gwnn Posted April 25, 2014 Report Share Posted April 25, 2014 edit: never mind Quote Link to comment Share on other sites More sharing options...
MrAce Posted April 25, 2014 Report Share Posted April 25, 2014 Who let the dogs out ?! http://www.bridgebase.com/forums/public/style_emoticons/default/tongue.gif Quote Link to comment Share on other sites More sharing options...
gergana85 Posted April 25, 2014 Author Report Share Posted April 25, 2014 Lreal = 13 - (tricks quitted my way) = tricks quitted their way. 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). Quote Link to comment Share on other sites More sharing options...
gergana85 Posted April 25, 2014 Author Report Share Posted April 25, 2014 Lreal = 13 - (tricks quitted my way) = tricks quitted their way.sorry double post Quote Link to comment Share on other sites More sharing options...
gwnn Posted April 25, 2014 Report Share Posted April 25, 2014 (edited) gergana85, if I see correctly, Lmax is 9 for 5530, right? P1, P2, P3 are all 0 for this distribution. Yet xxxxxxxxxxxxx- 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. Edited April 25, 2014 by gwnn 1 Quote Link to comment Share on other sites More sharing options...
gergana85 Posted April 25, 2014 Author Report Share Posted April 25, 2014 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". This can now be considered proven by me. This was supposed. However, the evidence was not there. Until now ... :rolleyes: Quote Link to comment Share on other sites More sharing options...
Trinidad Posted April 25, 2014 Report Share Posted April 25, 2014 This can now be considered proven by me.I need to remember that phrase for future discussions... Rik 2 Quote Link to comment Share on other sites More sharing options...
MrAce Posted April 25, 2014 Report Share Posted April 25, 2014 I need to remember that phrase for future discussions... Rik http://www.bridgebase.com/forums/public/style_emoticons/default/biggrin.gif Quote Link to comment Share on other sites More sharing options...
gergana85 Posted April 29, 2014 Author Report Share Posted April 29, 2014 This can now be considered proven by me. This was supposed. However, the evidence was not there. Until now ... :rolleyes:Somewhere I was told that I not responsible for those which cannot to calculate :huh: . 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 S1,2 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 = 0P2 = (|3 – S2| - (S2 – 3))/2 = (|3 – 2| – (2 – 3))/2 = (1 – (-1))/2 = (1+1)/2 = 1P3 = (|3 – S3| - (3 – S3))/2 = (|3 – 2| – (3 – 2))/2 = (1 – (1))/2 = (1 – 1)/2 = 0/2 = 0 Тtherefore: Lmax = 19 – S1,2 – (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. :) Quote Link to comment Share on other sites More sharing options...
gwnn Posted April 29, 2014 Report Share Posted April 29, 2014 Could you answer my post? Lmax=13 for the majority of distributions. Quote Link to comment Share on other sites More sharing options...
Vampyr Posted April 29, 2014 Report Share Posted April 29, 2014 OP, why did you quote yourself and attribute the quote to other forum members? This is a serious no-no. The hand in your most recent post will usually not take a trick, no matter what calculations you make that demonstrate (according to you) that it will. Quote Link to comment Share on other sites More sharing options...
gwnn Posted April 29, 2014 Report Share Posted April 29, 2014 Yes, Lmax=9 according to your formula but your opponents can take as many as 12 tricks if spades split very poorly (did you read my post?). So your formula is off by 3. Hence my question: what do you mean by maximum? Are you just discounting the possibility of losing a trick in the 4th round of playing a suit? If you are, please say so. Then we are back to basic losing trick count in which we take every card a loser except 4th and subsequent cards. If you mean something else by maximum possible loser, please define what you mean. 1 Quote Link to comment Share on other sites More sharing options...
diana_eva Posted April 29, 2014 Report Share Posted April 29, 2014 Are you just discounting the possibility of losing a trick in the 4th round of playing a suit? If you are, please say so. Then we are back to basic losing trick count in which we take every card a loser except 4th and subsequent cards. I think you cracked the code :) Quote Link to comment Share on other sites More sharing options...
