[tnevolin]

jogs, on 2016-July-15, 08:22, said:

I believe systems should attempt to estimate tricks. All point count systems are artificial. Tricks are real. Losing trick count attempts to estimate tricks.
...
I believe all systems should stop adjusting points for distribution. These are parameters of separate independent random variables. Adjust the estimated tricks directly. That makes it easier to detect duplication of values.

Exactly! Thanks, jogs. These are exactly two messages I am trying to deliver. And I explicitly founded my method on these concepts with my eyes open. The model predicts tricks even though it uses 1 point = 1/3 trick for the sake of finer grained computation and ease of memorizing. It is also treat all features same for the computational purpose.

[tnevolin]

Stefan_O, on 2016-July-15, 08:28, said:

Out of curiosity, another idea/suggestion...

If you use double-dummy-analysis results on the deals, instead of actual play results,
would your eval-system come out significantly different, or would it be mostly the same?

Don't know. I would speculate it may be coincidentally close especially in HCP values. I wouldn't be bothered by it much because double dummy and duplicate bridge are different games!

[tnevolin]

jogs, on 2016-July-15, 08:30, said:

Let's look at joint partnership patterns.

a) 4324 // 4324
b) 4324 // 4234

The doubletons in different suits makes pattern b stronger than pattern a. Only how can we learn this quickly during the auction?

a) AKxx xxx xx QJxx // QJxx xxx xx AKxx
b) AKxx xxx xx QJxx // QJxx xx xxx AKxx

5 losers in a. Only 4 losers in b.

90+% of bidding systems won't bother finding it.

[Stefan_O]

tnevolin, on 2016-July-15, 08:48, said:

90+% of bidding systems won't bother finding it.

I would think 99.99% or so of all pairs, generally dont bother or have methods to find it

Some 25+ years ago, I played a relay-system with a guy very smart in bidding-design,
where a strong-enough hand (both as opener and responder) could always initiate a relay-sequence to ask for partners exact honour-strength (A/K/Q's) and suit-pattern.
Also, a dominating principle was to always try to place the stronger (often totally unknown) hand as declarer.

It was fun and we had excellent results with it for our level.
But after all the crackdowns and suppression against "highly unusual methods" over the years internationally,
today it's virtually impossible to even practice such systems in real, duplicate-bridge, which is obviously a big reason why so few play such methods.

[tnevolin]

Stefan_O, on 2016-July-15, 11:30, said:

Some 25+ years ago, I played a relay-system with a guy very smart in bidding-design,
where a strong-enough hand (both as opener and responder) could always initiate a relay-sequence to ask for partners exact honour-strength (A/K/Q's) and suit-pattern.
Also, a dominating principle was to always try to place the stronger (often totally unknown) hand as declarer.

It was fun and we had excellent results with it for our level.
But after all the crackdowns and suppression against "highly unusual methods" over the years internationally,
today it's virtually impossible to even practice such systems in real, duplicate-bridge, which is obviously a big reason why so few play such methods.

Interesting to know you like(d) to experiment with the bidding design. I've built a system based that utilizes full potential of this calculation model. Comparing to calculation model itself that is a secondary project and do not even pretend this sketch bidding system is a piece of art.
Yet I will be thrilled if someone would like to try it out to test this system together with the evaluation model. If not to play it out then, at least, review and criticize. Do you think it make sense to start parallel discussion or continue in this topic?

[Stefan_O]

Hi Tim,

On page 3 in the pdf, it says

tnevolin, on 2016-July-15, 14:56, said:

4. If you have 10+ cards, estimate your tricks directly and don't evaluate.

I can probably estimate the tricks in my own hand...

But how to estimate the combined tricks with partners hand?

Should we just assume pard has 0 tricks, and place the contract?
Or what is the idea here?

[tnevolin]

Stefan_O, on 2016-July-15, 15:11, said:

Hi Tim,

On page 3 in the pdf, it says

Quote

4. If you have 10+ cards, estimate your tricks directly and don't evaluate.

I can probably estimate the tricks in my own hand...

