BBO Discussion Forums: How to calculate the distributive strength of the hand? - BBO Discussion Forums

gergana85, on 2014-April-22, 00:49, said:

I doubt that the formula is causing anyone much trouble.

The lack of interest probably reflects the fact that the forums spent a better part of a year trying to explain why Zar Points were badly flawed and no one wants to revisit the same discussions.

If you're genuinely interested, i recommend reading the (extensive) discussion of Zar points and consider some of the comments there (especially the ones involving how best to measure the accuracy of a hand evaluation method)

gergana85, on 2014-March-04, 07:46, said:

VERY GOOD ARTICLE ! !H!H!H

Don't get discouraged by comments by the BBO Experts.
They do that with every newcomer.

hrothgar, on 2014-April-22, 14:39, said:

1. I was not involved in the discussion of the Zar Points method and it had no effect on me.
2. I think that Zar Points method is incorrect and it is easy to demonstrate. See the article by Mr Petkov (http://www.bridgeguy...f/ZarPoints.pdf):

- In the Mr. Petkov’s formula is included the sum of the two longest suits. This is true, but why he did it, the author does not say. Perhaps the experience and the feeling are helped him. No evidence;
- Mr. Petkov completely frivolous includes in his formula the sum of the difference between the longest and shortest suits. To prove his point, he using some strange mathematical formula that does not even deserve a comment. And again, no evidence. I'll just say that if I use his logic, I can to prove that the difference between the longest and shortest suits can be replaced by the difference of any other two suits. For example let a, b, c and d are the lengths of the suits in the hand. I argue that instead of (a-d), may be included (b-d), as (b-a) + (a-c) + (c-d) = (b-d). Or (a-b), because (a-c) + (c-d) + (d-b) = (a-b). These examples show that the purpose was not to find the truth, but to get a "beautiful" formula.

3. Unlike Mr. Petkov, I do not propose a model to determine the general strength of the hand. I only argue that the distribution strength of the hand depends on the sum of the two longest suits. I also show that in rare cases (wild distributions) the three longest suits affect on the distributive strength of the hand. Based on this evidence I do some conclusions. Nothing more.
If you would like more information email me at: bogev53@abv.bg

Lurpoa, on 2014-April-22, 23:12, said:

Thanks.
They cannot scare me because I speak proven truths. Moreover - they help me a lot.

I might be able to prove that the max number of losers in a hand is 13, but it is tough.

gergana85, on 2014-April-10, 05:21, said:

Lreal = 13 - (tricks quitted my way) = tricks quitted their way.

Wow, you're better than I am matmat - Lreal = 13 - (tricks quitted my way) + 1 for the misguess + 1 for bad play - 1 for so bad play that the defence misdefended on the "that can't be right" ticket...

gergana85, on 2014-April-24, 07:59, said:

I couldn't care less whether your method is related to Zar Points.

I referenced that discussion thread because there are some good postings that describe how one might validate a claim that "the distribution strength of the hand depends on the sum of the two longest suits".

Am I the only one who thinks that it almost irrelevant that the Maths makes sense? Putting all the parts of the equation together, you get (with a little simplifying):
Lmax = 19 – S1 - S2 – ((|10 – S1| - (10 – S1)) - (|3 – S2| - (S2 - 3)) - (|3 – S3| - (3 – S3))/2

The good players will already know what their hand is worth as far as distribution goes, and the vast majority of the rest would not be willing to learn that, let alone apply it. While it does seem to work, it all seems accurate in the limited distributions I did check it with, it all seems pretty academic.

matmat, on 2014-April-24, 08:32, said:

It is not trouble. Do not even needs to prove. Take this hand:

♠xxxx
♥xxx
♦xxx
♣xxx

How many losers in it? Obviously they are thirteen. But the question is another, and that is that each distribution has its own maximum number of losers and it varies from zero (for hand AKQJ1098765432- void-void-void) to 13 (for hand xxxx- xxx-xxx-xxx).

edit: never mind

Who let the dogs out ?!

matmat, on 2014-April-24, 08:48, said:

Leaving aside the humor, the error is obvious. You accept that any distribution is characterized by a constant number maximum number of losers (Lmax) and that it is equal to thirteen (Lmax = 13). But this is not true. Take the distribution:

♠J1098765432
♥432
♦void
♣void

Let us assume that trump is SP. What is the maximum number of losers that you can lose without getting any help from your partner? The answer is six - three spades and three hearts. This hand cannot lose more than 6 tricks (Lmax), i.e. it will always win seven tricks.
Yes, in some cases, it may lose a less than 6 tricks (Lreal), but this does not depend on the distribution. This depends by other factors (eg location of the spade honors and heart honors in the other hands).

matmat, on 2014-April-24, 08:48, said:

sorry double post

gergana85, if I see correctly, Lmax is 9 for 5530, right? P1, P2, P3 are all 0 for this distribution. Yet

xxxxx
xxxxx
xxx
-

Can have as much as 13 losers. LHO has AKQJT AKQJT AKQ -, draws trumps, and claims. Also for xxxx xxx xxx xxx you said that the real number of losers is 13 but the maximum is 12 (again, P1, P2, P3 are all 0 if I see it correctly). So the maximum number of losers is not given by your formula.

The most balanced distribution I can think of that does not have a maximum number of losers=13 is 7222, since you will make at least one of your long suit, if that suit will be trump (Lmax=9 according to your formula, actual maximum losers=12 according to me - LHO has AKQJT9 AK AK AKx, draws trumps and plays AK AK AK).

What exactly do you mean by maximum number of losers, if it is not what we normally mean by maximum and losers? Could you define these terms so that we can understand what your formulas refer to?

edit: changed Lmax=9 because 7222 has P2=1 apparently.

This post has been edited by gwnn: 2014-April-25, 05:00

hrothgar, on 2014-April-24, 10:40, said:

This can now be considered proven by me. This was supposed. However, the evidence was not there. Until now ...

gergana85, on 2014-April-25, 05:31, said:

I need to remember that phrase for future discussions...

Rik

Trinidad, on 2014-April-25, 10:59, said:

Trinidad, MrAce said:

Somewhere I was told that I not responsible for those which cannot to calculate . Again, there are formulas - check them. For now I give only an example. Let's look at the distribution:

♠ххххххх
♥хх
♦хх
♣хх

For this distribution the sum of the two longest suits S_1,2 is equal to 9.
According to the article:

P₁ = (|10 – S₁| – (10 – S₁))/2 = (|10 – 7| – (10 – 7))/2 = (3 – (3))/2 = (3 – 3)/2 = 0/2 = 0
P₂ = (|3 – S₂| - (S₂ – 3))/2 = (|3 – 2| – (2 – 3))/2 = (1 – (-1))/2 = (1+1)/2 = 1
P₃ = (|3 – S₃| - (3 – S₃))/2 = (|3 – 2| – (3 – 2))/2 = (1 – (1))/2 = (1 – 1)/2 = 0/2 = 0

Тtherefore:

Lmax = 19 – S_1,2 – (P₁ + P₂ + P₃) = 19 – 9 – (0 + 1 + 0) = 10 – (1) = 9

Now let we to check the bills. Obviously the losers in the spades are three (we have assumed that each fourth and the next card are winners). The losers in the other three suits are generally six (two in each suit). Therefore, the maximum number of the losers for this distribution is 9. This is showing and the calculation.
Try to make similar calculations for the remaining 38 distributions and you will see that there is no discrepancy.
If something still bothers you, email me at bogev53@abv.bg I will gladly help you.

Could you answer my post? Lmax=13 for the majority of distributions.