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. 7476).
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 innerboard point
occupied. E.g. if X had one gap and O had none, you'd subtract 1 from
O because O occupied 6 innerboard points and X occupied 5. Caveat:
Ward says gaps must be treated on a casebycase 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 redoubletake, whereas highpower racecrunching
software evaluates the position as redoublepass, 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 gapadjustment
(Ward would recommend subtracting 0 or 1 because the high gaps
are unlikely to hurt O much; maybe he'll end up with a fastdecaying
5point 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