But how to estimate the combined tricks with partners hand?

Should we just assume pard has 0 tricks, and place the contract?
Or what is the idea here?

I don't have statistics beyond 9. "estimate your tricks directly" means exactly what it is: don't use the system but just count your tricks. Like if you have ten cards headed with AKQ - you have 10 tricks = 30 points. That's it. You don't need to assume anything about your partner just keep bidding as you would normally bid with 10 cards suit and 30 points.
Look for similar bids in other systems. Like American Standard 2C or 4M. They are based on number of tricks in your hand. That's what it is.
In fact, extreme distribution (of cards or points) cannot be covered by any calculation system at all. Just because they are extremely rare and no evaluation would care if they are good match or not. I think players wouldn't too.

https://en.wikipedia...e_probabilities
10-2-1-0 0.000011 = one in 100 thousands!!! I bet you won't ever encounter such hand ever.

[Stefan_O]

tnevolin, on 2016-July-15, 18:22, said:

https://en.wikipedia...e_probabilities
10-2-1-0 0.000011 = one in 100 thousands!!! I bet you won't ever encounter such hand ever.

OK, got it

[Stefan_O]

Hi Tim,
I wrote a program to calculate "Evelyn-points" under the various denominations, and applied to some random hands...
Does this look correct?

Deal   1:  AT32.KT.975.Q743   NT => 11   S => 11   H =>  9   D =>  9   C => 12
Deal   2:  J872.9.KQJ4.JT83   NT =>  9   S => 10   H =>  4   D => 11   C => 10
Deal   3:  AJ5.J64.T75.AT76   NT => 11   S => 11   H => 11   D => 10   C => 11
Deal   4:  KJ98.AK.864.Q874   NT => 13   S => 15   H => 12   D => 12   C => 15
Deal   5:  Q82.K42.AQJ8.743   NT => 13   S => 11   H => 11   D => 13   C => 10
Deal   6:  965.Q3.72.T97642   NT =>  3   S =>  3   H =>  3   D =>  2   C =>  9
Deal   7:  KQJ4.943.A4.JT86   NT => 12   S => 14   H => 10   D =>  9   C => 13
Deal   8:  A532.JT632.8.A53   NT =>  9   S => 13   H => 15   D =>  8   C => 12
Deal   9:  A8.KT9.Q4.AK8653   NT => 18   S => 14   H => 16   D => 15   C => 22
Deal  10:  54.64.KT9.QJ7632   NT =>  8   S =>  5   H =>  5   D =>  7   C => 13
Deal  11:  K9.AT7.93.QT8754   NT => 12   S =>  9   H =>  9   D =>  8   C => 16
Deal  12:  875.Q8.KQ52.AJ72   NT => 12   S => 11   H => 11   D => 14   C => 14
Deal  13:  A6.Q.T7642.KT753   NT => 10   S =>  9   H =>  8   D => 14   C => 15
Deal  14:  AT73.A3.J965.K73   NT => 13   S => 14   H => 11   D => 15   C => 13
Deal  15:  T9865.J86.A9.T73   NT =>  5   S =>  9   H =>  7   D =>  5   C =>  6
Deal  16:  A74.K32.KJ2.AJ97   NT => 16   S => 15   H => 16   D => 16   C => 17
Deal  17:  73.JT5.AQT843.J3   NT => 10   S =>  6   H =>  8   D => 14   C =>  7
Deal  18:  5.KJ96.AQT2.J742   NT => 12   S =>  8   H => 14   D => 14   C => 14
Deal  19:  A8632.9432.632.8   NT =>  4   S => 10   H =>  9   D =>  8   C =>  4
Deal  20:  K7653.5.T65.A873   NT =>  7   S => 13   H =>  6   D => 10   C => 11

