LoTT is a parabola

[jogs]

 

 

For example. With two 8 card fits, if one is 4-4 and trumps = 5+ possible tricks. What about the other one - a side suit? If it is also 4-4, not so hot. If it is 6-2 and solid = six tricks.

 

I certainly think it's true.

8-0 fit > 7-1 fit > 6-2 fit > 5-3 fit > 4-4 fit

The skewed fits generate more tricks than the flatter fits.

 

 

Modeling Bridge is really, really tough, and I have little confidence anyone can get very far with it. Again, there is the underlying problem is that Bridge is not played double dummy.

I like to use observed results from BBO minis. Only it is really hard to find boards which fit the desired conditions.

[jogs]

Our trump length correlates with opponents trump length. Basically if e have 9 card fit opponents have it too or they have two times 8 card fits.

 

The correlation is weak.

 

We have 8 trumps. Think in terms of trumps for the partnership. They have 5 in that suit.

5=7=7=7

They can have 7 as their longest suit.

We have 8=8=8=2. They can have 11 as their longest suit.

Therefore when we have 8, they have from 7 to 11 trumps.

 

We have 9 trumps. They have 4 in that suit.

4=8=7=7

They can have 8 as their longest suit.

We have 9=9=8=0. They can have 13 as their longest suit.

Therefore when we have 9, they have from 8 to 13 trumps.

[navahak]

The correlation is weak.

 

I knew about multiple fits increasing the opponents fit. But if you take into account probabilities of different distributions the correlation is a lot stronger than if you just look extreme examples. This correlation is very similar to LoTT and how accurate it is on average.

 

Same as Mike explained in long detailed post: I follow law only loosely and base my decision in many other factors too.

[jdeegan]

I like to use observed results from BBO minis. Only it is really hard to find boards which fit the desired conditions.

I am with you all the way, but consider:

 

The original Verne analysis modeled world championship Bridge circa the late 20th century. Your approach is, no doubt, better and more sophisticated than this original work, but you are left with the undeniable fact that you are 'just' modeling BBO minis. This should help you beat BBO minis, but what else?

 

I am afraid that the best you can really do with modeling is to define the kind of game you are trying to beat - both in a time and a place - and figure out the best way to beat it. Social science modeling is not like modeling in physics. The bastards you are trying to model react and keep moving.

 

Put another way, my credit card default model worked like a champ in its time. It even made me some pretty good money for a while. I would not advise using it today even reestimated using current data.

 

As far as selecting just the right kind of hands as a sample for any kind of valid statistical analysis. Welcome to the real grunt work involved in modeling Bridge. Computer screens have to be loose enough not to truncate your sample too much, if at all, and how much is too much, anyway? I think you need multiple screens customized for your specific application. In other words, screen the sample, then screen the rejects using a looser screen, etc., etc. I don't know of any canned programs you might use, but by now, they may exist.

 

Good luck. I am afraid a certain amount of hand selection may end up being called for.

 

P.S. There may a way to use BBO data from 'better' players. Several 'salons' featuring high level players have emerged on BBO. They feature sponsors like JEC and hosts like Susina. Those hand records are available to anyone using BBO software back for 2 or 3 months. You might be able to get more history if you asked BBO and reassured them you would keep it confidential. I can't see why they would object, assuming they trusted you.

[campboy]

The correlation is weak.

FWIW the correlation coefficient is 0.58645 by my calculations.

[whereagles]

Do you have a copy of the Ginsberg article? There's a chart of the data in its raw form on page 10. The tricks are occasionally +/- 4 from trumps.

 

No, I don't. Where can I find that chart? I could use it for the analysis.

[Trinidad]

FWIW the correlation coefficient is 0.58645 by my calculations.

Just out of curiosity: Did you solve this analytically or did you just run a sim?

 

Rik

[campboy]

Just out of curiosity: Did you solve this analytically or did you just run a sim?

Analytically.

[RMB1]

Actually, you can fit a non linear regression model to Ginsberg's data and run significance tests on c, c1, c2. I might set c = 0 from the start, though.

 

I don't think c=0 is safe.

 

A naive fit to the data looks something like: trumps - (trumps - 16.5)^2/24, where c <> 0.

[whereagles]

Fitting with c <> 0 is better/easier than c = 0. But if you can do with c = 0, that seems more logical.

[jogs]

The frequency tables.