Vampyr Posted April 29, 2014 Report Share Posted April 29, 2014 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? Yes, that would be helpful. Anyway I think that gwnn is probably correct that we are just arriving at basic LTC in a rather roundabout fashion. LOL EDIT: crossed previous post Quote Link to comment Share on other sites More sharing options...
gwnn Posted April 29, 2014 Report Share Posted April 29, 2014 LTC_max=min(L1,3)+min(L2,3)+min(L3,3)+min(L4,3) is equivalent to your formula, assuming no 10-card or longer suits (the cutoff of 10 cards as opposed to 9 cards seems arbitrary anyway). You just discount any card beyond the 3rd of any suit. Quote Link to comment Share on other sites More sharing options...
WellSpyder Posted April 29, 2014 Report Share Posted April 29, 2014 I decided early in the life of this thread that I wouldn't bother reading it since it didn't seem to be addressing a very interesting question, or at least not in a way that added anything. Today I noticed that it had got quite long and thought perhaps people had found something of interest after all, so I took another look. Oh well, it won't be the only mistake I make today, I guess. :( 1 Quote Link to comment Share on other sites More sharing options...
Zelandakh Posted April 29, 2014 Report Share Posted April 29, 2014 Rather than wading through a bunch of pseudo-maths, could you post your evaluation nunbers for the first 10 or 15 distributions. I have previously posted equivalent numbers for other evaluation methods and this makes for an easy comparison. In addition you need to provide some indication as to what weighting the hcp side will have so that the numbers can be properly scaled. The truth is that the distributional sides of ZP and the 531 scale work pretty well when analysed. It might be that your formulae are just an equivalent transformation of one of these, or perhaps you have changed the scalings somewhat. If the former then the work is of no practical interest; if the latter then we can consider whether the changes represent an improvement or not. My intuitive guess would be that the distributional numbers represent an improvement over MLTC but no improvement over ZP or 531; but it would be nice to see the numbers to know for sure because intuition is often wrong. PS: as an aside, I stand by my comment in the mod thread that works of this type would be of much greater benefit to I/A players and suggest that you post there next time if you successfully improve upon your ideas. Quote Link to comment Share on other sites More sharing options...
gergana85 Posted April 30, 2014 Author Report Share Posted April 30, 2014 gergana85, if I see correctly, Lmax is 9 for 5530, right? P1, P2, P3 are all 0 for this distribution.....I have accepted (see post № 1) that each fourth and next card is winner. This is a probabilistically assumption, based on the practice. It could be assumed that the 5th and the subsequent cards are winners or the every 6th and next cards are winners or as you say - every 7th and the next card are winners, etc. But these assumptions are not acceptable in terms of probability theory. The reason is that the assumption "the fourth and each subsequent card is a winner" is sufficiently likely to be accepted as true. The others also may be accepted as truth. But in this case some distributions remain outside the analysis (4-3-3-3, 4-4-3-2, 5-4-2-2, 5-4-3-1, 5-4-4-0, etc.).As everyone knows, bridge is a game of probabilities. That is why it obeys the laws of the theory of probability - аs someone said, the bridge is not "rocket science" and everything is based on assumptions. Maybe you know that the main instrument for some sciences (eg quantum mechanics) is the theory of probabilities. Any conclusions that are made in them are based on the theory of probability. Moreover, they are proved by practice. For this reason, no one rejects them and accepts them as useful. Are your examples useful? I would say no! And not because they can not happen, but because there is no way the player with these hands to be an active player.I ask you, would you open the auctions with distribution: ♠ххххх♥ххххх♦ххх♣void If someone did, it would be too adventurously. This hand has a value only if your partner opened the action with 1♥ or with 1♠. Assuming that the trump is the spade, the losers in this hand are actually equal to 9. In all likelihood, the 4th and 5th trumps will be the winners. Also winners will be the 4th and 5th hearts (if timely develop this suit). And notice we are talking about most likely event, but not talking about possible event. In connection with this I mean that there is no need to invent any possible events, refuting the probable events. It's not difficult, but it is incorrect. Consider the following hand: ♠АКQJ109♥AKQJ♦AКQ♣void Should be appoint contract 7♠? Or maybe to prevent any hazards you will have to pass (because any of the opponents might have ♠8765432 ♥void ♦void ♣765432) and probably cannot win more than 6 tricks when the lead is ♣A? Probably you understand the absurdity of the situation! I could go on with examples, but this is not necessary. I will just say that you do attempt to put me in an absurd trap. :rolleyes: I apologize for the error. It does not change the meaning of what was said. Quote Link to comment Share on other sites More sharing options...
gwnn Posted April 30, 2014 Report Share Posted April 30, 2014 That hand has 10 cards, I think you just beat the record for the worst example hand on the forums (BBF has many 12- and 14-card example hands but I've never seen 16 and 10 carders). So you accept that your method is simply equivalent to LTC up to 10-card suits. Why should anyone use your method and not LTC? Even a trained monkey can apply the LTC formula (consider every 4th card and beyond a winner), while your formula is much longer. Quote Link to comment Share on other sites More sharing options...