(Voids:)
Deal  94:  Q963.KQ98.KJT54.   NT => 12   S => 15   H => 15   D => 17   C =>  7
Deal 104:  AQ8532..Q987.Q94   NT => 12   S => 18   H =>  6   D => 14   C => 12
Deal 183:  QJ86.J87642..A96   NT =>  9   S => 13   H => 17   D =>  5   C => 10

(8+card suits:)
Deal 539:  AKJT9832.A832.A.   NT => 22   S => 31   H => 22   D => 14   C => 13
Deal 759:  AQJ98763.JT..KJ2   NT => 18   S => 24   H => 11   D =>  7   C => 13
Deal 774:  .87.AQJ87542.T52   NT => 13   S =>  4   H =>  7   D => 21   C =>  9
Deal 816:  A9.A2.AKQT8765.J   NT => 24   S => 17   H => 17   D => 30   C => 16
Deal 861:  QT72.6.KQJT9864.   NT => 15   S => 14   H =>  5   D => 22   C =>  4

[Stefan_O]

wikipedia said:

https://en.wikipedia...e_probabilities
"10 card | 10-1-1-1, 10-2-1-0, 10-3-0-0 | 0.000017"

Looks like the wiki-stats are correct -- I found exactly 17 hands in 1 million!

Deal 4493:  K42...AKQJT76542   NT => 24   S => 17   H =>  9   D =>  9   C => 31
Deal 58287:  AKQJT986432...T4   NT => 23   S => 31   H =>  7   D =>  7   C => 11
Deal 131106:  AKQJT97432.3.5.7   NT => 21   S => 27   H =>  8   D =>  8   C =>  8
Deal 187123:  Q75.AKQJ987654..   NT => 22   S => 16   H => 30   D =>  8   C =>  8
Deal 325519:  A.KJT8765432.82.   NT => 18   S =>  7   H => 26   D => 10   C =>  6
Deal 371358:  .Q43..AQJT987532   NT => 19   S =>  5   H => 13   D =>  5   C => 27
Deal 440790:  AKQJ985432..53.5   NT => 20   S => 28   H =>  7   D => 11   C =>  8
Deal 512579:  QT..AKQJT98543.9   NT => 24   S => 13   H =>  8   D => 29   C =>  9
Deal 697433:  53.AKQT987632.5.   NT => 19   S => 10   H => 27   D =>  7   C =>  6
Deal 717414:  A.Q7..AKQJT97432   NT => 27   S => 13   H => 17   D => 12   C => 33
Deal 819367:  AKQJ876532..5.J8   NT => 21   S => 28   H =>  7   D =>  8   C => 12
Deal 881121:  KQJT987542..J6.Q   NT => 19   S => 25   H =>  4   D =>  9   C =>  6
Deal 898986:  AKQT976543.7.9.5   NT => 19   S => 26   H =>  7   D =>  7   C =>  7
Deal 913494:  6.6.2.AQJT987532   NT => 17   S =>  5   H =>  5   D =>  5   C => 24
Deal 939445:  AKQJT96542.K4.4.   NT => 24   S => 30   H => 14   D => 10   C =>  9
Deal 961699:  7.4.5.AKJ9865432   NT => 18   S =>  6   H =>  6   D =>  6   C => 25
Deal 994598:  AKQT765432..T4.J   NT => 20   S => 27   H =>  6   D => 10   C =>  8

[m1cha]

tnevolin, on 2016-July-15, 06:05, said:

"Evolin" is cool but I'm afraid it won't stick as a new word. "Evelyn" is more promising. Same word already exists and people would just transfer the meaning and remember it easier. What do you think?

I think that either should work. "Evelyn" seems a bit easier but doesn't carry the other ideas as easily as "Evolin". If your system is good, any name will do. People even remember "Cappelletti" . Though in that case misspellings are frequent, understandably.

tnevolin, on 2016-July-15, 06:05, said:

That is the same story again - statistics. I load pile of games into the machine and it gives out some coefficients those it thinks are best fit. I just may round them to the nearest whole number, that's all.

