jogs Posted July 9, 2013 Author Report Share Posted July 9, 2013 POWER Power is one of two multi-dimensional independent random variables used for estimating tricks. High card point count is the main componentof this estimator. Most systems treat high card points as if it were theestimator, power. The location of the HCP and how honors interact are also components of the power estimator. Honors in long suits are worth more than honors in short suits. Honors working together are worth more than honors standing alone. Point count: ace=4, king=3, queen=2, and jack=1♠ AKQJ ♥ 432 ♦ 432 ♣ 432♠ A432 ♥ K32 ♦ Q32 ♣ J32♠ AK32 ♥ QJ2 ♦ 432 ♣ 432Each is an example of a 10 point hand. They are obviously notof exactly equal value. It is difficult to measure the exact effects of the honors in each hand. 10 is the approximately correct value for each hand. In statistics there is the error component in every model to account for the deviations.Power is the independent random variable. High card points isthe dependent variable of power which is proportional to tricks.Therefore it is easier to use HCP to estimate tricks than theactual independent random variable, power. Quote Link to comment Share on other sites More sharing options...
Scarabin Posted July 10, 2013 Report Share Posted July 10, 2013 :) POWER Power is one of two multi-dimensional independent random variables used for estimating tricks. High card point count is the main componentof this estimator. Most systems treat high card points as if it were theestimator, power. The location of the HCP and how honors interact are also components of the power estimator. Honors in long suits are worth more than honors in short suits. Honors working together are worth more than honors standing alone. Point count: ace=4, king=3, queen=2, and jack=1♠ AKQJ ♥ 432 ♦ 432 ♣ 432♠ A432 ♥ K32 ♦ Q32 ♣ J32♠ AK32 ♥ QJ2 ♦ 432 ♣ 432Each is an example of a 10 point hand. They are obviously notof exactly equal value. It is difficult to measure the exact effects of the honors in each hand. 10 is the approximately correct value for each hand. In statistics there is the error component in every model to account for the deviations.Power is the independent random variable. High card points isthe dependent variable of power which is proportional to tricks.Therefore it is easier to use HCP to estimate tricks than theactual independent random variable, power.Of the 3 example hands 1&3 clearly have less losing tricks than 2, because the high cards are concentrated. Culbertson honour tricks, Bissell point count, and surely all losing trick counts reflect this. Milton Work HCP are only good for:ease of calculationbalanced handslocating missing honours. I feel I must be missing some deep point, otherwise what is the problem? :) Quote Link to comment Share on other sites More sharing options...
FM75 Posted July 10, 2013 Report Share Posted July 10, 2013 Estimating Our Tricks <snip>The estimate of our tricks for the general case. E(tricks) = trumps + (HCP-20)/3 + e Trumps is the total combined trumps of the partnership.(HCP-20)/3 means for every HCP over 20 assign another 1/3 trick to the estimate.The curve generated by this equation is approximated by the normal curve(bell curve).'e' is the error of the observations. E(e)=0. For large samples the average error will approach zero.<snip> Strain = NT, HCP = 29, E(tricks) = 3 Huh? Does that mean we get 9 tricks, 3 tricks, or the formula is garbage for NT? (The assumption is that e = 0, because this is a large sample case). Or does it mean we expect 6.5 +3 = 9.5? Strain NT, HCP = 40, 6.23 trick, 12.73 tricks?Strain some suit, HCP=40, trumps = 13, 19.23 tricks or 25.73 tricks? Strain some suit, hcp =11, trumps = 13, 16 tricks, 22.5 tricks? I suspect that the "average" is closer to making 7 tricks (1 suit) than 16 tricks. If the formula does not predict the simple cases, toss it in the garbage bin. (Ok, I studied physics, and expected the friction free limit of a formula including friction to reduce to the friction-free formula.) Quote Link to comment Share on other sites More sharing options...
