## Ł.Slawinski 1982

PREX (2) = Play Result EXpectation

from Pikier 9

Ending the first part, I remarked that more accurate honor–point counts  than the 4–3–2–1 count exist. Now, I present one of them.

 Honor equations

Let's assume that point–count is to be done by WE pair, and (for the time being) – each hand is totally balanced (4333). In each suit one may receive one of the following 15 combinations:

 N N = Nothing (only small cards) A   K   Q   J AK   AQ   AJ   KQ   KJ   QJ Of course, each combination has its own HT value. The problem is to determine it. AKQ   AKJ   AQJ   KQJ

The values to be determined should fulfill the following condition:

For each suit:

( W's points) + (E's points ) = the number of HT really possessed by WE pair

In other words, the following system of 40 equations should be fulfilled:

 W + E = WE tricks 40 (equations) is much more than 15 (honor combinations). Algebra tells us that in such a case the system has no solution. So is in the case; it may be seen, for instance, from the subsystem of the first four equations. So, what to do ? N + N = 0 A N = 1 N + K = 1/2 A K = 2 AJ K = 2 1/2 AQ K = 3 ...and so on

 The solution !

As the system is contradictory, we should do what is frequently done in bridge:– abandon accuracy and be satisfied with an approximate solution (permitting some, possibly small, difference between the left and the right side of each equation).

In addition (to simplify the calculations) we'll try to obtain a rounded solution (a quarter of a trick (1/4) will be the unit).

Two hours of elementary calculations lead us to three, almost equivalent, approximate solutions. Here's the simplest one and somewhat better than two others (the values are expressed in quarters of trick):

 A = 4 The „HH = +2” means that for a two–honor combination 2 quarters should be added to the total value of the honors. Similar addition is in order for a three–honor combination – ie 3 quarters should be added. K = 3 HH = +2 Q = 2 HHH = +3 J = 0

Examples:

 KQ 3 for King AKJ 4 for Ace 2 for Queen 3 for KIng 2 for HH 0 for Jack total 7 quarters = 13/4 HT 3 for HHH total 10 quarters = 21/2 HT

 Errors

Since the approximate evaluation has been applied, some equations are burdened with error: positive – if the HT result is greater than the real strength of the suit (left side > right one), negative – in opposite case.

Here is the statistical breakdown:

 Error No of equations The value of error doesn't exceed one quarter (except of „AQ + N = 0” equation) –1 10 0 17 +1 12 +2 1

 Anihilation of errors

Each deal has four honor equations for Îë pair – one for each suit. The existence of positive and negative errors (they are almost evenly distributed) gives an additional chance:–  although each suit would have an error, the total error may be closed to zero.

For example:

 West East Count Real Error xxx QJxx 4 3 +1 AKx xxx 9 8 +1 Kxx AQx 11 12 –1 Kxx Axx 7 8 –1 31 31 0

 The 10 and the 9

Combinations with the 10 or the 9 can be evaluated by solving an auxiliary system of equations.

Eg for Q109:

 W E WE Thus, the optimum value of D109 is 3 quarters (with maximum error = 1 quarter) Q109 + N = 2  Q109 = 2 Q109 + J = 4  Q109 = 4 Q109 + K = 6  Q109 = 3 Q109 + A = 7  Q109 = 3 Q109 + KJ = 8  Q109 = 3 Q109 + AJ = 10  Q109 = 4 Q109 + AK = 12  Q109 = 3 Q109 + AKJ = 12  Q109 = 2

Exercise: Calculate the values of remaining 10–or–9 combinations.

 Short honors

Singleton or doubleton honors have usually somewhat smaller value.

Here's an example of calculating the value of the singleton king:

 W E WE Thus, the optimum value of singleton king is 2 quarters (with maximum error = 2 quarters). K + N = 0  K = 0 K + J = 0  K = 0 K + Q = 4  K = 2 K + A = 2  K = 4 K + QJ = 8  K = 4 K + AJ = 8  K = 2 K + AQ = 12  K = 4 K + AQJ = 12  K = 3

Exercise: Calculate the values of remaining singleton and doubleton combinations.

 Historicalities

The idea of honor equations has been originated by Pierre Collet

in his „Introduction au bridge scientifique”, ..1963 ]

His solution was:

 A = 4 N = –1 N = Nothing (only small cards) K = 3 Q = 2 HH = +2 J = 0 HHH = +4

Pikier’s solution is easier and only a very little bit worse.

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