No, I disagree here. These coefficients don't come up randomly, they come up for a reason. For example, if you hold AKQ opposite xxx you can expect that those three honors in one hand will cover the three losers in the other hand. But if you hold AKQ opposite x, there is only one loser in this suit to be covered and the other two honors will make a trick only if you have losers in other suits. But sometimes your opponents will take their tricks in those other suits, so your honors become worthless. Or you may have to guess which card will become a loser and you discard the wrong card. This is why AKQ opposite x gets -1 point, it is worth 1/3 trick less than opposite xxx. Even worse opposite a void, you may not be able to access the honors when you need to because you cannot play to them from the other hand. This is why here you get -3 points for AKQ opposite a void. I would never be able to predice the coefficients correctly, of course, but it should usually be possible to predict if they are positive or negative and if they are high or low.

There is always a reason. But I admit that sometimes the things may be so complex that we cannot understand the reason easily.

tnevolin, on 2016-July-15, 06:05, said:

Now here are my speculation about K against a singleton. We are talking about controls duplication.

Not really. We are talking about the trick-taking probability of certain cards under certain circumstances (that is, opposite short suits). That is what the computer calculates. You are calling them "duplications" for good reasons but the computer doesn't know that word.
I know I appear fussy here but it will become clear in a moment why I am doing this.

tnevolin, on 2016-July-15, 06:05, said:

First round controls are Ace and void. Second round controls are King and singlet. Practically A against a singleton is not a duplication as A controls first round and singleton - second.

Correct. In other words: The ace covers the singleton and the singleton covers the next round(s). This is why the ace opposite a singleton gets its full value while an ace opposite a void gets -1.

tnevolin, on 2016-July-15, 06:05, said:

Q against singleton is not a duplication either as Q doesn't control either first or second round but singleton does. While K against a singleton is a duplication by definition as they both control second round. Same story about void. A against void is a duplication while K or Q is not and you see it in the table.

A void does not only control the first round, it also controls all following rounds. That makes the "duplication" thinking difficult for K/Q opposite a void.

Back to that king: 0 opposite a void while -1 opposite a singleton means that the king opposite a void is more likely to take a trick than a king opposite a singleton - and I cannot believe that. I could believe 0 for both and I could believe -1 for both but not the combination. Also +1 for that queen would mean that a queen opposite a void is more likely to generate a trick than a queen opposite more cards, and I cannot believe that either; and I don't think that other people will find that easy to believe.

400k boards is a huge amount of data. I am honestly respecting this. Yet voids are quite rare and you are optimizing many parameters at the same time. I can imagine that these numbers are statistical errors but I cannot be sure. If you want to make sure, you might divide the boards into 4 packages of 100k each, make 4 separate evaluations and compare the coefficients: Do they fluctuate or are they stable? I am aware this is a lot of work. You don't have to do this for me and you don't have to do it fast.

[m1cha]

tnevolin, on 2016-July-15, 06:12, said:

This is my pain. I couldn't get any freely available source anywhere in the internet. There are plenty tournament results but they do not include each pair score! Damn. I got part from swan games and part from one of the online bridge web site. Even there they are not prepared for download. I had to write my own crawler to scrape screens.
If you find any web source where pair results are at least visible on a page - let me know.

Hm, last month the 53rd European Team Championships took place in Budapest, the results are here:
http://www.eurobridg...ite/Results.htm

and if you click on a round and then on a table you get the individual scores such as here:
http://www.eurobridg...?qmatchid=34985

This is high-class bridge, definitely good data but probably not sufficient quantity for you, I am afraid, even if you check for earlier years.

Another possibility perhaps if you write to the BBO people, they might give you the accumulated data of these new "Free Daylong tournaments". That is ~ 10000 participants @ 8 boards each EVERY DAY. Worth writing another crawler for it, I guess . Not all of this is high-class bridge though. (Edit: and three of the players are always robots.)

[jogs]

Stefan_O, on 2016-July-15, 15:11, said:

Hi Tim,

On page 3 in the pdf, it says



I can probably estimate the tricks in my own hand...

