Backgammon Cube Handling in Races

Cube Handling in Races

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Ward's racing formula

From:   Marty Storer
Address:   martys@juliet.ll.mit.edu
Date:   24 January 1992
Subject:   Ward's racing formula
Forum:   rec.games.backgammon
Google:   MARTYS.92Jan24103635@escalus.juliet.ll.mit.edu

A few people have asked me to post Jeff Ward's racing formula (from Jeff Ward, _The_Doubling_Cube_In_Backgammon_Vol._I_, Aquarian Enterprises, San Diego 1982, pp. 74-76). Here it is: Let's say the side on roll has the X checkers, the other side the O's. Ward's formula gives predictions as to whether X should double or redouble, and whether O should take. First take the raw pip counts: call X's pip count X, O's pip count O. Next adjust X and O as follows: add 2 for each checker on the 1 point beyond the second, and add 1 for each checker on the 2 point beyond the second. I.e. "extra" checkers beyond the second (the third checker, fourth, etc.) count for 3 pips in the adjusted count. Then, for either X or O, subtract 1 for each extra inner-board point occupied. E.g. if X had one gap and O had none, you'd subtract 1 from O because O occupied 6 inner-board points and X occupied 5. Caveat: Ward says gaps must be treated on a case-by-case basis, because some gaps are more harmful than others. For example, he wouldn't adjust for a gap on the 3 point with the 6, 5, and 4 points evenly filled, because 3's play well to fill the gap anyway. Then add 2 for each extra checker on the board. Then add 1/2 for each extra checker outside the inner board. Then multiply X's adjusted count by 1.1. Now you have both sides' final adjusted counts: call them X' and O'. If X' <= O'+1, X should redouble. If O'+1 < X' <= O'+2, X should double but not redouble. If ((X+O) / 2) >= 100, O takes if and only if O' <= X'+4. If ((X+O) / 2) = 75, O takes if and only if O' <= X'+3. If ((X+O) / 2) <= 50, O takes if and only if O' <= X'+2. Ward doesn't say anything about the cases 75 < ((X+O) / 2) < 100 and 50 < ((X+O) / 2) < 75; I guess you linearly interpolate in the absence of any information. At an average raw pip count of 62.5, O can take if and only if O' <= X'+2.5, I suppose. There are some problems with Ward's formula, as with any other: (1) it's not clear how to adjust for gaps (Ward gives a couple of examples, but not enough); (2) the adjustment for extra outside checkers doesn't penalize the side with the extra outside checkers enough. For example, applying Ward's formula to Problem 24 from the last two issues of Inside Backgammon predicts redouble-take, whereas high-power race-crunching software evaluates the position as redouble-pass, worth about 1.08 x cube for the strong side (cited by Goulding, "Inside Backgammon", Vol. 2 No. 1, p.17): 01 02 03 04 05 06 | | 07 08 09 10 11 12 O O O O | | O O O O O O | | O O O | | O | | O | | | | | | X | | X X | | X X X X X X | | X X X X X X | | 01 02 03 04 05 06 | | 07 08 09 10 11 12 X on roll: cube action? Even if you subtract 2 pips from X's count for the gap-adjustment (Ward would recommend subtracting 0 or 1 because the high gaps are unlikely to hurt O much; maybe he'll end up with a fast-decaying 5-point gap), we get X' = 58.3, O' = 59, predicting a take. Recent work by Kleinman has shown that extra crossovers hurt very much, in general. In using Ward's formula, I used to add 1/2 pip per checker, plus another 1/2 pip for each 4 pips away from the home board each outside checker was: even this didn't penalize enough for extra outside checkers. But Ward's formula is pretty good otherwise. Seems to be better than Thorp's, though a little harder to use. And any formula is much better than nothing. Hope this helps. Play well, Marty

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