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Carol Joy Cole
3003 Ridgecliffe Drive
Flint, MI  48532
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Dear Carol:

The following is for your invaluable little newsletter.  I hope it is worth
another year's free subscription.

\noindent{\it The Ultimate Match Equity Formula}

I don't know of any published formulas that give accurate equities in
long matches.  Here is a formula (well, really a formula and a short
supplementary table) that agrees with Robertie's computer-generated tables and mine
within about 1\% for all situations arising in 25-point matches, except for
match-point situations and one other score.  I have reason to believe it is
equally accurate for much longer matches.  The formula is illustrated
with a running example of a 4-0 score in a 15-point match.

{\obeylines
Let $A$ be the number of points the leader needs.
$(A=11)$

Let $B$ be the number of points the trailer needs.
$(B=15)$

Let $S = A+B-1$, the maximum number of points still to be played.
$(S=25)$

Let $L=B-A$ be the lead.
$(L=4)$

Let $R=L/\sqrt S$ be the lead relative to the length of the match.
$(R=4/5=0.8)$

If $R$ is less than 1, the leader's chance of winning the match is $W=0.5+R/4$.
$(W=0.7=70\%$; Robertie's tables and mine agree).}

If $R$ is greater than 1, estimate $W$ by rough interpolation in  this table:
\vskip 1in
For example, at 8--0 in a fifteen point match,
$$A=7, B=15, S=21, L=8, R=8/\sqrt 21 = 1.75,$$
which is a little closer to 1.6 than to 2.0, so $W=87\%$.  Robertie's tables and 
mine say 88\%.

To mentally calculate the square root of $S$, start with a whole number 
estimate; if $S=21$, the estimate is 5.  Then average 5 and 21/5; (5 + 4.2)/2 = 4.6,
which is close enough.  If you can't handle the mental arithmetic, just use the
nearest whole number to the square root of $S$; the resulting error is
about 1\% of equity.

There is no way to avoid the square root, the division, and the table.  Any formula
good for very long matches has to agree with the statistical laws of large
numbers, and these laws have square roots in denominators.  They also involve
the dreaded {\it error function}, for which there is no elementary formula;
thus the table.

When $A=2$, $B=3$, the formula gives 62\%; the correct figure is probably 59\% or
60\%.  When $A=1$, post-Crawford, let $N$ be the number of games to reach double
match point; $N$ is $(B-1)/2$ with remainder discarded.  Then, assuming 20\%
of the trailer's wins are gammons, his change is $0.6↑N/2$.

For example, if $A=1$, $B=8$, then $N=3$, $W=0.6↑3/2 = 10.8\%$.

Here are the equities, $W$, used in the analysis of a Sylvester/Glaeser game in 
the November 1988 FAB, along with the equities, $u$, computed with the 
Underwood sequence.
\vskip 3in
The 2\% discrepancies are generally quite significant in such an analysis.  However,
in this case their effects largely cancel out, and the conclusions stand.  I
suspect the Underwood sequence is inaccurate in long matches.

\closing
Sincerely,
Robert W. Floyd
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