But how to estimate the combined tricks with partners hand?

This is the method I use for estimating tricks.

E(tricks) = Trumps + (HCP-20)/3

Usually by the 2nd or 3th bid of the auction one or both partners will know the partnership combined trumps and the combined HCP +/- 1.

Then if the auction precedes further.


E(tricks) = Trumps + (HCP-20)/3 + SST + SS

Adjust the count from shortness of the side suits and source of tricks(usually from the second suit). I know my(hand's) contribution towards SST and SS. We must find ways to learn partner's contribution to these parameters.
Also it would be nice if it were possible to exchange this information starting from the three level.

Playing a natural 5-card major system(due to the importance of combined trump length, I'm convinced 4-card major systems are inferior) play fit jumps even in uncontested auctions.

Example:

AKxxx Axx x Qxxx // QJxx x xxx AKJxx

West deals. Opponents silent.

1♠ - 3♣
3♥ - 3♠
4♦ - 6♠

3♣ is a fit jump and forcing to game.
3♥ heart ace.
3♠ forcing
4♦ second round control of diamonds, usually a singleton.

[tnevolin]

Stefan_O, on 2016-July-15, 18:50, said:

Hi Tim,
I wrote a program to calculate "Evelyn-points" under the various denominations, and applied to some random hands...
Does this look correct?

Deal   1:  AT32.KT.975.Q743   NT => 11   S => 11   H =>  9   D =>  9   C => 12
Deal   2:  J872.9.KQJ4.JT83   NT =>  9   S => 10   H =>  4   D => 11   C => 10
Deal   3:  AJ5.J64.T75.AT76   NT => 11   S => 11   H => 11   D => 10   C => 11
Deal   4:  KJ98.AK.864.Q874   NT => 13   S => 15   H => 12   D => 12   C => 15
Deal   5:  Q82.K42.AQJ8.743   NT => 13   S => 11   H => 11   D => 13   C => 10
Deal   6:  965.Q3.72.T97642   NT =>  3   S =>  3   H =>  3   D =>  2   C =>  9
Deal   7:  KQJ4.943.A4.JT86   NT => 12   S => 14   H => 10   D =>  9   C => 13
Deal   8:  A532.JT632.8.A53   NT =>  9   S => 13   H => 15   D =>  8   C => 12
Deal   9:  A8.KT9.Q4.AK8653   NT => 18   S => 14   H => 16   D => 15   C => 22
Deal  10:  54.64.KT9.QJ7632   NT =>  8   S =>  5   H =>  5   D =>  7   C => 13
Deal  11:  K9.AT7.93.QT8754   NT => 12   S =>  9   H =>  9   D =>  8   C => 16
Deal  12:  875.Q8.KQ52.AJ72   NT => 12   S => 11   H => 11   D => 14   C => 14
Deal  13:  A6.Q.T7642.KT753   NT => 10   S =>  9   H =>  8   D => 14   C => 15
Deal  14:  AT73.A3.J965.K73   NT => 13   S => 14   H => 11   D => 15   C => 13
Deal  15:  T9865.J86.A9.T73   NT =>  5   S =>  9   H =>  7   D =>  5   C =>  6
Deal  16:  A74.K32.KJ2.AJ97   NT => 16   S => 15   H => 16   D => 16   C => 17
Deal  17:  73.JT5.AQT843.J3   NT => 10   S =>  6   H =>  8   D => 14   C =>  7
Deal  18:  5.KJ96.AQT2.J742   NT => 12   S =>  8   H => 14   D => 14   C => 14
Deal  19:  A8632.9432.632.8   NT =>  4   S => 10   H =>  9   D =>  8   C =>  4
Deal  20:  K7653.5.T65.A873   NT =>  7   S => 13   H =>  6   D => 10   C => 11