jogs Posted July 10, 2013 Author Report Share Posted July 10, 2013 Huh? Does that mean we get 9 tricks, 3 tricks, or the formula is garbage for NT? (The assumption is that e = 0, because this is a large sample case). Or does it mean we expect 6.5 +3 = 9.5? This model is for suit strains and the general case.E(e) = 0, The expected error is zero.The standard deviation of the error is approximately1.25 tricks/boards. Turns out E(e) <0 for combined trumps >= 10 or expected tricks >= 10.This model works best when the expected tricks is between 3 to 10.Our expected tricks fluctuated wildly depending on whether wedeclare in our suit or defend in their suit. When trumps >= 10 or expected tricks >= 10 it requires a much more complex polynomial model. For a specific board in high level auctionswe should attempt to count the tricks. Quote Link to comment Share on other sites More sharing options...
jogs Posted July 11, 2013 Author Report Share Posted July 11, 2013 test: trying to post west/east hands [hv=pc=w&e=st842hatdkq4ck732&n=shk8732dajt97caj4]266|200[/hv] Quote Link to comment Share on other sites More sharing options...
jogs Posted July 11, 2013 Author Report Share Posted July 11, 2013 retest: trying to post west/east hands [hv=pc=w&e=st842hatdkq4ck732&w=shk8732dajt97caj4]266|200[/hv] Quote Link to comment Share on other sites More sharing options...
jogs Posted July 11, 2013 Author Report Share Posted July 11, 2013 :) Of the 3 example hands 1&3 clearly have less losing tricks than 2, because the high cards are concentrated. Culbertson honour tricks, Bissell point count, and surely all losing trick counts reflect this. Milton Work HCP are only good for:ease of calculationbalanced handslocating missing honours. I feel I must be missing some deep point, otherwise what is the problem? :) The point is the estimator isn't linear. It isn't a straight line. It is more likemulti-dimensional blob. It isn't only our honor count that matters. The location of those honors and whether they are working together alsomatter. During the initial evaluation of the hand there is no need to beprecise.Bissell points and workcount are just loose guidelines for valuation.I use Milton Work HCP because of ease of calculation.Other methods are much harder to calculate. Have never seen evidenceany of those methods are more than a small marginal incrementalimprovement. The variance of the calculations from those methods arejust as large as the variance of workcount.Assuming we are dealer, what do we need to know? First, whether ornot to open. If we decide to open, what to open. Why do we needa more precise valuation of our hand?During the auction when we learn more about how partner's hand fits with our hand the valuation can change dramatically. This negates theefforts of a complex initial valuation. This model is in terms of our tricks rather than my points. Point count: ace=4, king=3, queen=2, and jack=1♠ AKQJ ♥ 432 ♦ 432 ♣ 432♠ A432 ♥ K32 ♦ Q32 ♣ J32♠ AK32 ♥ QJ2 ♦ 432 ♣ 432 These hands are 10 +/- e points and this error is large. As we learnmore about partner's hand the error will be reduced. During theinitial evaluation, it is sufficient to know to pass these hands as dealer. E(tricks) = trumps + (HCP-20)/3 + eStd dev is about 1.25 tricks/board.E(tricks) = trumps + (HCP-20)/3 + SF + eFor flat hands the std dev can be as low as 1 trick/board.This SF term is skewness/flatness as suggested by Lawrence/Wirgren.Flat boards reduce the trick estimates. Skewed boards increase thetrick estimates. These two models are for fitting partner's hand with our hand.With luck we will know the value of the terms by the secondor third call of the auction. These models are best for assisting the contested auctions 3 over 2 and 3 over 3. In general the flat hand should not bid 3 over 3.5332 should rarely compete 3 over 3..................In another post I will introduce the other independent randomvariable, pattern. Quote Link to comment Share on other sites More sharing options...