 

		Error Analysis of the Law of Total Tricks

Length	  -4	  -3	  -2	  -1	   0	   1	   2	   3	   4

14	0.001	0.006	0.048	0.270	0.466	0.182	0.024	0.002	0.000
15	0.001	0.006	0.050	0.266	0.457	0.192	0.026	0.002	0.000
16	0.001	0.005	0.038	0.207	0.424	0.263	0.055	0.007	0.001

17	0.001	0.006	0.052	0.235	0.394	0.245	0.058	0.007	0.001
18	0.001	0.008	0.071	0.250	0.360	0.228	0.069	0.012	0.001
19	0.001	0.013	0.107	0.289	0.338	0.188	0.054	0.010	0.001

20	0.002	0.031	0.164	0.315	0.299	0.137	0.044	0.007	0.000
21	0.007	0.074	0.240	0.319	0.233	0.101	0.023	0.003	0.000
22	0.027	0.137	0.286	0.302	0.177	0.054	0.016	0.001	0.000

23	0.042	0.257	0.291	0.262	0.122	0.021	0.000	0.000	0.000
24	0.111	0.333	0.289	0.222	0.022	0.022	0.000	0.000	0.000
Totals	0.001	0.009	0.062	0.245	0.400	0.224	0.051	0.007	0.001

[whereagles]

Those are percentages, right?

 

Still need to think how to test this. A linear regression seems too much of a stretch here, given the data is highly non-normal.

[jogs]

Each row sums to 1.

 

Most one way layout analysis of variance examples I've seen were testing if means of different groups are equal.

In this study the group means are known to be increasing as total trumps increase.

Also the group variances are not equal. Group variances increase as total trumps increase.

 

It is probably not that useful to know the exact equation for this parabola.

Total trumps is 21 or more only about 1.5% of the time. That is about 3 times every 8 sessions.

For trumps =14,,,18: tricks = trumps.

For trumps = 19 and 20: tricks are nearly equal to trumps.

For trumps > 20: tricks are clearly less than trumps.

That's probably all we need to know.

[whereagles]

Interesting stuff nonetheless. I'll see if I can dig up something. (Time constraints apply...)

 

In any case one thing is clear: the LOTT, in its E(tricks) formulation, breaks down for 20+ trumps, with a shift towards less tricks. Whether or not the shift is parabolic is another story, though in 1st order it should be so.

[jogs]

I'm in agreement with Lawrence/Wirgren that we should be attempting to estimate our tricks. LoTT is estimating total tricks. Too many unknowns beyond our control. Too difficult to enumerate them. Our tricks is loosely independent of their tricks. Tricks are mostly dependent only in the 4333 suits.

[jdeegan]

Lawrence/Wirgren seems to work fairly well even with weird hands.

[jogs]

Let's clarify what I meant. Estimate our side's tricks. Not necessary by L/W methods.

Working points is more comprehensive than HCP. But too abstract for any of us to use without being allowed to view all 26 cards of our partnership.

The other independent variable is the joint suit pattern of both hands. This is best represented by combined trumps with a SST displacement adjustment.

 

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

 

With a 4-4 trump fit and SST=4 the adjustment is + 1/6.

 

With a 5-4 trump fit and SST=4 the adjustment is - 1/3.

 

There is no fix adjustment for each SST value. It depends on the trump fit.

[campboy]

The frequency tables.

Thanks for posting the full tables. In that case, whatever the figure after the +/- in the summary table posted upthread are supposed to be, they are not the standard deviations (or the variances). The standard deviation for 24 total trumps, based on Ginsberg's frequencies, is 1.113 -- almost identical to the result of my sim (and the variance -- the square of the standard deviation -- is 1.240).

 

The actual standard deviations arising from Ginsberg's table are as follows.

14	15	16	17	18	19	20	21	22	23	24
0.898	0.912	0.986	1.034	1.118	1.145	1.188	1.216	1.240	1.148	1.113

[jogs]

E(tricks) = trumpsi + (HCP-20)/3 + SSTi,m

 

for i = 7,8,....,13 and for m = 0,1,....,6

SST is a table of displacement constants from the estimates based solely on trumps and HCP.

 

For any specific board the SST displacement is a discrete integer. In the general case it is a fractional amount.

 

xxx ------- x: There may be two additional tricks from ruffs

AKQ ------ x: No additional tricks.

------------------------

Every trump combination may produce a different set of SST.

6-2 fit is different from 4-4

SST=3+1 may be different from SST=2+2

 

E(tricks) = trumpsi+j + (HCP-20)/3 + SSTi+j,m+n

 

for i+j = 7,8,....,13 and i>=j

for m+n = 0,1...,6 where m is from the long trump hand and n may be from either hand.

 

There may be over 100,000 permutations. Luckily it is only necessary for us to know the directional bias of the displacement. That should put us ahead of the crowd.

[navahak]

Actually as simple as tricks=(WP-20)/3+13-SST puts you very close to actual trick count available for your side.

[helene_t]

Maybe at some point I should stop repeating myself but I really don't see the relevance of Wirgren's formula in a thread about the lott, the misleading title of his book notwithstanding. The lott is about total tricks so counting the hcps for a particular side is obsolete. Unless it is really the tricks for a particular side you are interested in, in which case the lott is obsolete.