(Voids:)
Deal  94:  Q963.KQ98.KJT54.   NT => 12   S => 15   H => 15   D => 17   C =>  7
Deal 104:  AQ8532..Q987.Q94   NT => 12   S => 18   H =>  6   D => 14   C => 12
Deal 183:  QJ86.J87642..A96   NT =>  9   S => 13   H => 17   D =>  5   C => 10

(8+card suits:)
Deal 539:  AKJT9832.A832.A.   NT => 22   S => 31   H => 22   D => 14   C => 13
Deal 759:  AQJ98763.JT..KJ2   NT => 18   S => 24   H => 11   D =>  7   C => 13
Deal 774:  .87.AQJ87542.T52   NT => 13   S =>  4   H =>  7   D => 21   C =>  9
Deal 816:  A9.A2.AKQT8765.J   NT => 24   S => 17   H => 17   D => 30   C => 16
Deal 861:  QT72.6.KQJT9864.   NT => 15   S => 14   H =>  5   D => 22   C =>  4

Great idea. You doesn't cease to amaze me. Why didn't I think about it myself? I checked first three lines and they are correct. I think I'll follow your idea and create a complimentary Java program to distribute along with the document. Probably add some description as how each value is calculated.
...
Also checked first line from void and 8+ cards. Void is correct but for 8+ cards you probably can count this rule "High card combinations in side suit with 8+ cards on line (out of 3 top cards)" too because you have known 8 cards on line.

[Stefan_O]

tnevolin, on 2016-July-16, 08:47, said:

for 8+ cards you probably can count this rule "High card combinations in side suit with 8+ cards on line (out of 3 top cards)" too because you have known 8 cards on line.

Yes, well spotted...
I didnt implement the "combination-rules", and missed that one...

[tnevolin]

@m1cha
Thank you for great post. I'll try to comment on your points one by one.

m1cha, on 2016-July-15, 21:47, said:

Quote

That is the same story again - statistics. I load pile of games into the machine and it gives out some coefficients those it thinks are best fit. I just may round them to the nearest whole number, that's all.

No, I disagree here. These coefficients don't come up randomly, they come up for a reason. For example, if you hold AKQ opposite xxx you can expect that those three honors in one hand will cover the three losers in the other hand. But if you hold AKQ opposite x, there is only one loser in this suit to be covered and the other two honors will make a trick only if you have losers in other suits. But sometimes your opponents will take their tricks in those other suits, so your honors become worthless. Or you may have to guess which card will become a loser and you discard the wrong card. This is why AKQ opposite x gets -1 point, it is worth 1/3 trick less than opposite xxx. Even worse opposite a void, you may not be able to access the honors when you need to because you cannot play to them from the other hand. This is why here you get -3 points for AKQ opposite a void. I would never be able to predice the coefficients correctly, of course, but it should usually be possible to predict if they are positive or negative and if they are high or low.

There is always a reason. But I admit that sometimes the things may be so complex that we cannot understand the reason easily.


Not really. We are talking about the trick-taking probability of certain cards under certain circumstances (that is, opposite short suits). That is what the computer calculates. You are calling them "duplications" for good reasons but the computer doesn't know that word.
I know I appear fussy here but it will become clear in a moment why I am doing this.


Correct. In other words: The ace covers the singleton and the singleton covers the next round(s). This is why the ace opposite a singleton gets its full value while an ace opposite a void gets -1.


A void does not only control the first round, it also controls all following rounds. That makes the "duplication" thinking difficult for K/Q opposite a void.

Back to that king: 0 opposite a void while -1 opposite a singleton means that the king opposite a void is more likely to take a trick than a king opposite a singleton - and I cannot believe that. I could believe 0 for both and I could believe -1 for both but not the combination. Also +1 for that queen would mean that a queen opposite a void is more likely to generate a trick than a queen opposite more cards, and I cannot believe that either; and I don't think that other people will find that easy to believe.

