math every time. Just remember that you're approximately at break-even if c=n+d. If your RC is at that level, then bets you make to hold your place at the table have zero expectation. The only price you pay is the increase in your variance.
Using similar reasoning, I calculated the "safe at quarters" level, where I was below pivot and didn't want to go to $50 but had enough of an advantage that I was happy to bet $25. Rounded off for ease of use, the guideline I developed for 8-deckers was: With less than 2 decks in the discards, 13; with 3 or 4 decks in the discards, 14; with 5 or 6 decks in the discards, 15.
MathProf pointed out that a RC above the pivot is more valuable (i.e., shows a higher advantage) when it arises late in the shoe. A quirk of unbalanced systems is that a RC below the pivot is more valuable early in the shoe. Consider my "safe at quarters" example, where the pivot is 16: If the RC is 14, that's enough of an advantage for quarter betting near the beginning of the shoe, but if you've seen five decks or more, your advantage is so small that you don't meet that threshold.
The following generalization is true for all the standard unbalanced counts and for any number of decks: At the pivot, your edge is always the same, regardless of how many cards have been dealt. If you're not at pivot, the impact of the difference from pivot (positive or negative) is magnified as the number of undealt cards dwindles.
The approximation c=n+d assumes a one-level count, unbalanced by two per deck, and a starting house advantage of 0.5%. For other counting systems or other game conditions, the fudging formula would be different. If you want to go beyond these crude guidelines, there's a much more extensive discussion of use of a true-count equivalent in Blackbelt in Blackjack, by Arnold Snyder (2nd edition), available for sale at this very site (click here). It's keyed to the Red 7, but it will acquaint you with the principles of adjusting for deck penetration while using an unbalanced count.