How to calculate the distributive strength of the hand?

[hrothgar]

Anyone else flashing back to the Zar Points arguments last decade

[gergana85]

Once we look at this wonderful collection I hope it is painfully obvious that the sum of the 2 longest suits (8)

would appear to make the maximum number of losers in this hand 5----as in five hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm.

 

About post # 24.Interesting arithmetic! In your example you think that 19 – 8 = 5(?), but I thing that 19 – 8 = 11. And once we accept that every fourth card is winning, everything falls into place.

But I agree with one - to say that 19-8 = 5 is really terribly wrong. And that someone quotes him is funny.

I say that Lmax = 19 - S₁,₂ - (P₁ + P₂ + P₃).

In the example (2-4-4-3) S₁,₂ = 8, (P₁ + P₂ + P₃) = 0 and therefore Lmax = 11 (number of the losers). The fact that someone can't to calculate is at his expense.

I apologize!

 

[gergana85]

Anyone else flashing back to the Zar Points arguments last decade

 

Written in post № 1 has nothing to do with Zar Point. Moreover, it (Zar Point) is contrary to The law of the two longest suits.

[helene_t]

The maximum number of losers in a hand is inversely proportional to the sum of the lengths of the two longest suits.

Do you know what inversely proportional means? Y inversely proportional with X means specifically that

Y= c/X

where c is some constant.

 

If you mean to say "the longer the two longest suits, the smaller the maximum number of losers" then say that.

[Trinidad]

Dear gergana85,

 

There are quite a few mathematicians, statisticians, enigineers, and beta scientists on BBF. I am sure they are able to evaluate the function L_max given by:

 

L_max = lim s->0 (Sum i=1 to 3 ((3-N_i)*erf(N_i,pi,s)) + ((Sum i=1 to 3 (N_i))-10)*erf((13-Sum i=1 to 3 (N_i)),pi,s))

 

where

• 
   
  
• N_i is the number of cards in a suit, where i specifies the number of coughs needed to indicate the suit
  
• erf(x,m,s) is the error function (integral of the normal distribution) of x with parameters m and s
  
• pi is the ratio between the diameter and the circumference of a circle
  

I think that all expert bridge players (for whom this forum is meant) are perfectly capable of evaluating this function too, even if they are not mathematicians,... etc. . After all:

It's not rocket science! :P

Rik

[hrothgar]

Written in post № 1 has nothing to do with Zar Point. Moreover, it (Zar Point) is contrary to The law of the two longest suits.

 

I wasn't discussing the specific's of the equation.

[gszes]

I would advise learning what inversely proportional means.

 

If other factors remain constant, ???? ??? ??? ??? has more losers than ??????? ?????? - -

 

It's not rocket science! :P

 

I remain totally and utterly unconvinced that your contribution to this matter is anything but

a useless attempt to get in the last word. There is nothing remotely useful about this latest

response as it completely fails to address the mathematical issues involved. The fact that it

SOUNDS logical that the longer your 2 longest suits are the less losers you have does not mean

there is any practical advantage to using that theory when it comes to calculating losers in a

bridge hand.

AKQJxxxxxxxxx void void void

AKQJxxxxxxxx x void void

AKQJxxxxxxx xx void void

AKQJxxxxxx xxx void void

AKQJxxxxx xxxx void void

AKQJxxxx xxxxx void void

AKQJxxx xxxxxx void void

AKQJxx xxxxxxx void void

AKQJx xxxxxxxx void void

AKQJ xxxxxxxxx void void

 

ALL of these examples include 13 cards in the 2 longest suits the AKQJ are always in the spade suit

(so we aren't splitting them up making them less useful). If the theory of inverse proportion is a

reasonable premise to apply to the number of losers in a bridge hand then all of the above examples

would yield the same number of losers. The fact that it is totally obvious that this is incorrect

should make it simple for anyone to recognize that building an entire argument based on the idea

has to be wrong and not taken seriously. My apologies if I was unclear from the beginning and I hope

this somewhat expanded version will help lay to rest any further thought that using the principles

of this particular writing are useless from a practical standpoint. If any brilliancies are derived

from the work they are a matter of luck not logical assumption.