What I meant about coefficients is that there are different ways to get them. Computer and people solve same task of calculating coefficients (feature values). They do it differently. Sometimes they converge on coefficient and people are happy to see two different approaches match in the end. That is all to it. We cannot actually judge the way computer think if coefficients do not converge as good as we would like. Our attempt to "explain" it is just a rationalization of our own model that doesn't actually prove that we are right. The only way to judge it is to do it 10 different ways and if 9 of them converge but 1 stands out then we can presume these 9 are correct. When you have only two it's inconclusive.

I agree with you that we are estimating tricks not controls. I was just using term control trying to "explain" these results in bridge terms. This explanation is merely a mind game or speculation that doesn't actually prove these values are correct or not.

Now, if we continue our speculative mind game , keep in mind that "High card combinations in side suit with 8+ cards on line" and "Value duplication" feature are corrective ones. You can see "Optional. Count only if known." note for each of them. That means that even if you do not count them due to lack of partner's hand knowledge the result still be correct. These two coefficients allow you to do finer tuning in case you have information to use them. That's why they go to both positive and negative sides.

Back to the K-x. When we are analyzing the value of "Value duplication" coefficient we need to remember that there are other features at play. You and your partner are separately counted trick taking potential of king and singleton, correspondingly. If you don't know each other hand - that's fine. However, if you do know that your king and his singleton are occurred in the same suit, you can use this knowledge to further fine tune the result by applying "Value duplication" rule and see what difference does it make. So K-x = -1 doesn't say anything about king trick taking potential. It says that king trick taking potential and singleton trick taking potential clash and the result of this clash is that the combined trick taking potential of kind and singleton when they are in the same suit is 1 point less than if they were in different suits. I agree with you that this is a very complex dependency to grasp. That's why I used term of controls to try to explain (not to prove) this result. Think of this as a mnemonic rule that helps to remember this irregularity.

m1cha, on 2016-July-15, 21:47, said:

400k boards is a huge amount of data. I am honestly respecting this. Yet voids are quite rare and you are optimizing many parameters at the same time. I can imagine that these numbers are statistical errors but I cannot be sure. If you want to make sure, you might divide the boards into 4 packages of 100k each, make 4 separate evaluations and compare the coefficients: Do they fluctuate or are they stable? I am aware this is a lot of work. You don't have to do this for me and you don't have to do it fast.

You are right that some features occur more often than others. That's why I explicitly excluded unstable coefficient with insufficient statistics. Those included in the document are reliable!
In numbers, I excluded feature those occur less than 100-200 times overall. With 100 results statistical error for corresponding coefficient is about 10%. So if its numeric value is less than 5 then absolute error is less than 0.5 which is OK. This is only for very very rare features. I can tell you void is not rare. 10 cards suit is rare.

[tnevolin]

m1cha, on 2016-July-15, 22:05, said:

Hm, last month the 53rd European Team Championships took place in Budapest, the results are here:
http://www.eurobridg...ite/Results.htm

and if you click on a round and then on a table you get the individual scores such as here:
http://www.eurobridg...?qmatchid=34985

This is high-class bridge, definitely good data but probably not sufficient quantity for you, I am afraid, even if you check for earlier years.

Another possibility perhaps if you write to the BBO people, they might give you the accumulated data of these new "Free Daylong tournaments". That is ~ 10000 participants @ 8 boards each EVERY DAY. Worth writing another crawler for it, I guess . Not all of this is high-class bridge though. (Edit: and three of the players are always robots.)

Thanks, m1cha.
First link is useful. I can see both deal and pair results on the same page. Will see how much can I scrape from there.
BBO robots are useless, unfortunately. They are not playing like humans.

[tnevolin]

jogs, on 2016-July-16, 07:44, said:

This is the method I use for estimating tricks.

E(tricks) = Trumps + (HCP-20)/3

Usually by the 2nd or 3th bid of the auction one or both partners will know the partnership combined trumps and the combined HCP +/- 1.

Then if the auction precedes further.


