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All comments by Stig Holmquist
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Sorry I'm late responding. Please tell me what other websites have the data. The OccEnt is clearly wrong. My primary interest was finding two 8-fits. RP calculates it long ago but did not post it.
April 20
Stig Holmquist edited this comment April 20
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Nick,
I see you are back to your old bag of tricks. When logic fails, you resort to derogatory and insulting comments.
The standards for this forum asks you to be respectful.
Do you know what that means?
Five bridge ideas of mine have been printed in Bridge World and one in The Bulletin. What have you ever contributed?
You need to adopt common civil manners.
April 18
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Re-read Loeb's comment. If you can memorize L. Cohen's complex method, more power to you. Most imtermediate players would not be able or care to try. So they need a simpler method. Use your imagination and devise a bid asking for shortness.

Let me suggest that after opener accounts for controls and responder thinks there might be a slam,
he can 3NT to ask opener for a shortness. With none, opener rebids his major, while a new suit will
show the shortness. This would work just fine with the first example. It can be no simpler.
April 18
Stig Holmquist edited this comment April 19
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You state strain before game and game before slam. As far as I can tell, the J2N establishes both the strain and the game. Why would anybody look for an 8-fit when a 9-fit has been established?

The only remaining question is asking it slam can be made.
Knowing if opener has a shortness is of little value.
Thus all that responder needs to know is the primary strength of openers hand.
This he can find out if opener tells him how many controls he has.
With too few he signs off in 4 or 5 M. What else would responder care about?
April 18
Stig Holmquist edited this comment April 18
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How often have you found playing in an 8-fit better than in a 9-fit ?
April 17
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Returning to The post by Alex please note that by showing the number of controls opener has, it's possible for responder to determine the pair is missing 3 controls, but
will not be able to tell if it is A+K or 3 Ks. That can be determined by asking for Ases. If one or more Ks were QJs
then responder would not consider bidding more than 4S, because too many controls would be missing.
April 16
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You need to read D. loeb's posting. He rejects the standard
response to Jacoby, where a new suit shows a singleton or void. It was posted 4 years ago.
April 15
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So far you have failed to tell us what bidding method you propose.
The result is no surprise. A similar bidding problem was
illustrated by Jeff Rubens in his book "The secretes of winning bridge. On p.7-8 he shows four hands with equal cards but arranged so that they can take 4 -7 tricks.
Look them up.
April 15
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Cohen & Be4kowtz played it the national and international level Their method requires too much memory.
April 15
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By what bidding sequence would you get to 6 S?
April 15
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Jeff Rubens just told me that calculating the probability for having two 8-fit is a trivial task not worth posting
in Bridge World. Personally, I'm unable to do such calculations. Why didn't a BW member do it?
The Occasional Enthusiast did not get it right
April 13
Stig Holmquist edited this comment April 13
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I asked R. Pavlicek if he has any data bearing on this problem, and he responded with a print out from an old file.
His data show that a pair will have 4432 fit 0.1 % of all deals. Since there are 6 such combinations, viz. S+H, S+D, S+C, H+D, H+S, D+C one would expect it to occur 0.6 % of the time.

In table 9 the 4-4-3-2 and the 4-4-2-3 is missing, but must
be equal to 3-2-4-4 and 2-3-4-4 for a total of 28.66 %.
R.P. shows there is a ratio of 1 125 for 32/23 with 44. The OCCENT data show them to be equal.

Related data can be found in the article “Hand Evaluation in the Game of Bridge” by Richard Cowan in J.Royal Stat. Soc.
Vol.38, 1987. on p. 58-70. He too shows a 1.125 ratio.
April 12
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Borel & Chevron reported 5% for a generic 4432 match. How can you explain the 27.8% by CE?
April 10
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Sorry, I forgot to identify the web site. If is “Occasional
Enthusiast”. It has two parts..Look at the 2nd one.
Then compare it with the table 21 of Borel& Chron with 6%.
April 10
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I just discovered a web site that has a lot of bridge math tables. Thee is one that seems to show the probability of
two hands having the same generic hand pattern.
For 4432 in Table 3 the value is 27.8%. And Table 9 shows
the 12 combinations for 4432 matching a fixed 4432.
But if you look a print out for 3 sessions you will not find
that many. So what do the data mean?
April 10
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I was unable to find any table showing the probability of both both hands being 4333. What value do you see?
April 7
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Control response after a strong NT bid could be based on a 4C asking bid by partner. The rebid would include any K control,and thus show only Ace controls.
April 4
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One could use a control asking bid after a strong NT bid..There are a total of 12 controls (A=2, K=1) per deal.
Opener will have at least 3 controls and thus needs to show any extra controls in steps after responder bids 4C.
This rules out any other use for 4C in NT bidding.
G.Rosenkrantz devised a table for “Expected Controls” in balanced hands. But I'm unable to find a table showing the
probability of having 0-12 controls together.Is thee one?
April 2
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There is no trump Queen in a NT contract. My new convention
is intended to be used when looking for a NT sam.
March 30
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Would you also open 1 S if the hand had the Q of S?
I'm trying to question the traditional wisdom that a 2C opening bid should be based on 22+ HCP.
The 2C opening bid should be based on trick taking potential, rather than HCP. Having more quick
tricks than LTC is a better basis for opening 2C.
March 16
Stig Holmquist edited this comment March 16
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