[MrAce]

My scale is the best. I am naturally born with a hidden scale in my hands.

 

I pull the cards from the board. I move my hand holding the cards, up and down 2-3 times. It auto scales the value of the cards and i bid accordingly.http://www.bridgebase.com/forums/public/style_emoticons/default/tongue.gif

[gergana85]

I remain totally and utterly unconvinced that your contribution to this matter is anything but

a useless attempt to get in the last word.....

 

You are mistaken. The reason is that you mix (probably unintentionally) terms Maximum number of losers (Lmax) and Real number of losers (Lreal) in the hand. For each distribution, Lmax is constant and does not change. At the same time the Lreal can be changed. It should be emphasized that Lreal is equal to Lmax minus the number of winning honors. In other words Lmax characterizes the distribution strength of the hand until Lreal shows general strength of the hand. Lreal may be equal to or less than Lmax.

For example in the distribution:

 

AKQJ1098765432-void-void-void

 

Lreal is equal to Lmax and they are nil, while in the distribution:

 

AKQJ109876543-x-void-void

 

Lreal is equal to 1, but Lmax is equal to 2. But in the hand:

 

KQJ109876543-x-void-void

 

Lreal is again equal to Lmax and they are equal to 2.

Else I'm not saying that players can’t determine the Lmax without these formulas. I say (and prove) that one of the main factors determining the general strength of the hand is Lmax and it directly depends on the sum of the two longest suits. Furthermore, I show what his influence is.

[Lovera]

I similarly gszes said , valute 1 point from 5th. card and so on ( 5th: 1p., 6th: 2p., 7th: 3p., 8th: 4p. 9th. 5 p,...) then with 8 cards ,such as, 4 points. Have a good work.

[Lord Molyb]

Someone tell me why I can't just do these things by how I feel or at least just a few simple calculations? :unsure:

[jogs]

Someone tell me why I can't just do these things by how I feel or at least just a few simple calculations? :unsure:

 

Don't feel bad. The calculations are pointless. The fifth card is a long suit is worthless if you end up defending with another suit as trumps.

[Lovera]

Someone tell me why I can't just do these things by how I feel or at least just a few simple calculations? :unsure:

I' m agree with you from the second half of your post

[gergana85]

Someone tell me why I can't just do these things by how I feel or at least just a few simple calculations? :unsure:

This article is proof that the distribution strength is determined directly from the sum of the two longest suits. Anyone can feel something, but not everyone can prove it. The feeling still does not prove anything. It may be true, but it cannot be true.

And ask how this formula (Lmax = 19 – S_1,2) is causing too much trouble?

[johnu]

This article is proof that the distribution strength is determined directly from the sum of the two longest suits. Anyone can feel something, but not everyone can prove it. The feeling still does not prove anything. It may be true, but it cannot be true.

And ask how this formula (Lmax = 19 – S_1,2) is causing too much trouble?

 

Fancy formulas B-) Do you have some examples where your valuation system leads to a better result than standard methods?

[hrothgar]

And ask how this formula (Lmax = 19 – S_1,2) is causing too much trouble?

 

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)

[Lurpoa]

This material aims to identify the factors...

 

DEPENDENCY OF THE MAXIMUM NUMBER OF LOSERS ON THE DISTRIBUTION

 

 

 

VERY GOOD ARTICLE ! !H!H!H

 

Don't get discouraged by comments by the BBO Experts.

They do that with every newcomer.

[gergana85]

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.

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.bridgeguys.com/pdf/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

[gergana85]

VERY GOOD ARTICLE ! !H!H!H

Thanks.

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

[matmat]

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

[matmat]

You are mistaken. The reason is that you mix (probably unintentionally) terms Maximum number of losers (Lmax) and Real number of losers (Lreal) in the hand. For each distribution, Lmax is constant and does not change. At the same time the Lreal can be changed. It should be emphasized that Lreal is equal to Lmax minus the number of winning honors. In other words Lmax characterizes the distribution strength of the hand until Lreal shows general strength of the hand. Lreal may be equal to or less than Lmax.

 

 

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

[mycroft]

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...

[hrothgar]

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

 

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".

[manudude03]

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.

[gergana85]

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

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).