[Trinidad]

I'm in agreement with Lawrence/Wirgren that we should be attempting to estimate our tricks. LoTT is estimating total tricks. Too many unknowns beyond our control. Too difficult to enumerate them. Our tricks is loosely independent of their tricks. Tricks are mostly dependent only in the 4333 suits.

Of course, it is important to be able to estimate our tricks. But it is also important to be able to estimate the total tricks.

 

These are two entirely different questions that are both important.

 

Estimating our tricks could e.g. tell us that we are have a 50% chance to make 3♠ and a 50% chance that it goes down. Does that tell us whether we should bid 3♠ over their 3♥? Not at all.

 

If we estimate 14 total tricks, meaning that they will take 5 or 6 tricks in 3♥, we should simply double 3♥. It will go down 3 or 4, which will be a lot better than making 3♠ or going down one.

If we estimate 20 total tricks, meaning that they will take 11 or 12 tricks in a heart contract, we should make them guess, and preferably make them guess wrong.

 

In these cases, it is fairly useless to know how many tricks we can take. It is more useful to know how many tricks the opponents can take.

 

I bought "I fought the Law" as soon as it came out. I was not amused when I read it. I expected an improved method for total tricks, better than the LoTT. That's what the title suggested and that is how Lawrence and Wirgren advertized it. Instead, I got yet-another-method-to-evaluate-hands-solely-for-offensive-purposes (after Milton Work count (4-3-2-1), Vienna count (7-5-3-1), zz points (3-2-1), LTC, Zar points, rule of 20, distribution points, fit points, shortness points, shortness points depending on fit, working points, ...). Frankly, I don't really care whether Lawrence/Wirgren works better than Zar points (or LTC or whatever). They promised me a book that would fight the Law, not Zar.

 

To me it felt like a book with the title "I fought McDonalds" where the author (perhaps even successfully, I don't care) argues why apple juice is tastier than orange juice (with somewhere a sentence "Burgers taste like $&%§% anyway"). But we all know that a book with the title "I fought orange juice" will not sell as well as "I fought McDonalds". (But perhaps "I fought OJ" would sell. ;) )

 

Rik

[jogs]

Of course, it is important to be able to estimate our tricks. But it is also important to be able to estimate the total tricks.

 

In uncontested auctions they aren't very helpful. In contested auctions we can't trust their bidding.

 

Normalize the boards where each side has 20 HCP, all in two suits.

S AKxxx		S QJxx
H xx		H xx
D AKx		D QJxx
C xxx		C xxx

 

We can estimate our tricks at 9. Use the SST adjustment and lower the estimate to 8.

We know virtually nothing about their tricks.

 

S xxx

H AKxxx

D xx

C AKx

 

S x

H QJxx

D xxxx

C QJxx

 

They can make 10 tricks in hearts.

 

S xx

H AKxxx

D xxx

C AKx

 

S xx

H QJxx

D xxx

C QJxx

 

Now they can only make 8 tricks.

Our chances of estimating total tricks is much less reliable than estimating our own tricks.

[GreenMan]

You can play with their and our side suits the same way and come up with 8-10 tricks for each side, averaging 9, for a total of 18, which is the same as the number of trumps.

 

We guess looking at our shape that we probably don't have 9, meaning the actual total averages 17. We don't know if they have 8 or 10 so we take the average for calculation purposes; in the long run it'll balance out to about that. That's what the LoTT is for, the long run, not figuring out This Hand.

 

We have to estimate; it's the nature of the game. That's why cards have backs. :P

[Trinidad]

Of course, it is important to be able to estimate our tricks. But it is also important to be able to estimate the total tricks.

Our chances of estimating total tricks is much less reliable than estimating our own tricks.

1) You don't give any decent arguments to support this statement. We just saw in this tread that a function describing the total number of tricks, only based on the number of trumps that both sides hold (no adjustments or anything), has a standard deviation of about 1 total trick. This scales to a standard deviation of 0.5 tricks for the number of tricks one side can take. I cannot think of a method that can predict the number of tricks with an accuracy of half a trick. Yet, you claim that estimating total tricks is much less reliable than estimating our own tricks. What magical method do you have that leads to a standard deviation of less than say 0.05 tricks? (I'll be nice to you and say that a factor of 10 is "much less".)

 

2) Suppose, for the sake or argument, that your statement would be true and estimating total tricks is indeed less reliable than estimating our own tricks, how does that mean that we should stop estimating total tricks when the situation clearly asks for a total trick estimate? If something is difficult, we should give up and throw our hands in the air? Remind me not to let you near my children. ;)

 

Rik