Zar Points by Zar Petkov Part 3

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What kind of weight to put on the components you consider valuable is not a matter of "expert judgment", but a…

By Zar Petkov
On 5 February, 2014 At 16:24

Category : Advanced @en, Hand Evaluation, Zar Points @en

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Zar Petkov
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Never miss a game again – Zar Points Bidding  Source: BridgeGuys Read Part 2

 Which HCP count is “mathematically correct”

What kind of weight to put on the components you consider valuable is not a matter of “expert judgment”, but a simple matter of solving a series of equations with unknown coefficients – an obviously overdetermined system of equations (you enter hundreds of equations based on the hundreds of boards you feed in, for finding the value of several of coefficients – the weights you are interested in).

It is a well-known fact that the standard 4-3-2-1 valuation IS the one that solves the system of equations when using any of the standard distribution-points systems (Goren, Bergen, etc.) and it is also known that the HCP Controls valuation (the 6-4-2-1) is NOT a solution with the standard distribution points.

How do you create the equations for a specific board in order to calculate the “right” weights? We’ll count HCP points and distribution points for void, doubleton, and singleton (kind of Goren style).

X_aces(a) X_kings(k) X_queens(q) X_jacks (j) X_void (v) X_singl(s) X_doublton(d) = X_total_points_for_game

where a is the specific number of Aces in both hands of this deal, k is the specific number of kings etc.

So for the board:

Q 10 x x
A x
x x
K Q x x x

K J x x x
K x x
x x x
A x

the equation will be= X_aces * 2 X_kings * 3 X_queens * 2 X_jacks * 1 X_doubleton * 3 = X_total_points_for_game

You make a collection of hundreds and hundreds of boards that have a game (4 in major) and solve the overdetermined system of equation to find the values of the unknown coefficients. Simple.

For the board above (4 Spades), if we consider the plain 4-3-2-1 Milton Works points, assigning 4 for A, 3 for K etc. and assign Goren distribution points (3 for void, 2 for singleton, and 1 for doubleton), we see that those ARE a solution for our first (and only for the time being) equation:

4*2 3*3 2*2 1*1 1*3= 8 9 4 1 3 = 25

so you have to “collect” 25 points to get a game with Milton / Goren points. The same way we have run the systems for the Zar Points which have much more variables to calculate.

While on the subject, you may come up with some “igneous idea” that in bridge it’s only the Kings and Jacks that count (because they are the “male cards” :-) and construct the corresponding equations. And you WILL find corresponding solutions for the coefficients – so the natural question is “Why not?” The answer is that such a solution will have much bigger deviation for the actual equations (well, I’ll stop here :-).

We will do an exhaustive comparison between Goren points (of Charles Goren), Bergen Points (of Marty Bergen), Drabble Points, and Zar Points in the second half of this article so you’ll get the picture.

You already might have guessed WHY the 4-3-2-1 valuation solves the equation system for the standard distribution-points systems while it does NOT solve the equations for the Zar points (the 6-4-2-1 is the one that does).

The reason is the relative weight of the distributional points vs. the weight of the honor points. As we have seen, the distributional Zar points range goes to 26 (for the extreme case of 13-0-0-0) while the standard distributional points range goes up casino to 13 at most (see below) – cannot compensate the weight of the 6-4-2-1.

Note also, that the experts know that the 4-3-2-1 is a “twisted” solution, meaning that it undervalues the A and K and overvalues the Q and J – that”s why they use fractions to “make it work” (they count A for 4 1/2 while Q for 1 1/2, which makes three Queens equal to one A, just like the 6-4-2-1 valuation).

Zar points allow the natural 6-4-2-1 honor count (which experts lean towards) to be the solution of the overdetermined system of equations.

Your judgment

Do you still need to apply your judgment and consider both hands of the partnership in the bidding process? You bet. Here is a simple example which covers both cases – your own hand and the combination of the two hands in the partnership.
Let”s consider two different hands in opening position, with the same 26 Initial Zar Points, 10 HCP, 3 Controls, 5-4-3-1 distribution. Which one do you like better?

Norte A
8 6 5 3 2
7 5 4 2
A Q J
K
Norte B
A J 10 8 4
K Q 10 9
10 9 6
4

I guess you have a preference here :-). Certainly hand “A” will be greatly downgraded from the initial 26 Zar Points while the second hand “B” will be upgraded for a number of reasons. BUT, let”s consider the Partner”s hand in both cases and how this dramatically changes the picture. In both cases the partner has 27 Zar points – 10 HCP, 4 controls, 5-4-3-1 distribution.

Sur A
A 10 9 7 4
9
10 8 6 5
A Q 6

Sur B
K
8 7 6
A 7 4 2
K 9 7 5 3

No comment needed – you”d reverse your “preferences” and you would prefer the set “A”. In bridge, you always need your head on your shoulders, at any stage of the game :-) This brings us to the next section where we consider the adjustments to the partner’s and opponents bidding.

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