E(tricks) = Trumps + (HCP-20)/3 + SST + SS

Adjust the count from shortness of the side suits and source of tricks(usually from the second suit). I know my(hand's) contribution towards SST and SS. We must find ways to learn partner's contribution to these parameters.
Also it would be nice if it were possible to exchange this information starting from the three level.

Playing a natural 5-card major system(due to the importance of combined trump length, I'm convinced 4-card major systems are inferior) play fit jumps even in uncontested auctions.

Example:

AKxxx Axx x Qxxx // QJxx x xxx AKJxx

West deals. Opponents silent.

1♠ - 3♣
3♥ - 3♠
4♦ - 6♠

3♣ is a fit jump and forcing to game.
3♥ heart ace.
3♠ forcing
4♦ second round control of diamonds, usually a singleton.

There are two different questions mixed in our conversation.

1. Evaluate your hand strength (in tricks or points).

a. By counting number of tricks you can take in you hand. You also can multiply it by 3 to express hand strength in points to standardize hand strength communications in bidding if needed.
b. By using evaluation method (summarizing all feature values).

2. Estimating combined partnership trick taking potential.

a. Each partner evaluate their hand and then you add two numbers. That happens in normal bidding.
b. One player tries to estimate partner's strength by observing bidding and assigning some average strength to partner. Used in blocks and strong openings to weigh the risk.

What I meant by "if you have 10+ cards in a suit ..." is that with extreme distribution you will get more precise evaluation by counting tricks in your hand directly and then multiplying by 3 rather than using evaluation method. Evaluation method helps you when you are not sure how many tricks you can take - when your values are all different type and scattered across suits, etc. When they are all concentrated in a single suit it is no brainier. You abandon complex evaluation method and switch back to trick counting. Then you continue bidding as usual to understand combining strength. Your partner may use evaluation model or something else depending on your agreements.
In short, my model is about 1-b that for extreme hands can be replaced with 1-a on player discretion. Then it ends up with 2-a. It has nothing to do with 2-b!

[tnevolin]

There is one concept common for all valuation methods. I didn't include it into document description as I thought it is quite obvious. Apparently I see many of you keep asking it over and over again in various forms. Let me try to clarify this confusion.

There are two ways to understand trick taking potential of your pair.

1. Have X-ray vision and just plainly count all tricks by solving double dummy problem.
2. Use some sort of valuation method, pass some encoded information to your partner, calculate combined strength, and evaluate critical game prospectus by some rule.

It goes without saying that first way is superior. It is much easier and is absolutely precise. Second way just sucks in comparison. Therefore, if you possess X-ray vision, by all means use it. However, sometimes dark forces cloud your vision of other hands and your hand alone doesn't give you a clue (99.99% of hands). Then evaluation system comes to the rescue. It is complex, cumbersome, and not 100% accurate but doable and better than educated guessing. Bridge battle is a battle of evaluation methods, bidding systems, and card play. Enhancing one of the components give you edge over others.

So please please don't ask me why this suit takes 10 tricks in NT while my model gives it only 21 points.

AKQJTxxxxx

[jogs]

tnevolin, on 2016-July-16, 10:28, said:

There are two different questions mixed in our conversation.

1. Evaluate your hand strength (in tricks or points).

a. By counting number of tricks you can take in you hand. You also can multiply it by 3 to express hand strength in points to standardize hand strength communications in bidding if needed.
b. By using evaluation method (summarizing all feature values).

2. Estimating combined partnership trick taking potential.

a. Each partner evaluate their hand and then you add two numbers. That happens in normal bidding.
b. One player tries to estimate partner's strength by observing bidding and assigning some average strength to partner. Used in blocks and strong openings to weigh the risk.

By summing partner's trumps with my own and summing partner's HCP with my own I can get a first estimate partnership trick taking potential. These are the two general case parameters. SST and source of tricks are parameters specific to the current board.
----
Also by estimating tricks AKQJTxxxxx is 10 tricks. Points are artificial. Tricks are real.

E(tricks) = Trumps + (HCP-20)/3

Notice that my trick estimate depends on which suit is trumps.