Backgammon Cube Handling in Races

Cube Handling in Races

Forum Archive : Cube Handling in Races

Kleinman count

From:   Chuck Bower
Address:   bower@bigbang.astro.indiana.edu
Date:   25 March 1998
Subject:   Re: Is this a take?
Forum:   rec.games.backgammon
Google:   6fbr4v$q98$1@jetsam.uits.indiana.edu

Dan Scoones wrote: > Last night I was playing a "money" session with JellyFish Level 7 and > got to the following position. > > +24-23-22-21-20-19-+---+18-17-16-15-14-13-+ > | O O O O O | | O O O | > | O O O O | | O | > | O | | | > | | | | > | | | | > | | | |[64] > | | | | > | | | | > | | | X | > | X | | X | > | X X X X | | X X | > | X X X X | | X X O | > +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+ > > X (rook) on roll. Cube action? > > The pip count is 84-96 in X's favour. X's lead is just over 14% of > his count. > > In Advanced Backgammon, Bill Robertie states that, in long races, the > trailer has a pass if the leader's racing advantage exceeds 12% of his > pip count. That is the case here. (snip) > Applying the Thorp count yields a different picture. Here, the > trailer's adjusted count minus the leader's adjusted count is exactly > 0, which indicates a take. (snip) > ...One of them must be yielding an incorrect estimate. The question is: > which one, and why? Good question. My experience with races has led me to use the percentage method and the Thorp Count to complement each other. That is, usually they don't both work for the same position. Thorp is best when both players have ALL their checkers borne in (and/or off), in other words-- no checkers in the outfield. The percentage method tends to work well in these longer, smoothly distributed races. As most of you know, I've converted both of these formulas into cubeless game winning chances. I repeat those modifications: W = 0.54 + 2*p (where 'p' is the percentage lead of player on roll as described by Dan above) W = 0.74 + 2*T (where 'T' is the Thorp Count) Here we have p = 1/7 and T = 0, as Dan has already said, giving 82.5% (I have a hangup rounding 0.5!) and 74% respectively. Clearly a disagreement (again as Dan observed). There is another slightly more complicated method which, I believe when fully implemented, works over the full range of both the percentage and Thorp Counts. That is the Kleinman Count. I have yet to learn the FULL Kleinman Count. (Anybody want to loan me some time? I promise to pay you back in my next lifetime....) In it's simple form (for long races like this one), K = (D + 4)^2 / (S - 4) where D is the pip count difference and S is the pip count sum. If you plug the above pip counts into this formula, you find K = 1.45. So what do you do with THAT?? Danny (Kleinman, that is) created a table based up the standard normal distribution. If you memorize the table, then you can convert K to cubeless winning chances. Some people just memorize the "key" money values of 68% (or 70%), 72%, 78%. In tournament backgammon, though, you often want to know winning chances over a MUCH wider range. How do you get THAT?? I've developed an algorithm (which isn't really THAT hard to use, especially if you're willing to learn a couple chapters of Art Benjamin's book "Mathemagics"). I "reveal" the algorithm here for the first time. (A roll of the drums, please!) if K < 1: W = 0.50 + 0.267*sqrt(K) This reproduces the Kleinman Table to 0.5% or better over the range 0.50 < W < 0.75. If you don't like 0.267, then 0.27 works even better except between 73% and 76%, but even there it is off less than 1%. Now, what about K >= 1? Well, I derived a logarithmic fit there (which works up to 95%) but even Art doesn't normally do logarithms in his head! (...though he could if he saw the need, I'm sure.) I then saw that it's not THAT hard to memorize the table if you see the following pattern: (approx.) W K delta K 0.76 * 1.0 --- 0.77 1.1 0.1 0.78 1.2 0.1 0.79 1.3 0.1 0.80 * 1.4 0.1 0.81 1.55 0.15 0.82 1.7 0.15 0.83 1.85 0.15 0.84 * 2.0 0.15 0.85 2.2 0.2 0.86 2.4 0.2 0.87 2.6 0.2 0.88 2.8 0.2 0.89 * 3.0 0.2 0.90 3.33 0.33 0.91 3.67 0.33 0.92 * 4.0 0.33 0.93 4.33 0.33 0.94 4.67 0.33 0.95 * 5.0 0.33 0.96 * 6.0 1.0 0.97 * 7.0 1.0 0.98 * 8.0 1.0 0.99 (does it matter???) (Note: * are "key" values of K in my algorithm.) As above, this part of the algorithm matches Kleinman's table to 0.5% or better. Notice that in the above problem the Kleinman Count says X will win just over 80%. I did 180 level-6 cubeless JF3.0 rollouts and it got X winning 79.7% with an S.D. of 0.2%. Are you impressed with the Kleinman Count? One problem doesn't mean a whole lot, but I've been amazed at how accurate it is in general. So where do you go from here? Well, as I mentioned, this is the original Kleinman Count. There are advances (which don't involved any more square roots!) which make adjustments to the Count for distributional anomolies (similar to the adjustments made in the Thorp Count). That is the part I still need to learn. Kleinman's many books (which are valuable to the serious student) cover the extensions. You can get them (and about everything else on the BG market) for Carol Cole ( cjc@flint.org ). Special thanks to Marc Gray (FlashGammon on FIBS) for urging me to learn the Kleinman Count. Chuck bower@bigbang.astro.indiana.edu c_ray on FIBS

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