jogs Posted July 14, 2013 Author Report Share Posted July 14, 2013 PATTERN Pattern is the ordered configuration of the four suits in one hand.Joint pattern is the joint pattern of the two partnership hands.Trumps is usually the suit with the longest combined length. LarryCohen has chosen trumps as his parameter from estimating tricks.SST(short suit totals) is the sum of the two shortest suit holdings of the partnership. Lawrence/Wirgren uses the shorter suit holding ofeach partner. Cohen and Lawrence/Wirgren are just using adifferent end of the same variable(pattern) to estimate tricks.Trumps is the coarse estimate. SST is the fine tuning. Usingboth gives us better estimates than either on a stand alone basis. Now for a naive test of joint pattern. Each side will be given20 points with all points in two suits. [hv=pc=n&s=sakt76h42daqtc642&w=s42hakt76d642caqt&n=sqj98h53dkj98c753&e=s53hqj98d753ckj98]399|300[/hv] This example has 18 trumps and only 16 tricks. [hv=pc=n&s=sakt76h2daqtc7642&w=s42hakt76d642caqt&n=sqj98h543dkj98c53&e=s53hqj98d753ckj98]399|300[/hv] Do not assume that when we hold flat patterns, opponents will alsohold flat patterns. In this example EW is 2434//2533, while NS is4342//5134. EW makes 8 tricks in hearts. NS makes 10 tricksin spades. 18 trumps producing to 18 tricks. This is compensatingerrors. LoTT is credited for being 'right' when it is 'right' for the'wrong' reasons. [hv=pc=n&s=sakt76h2d853caqt9&w=s42hakt76daqtc642&n=sqj98h543d7642ckj&e=s53hqj98dkj9c8753]399|300[/hv] In this example the points between the diamonds and clubs havebeen exchanged. EW is 2434//2533, while NS remains4342//5134. EW makes 8 tricks in hearts. NS makes 9 tricksin spades. 18 trumps producing to 17 tricks. [hv=pc=n&w=s42hakt76d642caqt&e=s53hqj98d753ckj98]266|200[/hv] On all three examples EW held 9 hearts and made only 8 tricks.So maybe the expected tricks are usually only dependent on our powerand pattern. Their pattern and trump length has little effect on our tricks. There are exceptions where their pattern reduces out tricks. Usually asingleton to an ace, followed by a ruff. It is rare. We should justignore their hands and use only ours to estimate tricks. E(tricks) = trumps + (HCP-20)/3 + SF + e -----------------Aside: this model applies mostly to when we expect to win 7 to 10tricks. Very useful for making contested part score auctions. At higher levels controls play a larger role. Examined hand records from BBO minis, most part score boards hada std dev between 0.5 to 1 when played in the same strain. Admittedlythe range of playing abilities on these minis was huge. Even in flight Aevents it was rare to see the std dev fall under 0.5.Could not find any evidence knowing their trump length assisted us inbetter competitive decisions. Knowing their joint distribution would have been useful. Quote Link to comment Share on other sites More sharing options...
jogs Posted July 19, 2013 Author Report Share Posted July 19, 2013 The objective of hand valuation should be to find the best contractfor the partnership. It is not to find the most precise valuation ofmy hand in a vacuum. The initial point count is just a transitoryvalue which will be readjusted with every successive bid in theauction.Power is the independent random variable used to estimate tricks. Point count is a dependent component of power. Example 1. I have an ace, two kings and a jack. My pattern is1=5=4=3. In work count I have 11 points regardless of thelocation of the honors. a) ♠2 ♥AK752 ♦KJ82 ♣642b) ♠K ♥J8752 ♦K852 ♣A42 Many would open hand 'a' 1♥. Few would open hand 'b'. Yetboth are 11 point hands. The location of the honors and whetherthey are working together does affect the ability of a hand to generatetricks. Hand valuation is about how honors interact with one another.Hand valuation is more complex than assigning a fixed value to each honor. Example 2.[hv=pc=n&w=s2hak752dkj82c642&e=s653hqj98daq53c98]266|200[/hv] West has 11 points. East has 9 points. 4♥ should make nearlyevery time. The joint pattern 1543//3442 fits well. No wastedhonors in the short suits. Example 3.[hv=pc=n&w=s2hak752dkj82c642&e=sqj98h64d975caq93]266|200[/hv] West has the same 11. East still has 9 points. 1543//4234 This pair of hands do not fit nicely. If everything goes wrong west may make only 3or 4 tricks in hearts. It is about power, points, location of those points,and how those points fit together. It is also about joint pattern. Too many systems go to great pains to evaluate individual hands to toomany decimal places. It isn't about points. Taking tricks depends onhow well those points are working together and how well the patternsof the partnership hands fit..............The main theme is this thread isThe objective of hand evaluation should be to find the best contractfor the partnership.The best method to achieve this objective is to estimate partnership tricks. Finding a more precise initial point count will do little to helpus achieve this objective. We need better methods to determinethe fit of the partnership, and thus would be better able to estimatethe partnership tricks. We must realize that expected tricks may varyradically with every successive bid. E(tricks) = trumps + (HCP-20)/3 + eStd dev is about 1.25 tricks/board.E(tricks) = trumps + (HCP-20)/3 + SF + eFor flat hands the std dev can be as low as 1 trick/board. There are four models for expected tricks, one for each suit. Expected tricks fluctuate wildly depending on strain of the final contract. We wantto be in the strain which maximizes our expected tricks. Quote Link to comment Share on other sites More sharing options...
