gergana85

Maybe 5♣ is better?! Or 6♠?

The question is incorrect. It is not clear what means opening 2♠. Some people assume that this bid shows 6-10 HCP without 4 hearts, while others assume that 2♠shows two suited hand with spades and minor. There is no problem. Obviously, you cannot dream to play a game. The rising 3♦ is not good. Probability to find a support in diamonds is not greater than 33% and should be played on the third level with a weak suit (opening 2♠ shows that the predominant strength is in the spades). Communications between the two hands are a problem too. A reply 3♥ creates even bigger problems. For these reasons you should to pass. To win 8 tricks in spades is much more likely than nine tricks in diamonds or hearts.

Hi Csaba. If I understood correctly, you ask me what are the sources that have helped me in my work. I do not have sources. I've only ever analyzed all 39 distributions and I have proven that the distribution strength depends on the sum of the two longest suits. I have not created a method to evaluate the hand. Pavel Bogev

I was not quite right with examples. I meant "meaningful comments." Your comments are indeed such and I will answer them soon.

Double post. Sorry.

It seems you are viewing the forum as a liability. Please do not take it that way. Especially in this case. Furthermore my criterias for interest in the topic are others - for example, one of them is the number of those comments topic. You are not interested in the topic. Well, say it clearly. If you specify a reason - it will be fine. If you can’t - it's OK. But you do not speak on behalf of others. It's too ambitious. Hardly you are authorized to represent all people. Pavel Bogev

Absolutely not understand people who take part in the forum, but did not discuss on the theme, just write sentences and dealing with conspiracy theory.

What you say is partly true. But here it comes to tricks, not for points (equivalent to HCP). I'm talking about the maximum number of losers, which means the maximum number of tricks that can be lost. Pavel Bogev

For me, the theme of the monkeys is complete and I will not more to comment on it. You have arguments, me too. They do not fully coincide. OK. Such is life. I don't agree with your argument that with LTC no need to count, but only to watch. According LTC you need to count the number of the three top honors in all suits and then removing them from 12. Based on your reply, your partner does the same, but he subtract of 24 (to determine the level of contract). And I repeat - you must remember that I do not give a method for evaluation. I want only to show what are the factors that influence and how they affect on the distribution strength. I dont mean nothing more. You ask me why you need to take a new formula when there are so many rules and the introduction of another would confuse the players. Well, some of those are unlikely to help in the evaluation of hand. On the other hand the new formula replaces some of them (e.g. rule 19/20/21 and rule 1/2/3). You say that there is no information that LTC assumed that Lmax is always 12. Here are some sources: http://www.bridgeguys.com/pdf/LosingTrickCount.pdf. http://www.bridgehands.com/L/Losing_Trick_Count.htm http://www.bridge-forum.com/i-trick-1b.htm https://www.mrbridge.co.uk/library/LTC_2.pdf http://www.phillipalderbridge.com/LTC.HTM There are many others. Just log into the internet. Pavel Bogev from Bulgaria (nick gergana85)

I say the same. Hands in which three of the suits have 2 or less cards must make an adjustment. In this case, the length of the longest second suit (S2) is determinant. If it is equal to 2 (S2 = 2, distributions 7-2-2-2, 8-2-2-1, 9-2-2-0, 9-2-1-1, 10-2-1 - 0 and 11-2-0-0) correction is -1. But if the second longest suit is only 1 card (S2 = 1, distributions 10-1-1-1, 11-1-1-0 and 12-1-0-0) or 0 cards (S2 = 0 , distribution 13-0-0-0), the correction is -2, or -3 respectively. For 3-suit distributions (4-4-4-1 and 5440) correction is always -1. Yes, the chance to get a hand with longest suit with 9 or more cards is almost zero and these cases can be ignored. Furthermore, the evaluation of such hands is very easy and requires no formulas. Pavel Bogev, Bulgaria

I accept your explanation of the trained monkey, but with some reservations. The reasons are 3: - I see that you understand that the example of the zoo insult me, but you are continuing with the same examples; - I accept that it is normal a person unintentionally to afflict someone without any intentions, but I also think it is okay to apologize. You do not apologize; - Following your logic why you insulted by my example with an animal from the zoo after you put me below level of the trained monkey and I like such (an untrained monkey) answered you? Why you not take my answer as a compliment to your exact definition? I want to say that in the future, when someone passes certain limit against me, he will receive immediate and equivalent reaction in the opposite direction. Until now, I did not allow myself this, even though I had enough reasons. I believe that the case is closed. Nevertheless, I am glad your questions (they are important for me) and I begin to answer them. 1. I never say that I created a method for а general evaluation of the hand (as LTC). I just tried to find the factors affecting on the strength distribution and found (and proved) that it depends of the sum of the two longest suits. It would be possible to imagine, but it has not been proven until now. I determined (and proved) that in some rare cases Lmax is influenced not only by the sum of the lengths of the two longest suits but there is an additional dependency by the lengths of the each three longest suits. Nothing more. 2. Continuing in this way, I do not reject the use of LTC. But it must undergo known corrections. Yes, LTC method gives correct results, but as you has noticed only for distributions with no more than 10-card suit. According to LTC, in the distribution 13-0-0-0 has 3 losers, which is obviously not true. My formula is true for all 39 possible distributions. 3. What do you find complicated in the formula: Lmax = 19 S1,2 (P1 P2 P3) This formula in more than 96% of cases is confined to the formula (when talking about percentages, I mean the probability to get such distribution): Lmax = 19 S1,2 To argue that this is difficult is not correct. It is no more difficult than using the formula: LTC_max=min(L1,3)+min(L2,3)+min(L3,3)+min(L4,3) Yes, this formula does the same job but it is not part of the method LTC. According to ypur formula, the maximum number of the losers is not equal to a constant and that each distribution has a different value of Lmax. I say the same. But LTC says otherwise. According to this method always Lmax is equal to 12. 4. Not every player uses the LTC. But each player evaluates the distributive strength of the hand. And every one of them must know the factors affecting on it. This also applies to those who use LTC. Pavel Bogev from Bulgaria. My nick in BBO is gergana85. Sorry for my bad English but.... no one here knows Bulgarian.

Even untrained monkey would have noticed that the method LTC is not true. I do not know about you. :P

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.

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 = 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 – 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. :)

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