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.


brydż, brydz, bridge, brydż sportowy, brydz
sportowy, bridge sportowy, Pikier, Sławiński, Slawinski, Łukasz Sławiński,
Lukasz Slawinski, 
All
restrictions on bidding must be destroyed