Scarabin Posted July 21, 2013 Report Share Posted July 21, 2013 OK so far. We can easily agree on the general principles but I expect problems only arise when we get to the details of your proposed complete evaluation system. ;) Quote Link to comment Share on other sites More sharing options...
jogs Posted July 22, 2013 Author Report Share Posted July 22, 2013 OK so far. We can easily agree on the general principles but I expect problems only arise when we get to the details of your proposed complete evaluation system. ;) There will be no proposed detailed complete evaluation system presentedhere. The theme of this thread is think in terms of tricks for our partnership rather than points in our hand.. These statistical models are for the general case. Often completed by the third bid of the auction. Our expected tricksis X +/- e. All estimates come with errors and this e >=1. There are three phases to the evaluation process. The first phase is the initial point count. The two our tricks models are the second phase.By using the model which consist of HCP, trumps, and skewness/flatnessone can improve his starting point for the third phase. The first twophases are for the general case. This third phase is counting the tricksfor the actual board. This will require uncovering the effects of addtional parameters. This thread will make no suggestions on how to accomplish this task of counting. Future posts will suggest how to utilize the models for better bidding decisions, often at the part score level. Quote Link to comment Share on other sites More sharing options...
jogs Posted July 25, 2013 Author Report Share Posted July 25, 2013 Power and Pattern Review of the material.Power and pattern are two multi-dimensional independent randomvariables used to estimate tricks.Power is high card points, the location of those HCP, and how theyinteract with each other. This would require a complex polynomialtoo difficult to solve at the table. HCP is the component of powerwhich is roughly proportional to tricks. HCP will be used as aproxy for power in the our tricks model.Pattern is the ordered configuration of the four suits in one hand.Joint pattern is the joint pattern of the two partnership hands.All four suits have effects on tricks. Combined trumps is the component of pattern which is roughly proportional to tricks. Trumps is the proxy for pattern. E(tricks) = trumps + (HCP-20)/3 + e Skewness/flatness. Singletons and voids increases the expectedtricks. The absence of singletons and voids(or flatness) reducesthe expected tricks. Singletons and voids are components ofpattern, not power. E(tricks) = trumps + (HCP-20)/3 + SF + e Interaction. There are effects due directly to each individualparameter. There are also effects due to the interaction of thoseparameters.Most systems assign a fixed value to a singleton. Singletons arecomponents of pattern, not power. The true value of a singletonis dependent on how it reacts with other parameters. Therefore asingleton should have a variable value expressed in expected tricks. 5431 // 3343The singleton is valuable when we play in spades. Singletons in the hand with the long trumps limits opponents to one trick in that suit. The ruffing in spades is probably tricks we already had. 5431 // 3523Singletons in the hand with the short trumps are additional trumptricks, provided there are sufficient trumps. Played in the 5-4heart fit we are able to ruff clubs with the short hearts. That ispotentially seven trump tricks with hearts as trumps. These examples show that singletons do have a variable valuedepending on how it interacts with other parameters and in the general case that value should be measured in expected tricks.On any particular board attempt to count the actual tricksduring the auction. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.