[gergana85]

This is not accurate. See post #20.

[diana_eva]

gergana85, on 2014-March-15, 05:23, said:

see post #1

The link doesn't work, gergana... When you start a discussion, it's usually a good idea to explain what it is that you are looking for? Feedback for that calculator, opinions on how experts evaluate hands, etc.

EDIT: ah, i didn't notice you edited OP

This post has been edited by diana_eva: 2014-March-15, 06:28

[Lovera]

Per errore ho parlato su questo argomento ma il mio contributo e' stato edito in Band evaluation a cui rimando chi fosse interessato, grazie.

[gszes]

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

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

I apologize for cutting your text short but it appears to be strongly based on the quote I kept. I feel if we
can disprove this one thought then we can safely ignore all that follows. Let us begin with a really
simple bridge hand (pretty much like every dummy I have ever laid down for partner in fact:)

87 8532 7643 854

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
Pretty much any theory that feels this hand has a maximum of 5 losers sounds like it was written by some government
bureaucrat with entirely too much time on his/her hands and I will bet they are at least partially responsible
for the economic policies of the USA. To be bluntly honest this hand pretty much has as many losers as it has
cards 13 the two longest suits are completely useless when determining the number of losers in this hand.

I apologize if I have improperly analyzed the work you presented but if I read it properly it is not a useful
work since one cannot build an successful argument based on such a horribly wrong premise.

[PhilKing]

gszes, on 2014-April-05, 11:50, said:

I apologize if I have improperly analyzed the work you presented but if I read it properly it is not a useful
work since one cannot build an successful argument based on such a horribly wrong premise.

I would advise learning what inversely proportional means.

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

It's not rocket science!

[hrothgar]

Anyone else flashing back to the Zar Points arguments last decade

[gergana85]

gergana85, on 2014-April-07, 00:51, said:

Quote

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]

hrothgar, on 2014-April-05, 13:37, said:

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]

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

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:

PhilKing, on 2014-April-05, 12:50, said:

It's not rocket science!

Rik

[hrothgar]

gergana85, on 2014-April-07, 03:45, said:

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]

PhilKing, on 2014-April-05, 12:50, said:

I would advise learning what inversely proportional means.

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

It's not rocket science!

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.

[gergana85]

gszes, on 2014-April-07, 11:58, said:

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?

[jogs]

Lord Molyb, on 2014-April-20, 08:49, said:

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

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]

Lord Molyb, on 2014-April-20, 08:49, said:

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

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

[gergana85]

Lord Molyb, on 2014-April-20, 08:49, said:

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

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]

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

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 Do you have some examples where your valuation system leads to a better result than standard methods?