gergana85

see post #1

This material aims to identify the factors which influence the distribution strength as well as to clarify the mechanism of their action. The solution of this problem will help us create a new, more accurate method for hand evaluation. To eliminate the influence of the honors strength on the common strength of the hand we decided to examine only hands which lack honors. The only indicator that is devoid of subjectivity (e.g. the ability to play well and the position of the honors are subjective) and solely reflects the distribution strength is the maximum number of losers in the hand (Lmax). Here we assume, the same way the LTC method does, that each fourth and next card in a certain suit is a winner. Based on this assumption we can say that Lmax varies from 0 (distribution 13-0-0-0) to 12 (distribution 4-3-3-3). DEPENDENCY OF THE MAXIMUM NUMBER OF LOSERS ON THE DISTRIBUTION The result of the analysis of all possible distributions clearly shows that the main indicator that determines the value of Lmax is the sum S1,2 = S1 + S2, where S1 and S2 are the number of cards in the two longest suits. Furthermore, we determined that in some cases Lmax is influenced not only by the sum of the lengths of the two longest suits but there is an additional dependency on the lengths of the three longest suits. We found that: Lmax = 19 S1,2 ( P1 + P2 + P3 ) P1, P2 and P3 are corrections that depend on the number of cards in the three longest suits. According to thise formula the sum S1,2 and the corrections P1, P2 and P3 decrease the maximum number of losers in a hand and therefore they increase its distribution strength. It is worth noting that the probability of getting a hand where (P1 + P2 + P3 ) ˃ 0 is no more than 3.8%. In the rest of the cases Lmax depends only on the sum of the lengths of the two longest suits. This dependency can be summarized in The Law of the two longest suits which states: The maximum number of losers in a hand is inversely proportional to the sum of the lengths of the two longest suits. It also proves the famous hypothesis: "The potential of a hand to win tricks is directly proportional to the sum of the lengths of the two longest suits." CORRECTIONS P1, P2 AND P3 For some unusual distributions Lmax is influenced by the length of the three longest suits. The correction P1 represents the presence of at least one of the top three honors in the longest suit, when it has more than 10 cards (S1 > 10). This correction has no real practical application, only in theory. Such distributions are too unbelievable and therefore we ignore them. P1 = (|10 S1| - (10 S1))/2 |10 S1| is the absolute value of the difference (10 S1). The correction P2 is positive only for S2 < 3 and it's a result from the decreased number of losers in the third and fourth longest suits in the hand. Of all distributions that satisfy this condition the only two that needs attention are 7-2-2-2 and 8-2-2-1, where P2 = 1. The other distributions that satisfy this condition (9-2-1-1, 9-2-0-0, 10-2-1-0) and 11-2-0-0 are ignored. It is still necessary to point out that when S2 = 1 (10-1-1-1, 11-1-1-0, 12-1-0-0) then P2 = 2 and when P2 = 0 (13-0-0-0) then P2 = 3. P2 = (|3 S2| - (S2 - 3))/2 |3 S2| is the absolute value of the difference (3 S2). The correction P3 refers to the 3-suited distributions (4-4-4-1 and 5-4-4-0) in which the number of cards in the third longest suit is exactly four (S3 = 4). Such distributions are stronger by one trick compared to the other distributions in which the sum of the two longest suits is the same. This is because the longer third suit decreases the number of losers in the fourth suit. P3 = (|3 S3| - (3 S3))/2 |3 S3| is the absolute value of the difference (3 S3). CONCLUSIONS The research shows that it is necessary to reconsider the factors that determine the distribution strength of the hand. We can make the following conclusions: The sum of the two longest suits (S1,2) is the most important factor that changes the distribution strength of a hand. In the 4-3-3-3 distribution (S1,2 = 7) Lmax = 12. With any increase of the sum S1,2 with 1 card we get a decrease in the maximum number of losers (Lmax) by 1 and therefore an increase in the potential of the hand to win by 1 trick; For some unusual distributions, in which the second longest suit has two cards, the maximum number of losers is decreased by 1 trick. When the second longest suit has 0 cards (distribution 13-0-0-0) or 1 card (10-1-1-1, 11-1-1-0 and 12-1-0-0) the potential of a hand to win tricks is increased by 3 and respectively 2 tricks. In practice though we have only the cases where the second longest suit has 2 cards (S2 = 2). This is because SS2 may have 1 or 0 cards only in distributions where the longest suit has 10 or more cards. However, such distributions most likely will never happen in a lifetime; 3-suited hands have a higher potential to win tricks. Under similar conditions it is a trick higher compared to other hands for which the sum S1,2 is the same. In practice this needs to be taken into consideration; It is necessary to reconsider the evaluation of the hand strength. It is the usual practice to first evaluate the honor strength and then correct it for a specific distribution. The analysis that was done shows that it is more proper to first estimate the maximum number of losers in the hand (calculate of the distribution strength) and only then to determine how many (and what) are the honors that would covers those losers. The difference between these two variables gives the potential of the hand to win tricks. The suggested model makes this evaluation quite accurate. For more information: bogev53@abv.bg

How to calculate the distributive strength of the hand? See that: http://bridge-law.hit.bg

How to calculate the distributive strength of the hand? See that: http://bridge-law.hit.bg