MrAce Posted April 30, 2014 Report Share Posted April 30, 2014 That hand has 10 cards, I think you just beat the record for the worst example hand on the forums (BBF has many 12- and 14-card example hands but I've never seen 16 and 10 carders). So you accept that your method is simply equivalent to LTC up to 10-card suits. Why should anyone use your method and not LTC? Even a trained monkey can apply the LTC formula (consider every 4th card and beyond a winner), while your formula is much longer. http://www.bridgebase.com/forums/public/style_emoticons/default/biggrin.gif Quote Link to comment Share on other sites More sharing options...
Trinidad Posted April 30, 2014 Report Share Posted April 30, 2014 OP, why did you quote yourself and attribute the quote to other forum members? This is a serious no-no.I fully agree that quoting yourself and attributing this to other forum members is a serious no-no. However, given that:- OP has little experience in these forums- probably meant to reply to the responses by MrAce and me to the text she did quote- probably didn't know how to nest quotes (how do you reply to a smiley?) we should be a little forgiving. I, at least, didn't think that (s)he wanted to spread the idea that MrAce and I were supporting the OP's views, and that we did that by miraculously writing a joint post on this subject. Rik Quote Link to comment Share on other sites More sharing options...
Vampyr Posted April 30, 2014 Report Share Posted April 30, 2014 - probably didn't know how to nest quotes (how do you reply to a smiley?) Actually I don't know how to quote nested quotes with the new software; can anyone help? Also is there a way to change the colour scheme? The current one makes me a little ill. Quote Link to comment Share on other sites More sharing options...
Trinidad Posted April 30, 2014 Report Share Posted April 30, 2014 You use "Multiquote" (button at right bottom of each post) to select every post that you want to include. Then you click "Add reply" (at the bottom of the page). The editor will show something like this: [ quote name=Trinidad' timestamp='1398874413' post='791557]- probably didn't know how to nest quotes (how do you reply to a smiley?)[/ quote][ quote name=Vampyr' timestamp='1398881838' post='791569]Actually I don't know how to quote nested quotes with the new software; can anyone help? Also is there a way to change the colour scheme? The current one makes me a little ill.[/ quote] This will look like this in your post:- probably didn't know how to nest quotes (how do you reply to a smiley?)Actually I don't know how to quote nested quotes with the new software; can anyone help? Also is there a way to change the colour scheme? The current one makes me a little ill. Then you cut and paste "[ quote name=Vampyr' timestamp='1398881838' post='791569]" to move it to the start of the post and you get: [ quote name=Vampyr' timestamp='1398881838' post='791569][ quote name=Trinidad' timestamp='1398874413' post='791557]- probably didn't know how to nest quotes (how do you reply to a smiley?)[/ quote]Actually I don't know how to quote nested quotes with the new software; can anyone help? Also is there a way to change the colour scheme? The current one makes me a little ill.[/ quote] This results in:- probably didn't know how to nest quotes (how do you reply to a smiley?)Actually I don't know how to quote nested quotes with the new software; can anyone help? Also is there a way to change the colour scheme? The current one makes me a little ill. I hope this helped with the nesting. I don't know how to fix the color scheme. Rik 1 Quote Link to comment Share on other sites More sharing options...
gergana85 Posted May 2, 2014 Author Report Share Posted May 2, 2014 Even a trained monkey can apply the LTC formula. Even untrained monkey would have noticed that the method LTC is not true. I do not know about you. :P Quote Link to comment Share on other sites More sharing options...
Vampyr Posted May 2, 2014 Report Share Posted May 2, 2014 I hope this helped with the nesting. I don't know how to fix the color scheme. tx :) Quote Link to comment Share on other sites More sharing options...
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