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All comments by Sergio Polini
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I've read 4, 6, 7, 13, 15, 18, 20, 22, 31, 34, 35, 38, 40, 42, 44, 49, 53, 56, 58, 60, 61, 63, 72, 76, 78, 81, 83, 84, 85, 87, 100.
Which should my next book be?
June 3
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@Max Schireson: as to DD vs real play in partscores vs games vs slams, you could look at http://www.rpbridge.net/9x29.htm
June 3
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As to double stepping stone squeezes, you can look at Kelsey, Double Squeezes, pages 66-69. There are also double vice and double winkle in that book.
April 20
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I'd say that it is a surplus winner strip-squeeze, because West must give up his spade length to keep the clubs control and the hearts fragile stopper:
- one suit (diamonds) has the squeeze card
- in one suit (spades) LHO (or whoever the squeeze is against) has winners
- there is a suit which you can set up tricks by knocking out controls (clubs)
- there is a suit in which you have a threat and LHO has to guard the threat (hearts).
See http://bridgesqueezes.info/sqwinnersqueeze.htm
April 12
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I think that the real difference is between 1NT-then-Pass (points, heart stopper, eventual diamond length) and 2-then-Double (points but nothing about red colors).
Aug. 4, 2017
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Yes. In the actual hand, at least.
West: AJxx xx Kxxx Jxx
East: Qxx Jxx xx Axxxx
Aug. 2, 2017
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Thank you.
However, you can only win if you finesse the Q (or J). West's hand was: AJxx xx Kxxx Jxx.

Was it just a strange deal? Yes, the bidding looks strange, but what about the opening lead? Can the winning plan be inferred from the opening lead? This is my real question.

In retrospect, the correct reasoning is something like this: West should try to get tricks in the side suits before I can discard them on the diamonds. If he has two black aces he can fear to blow tricks by leading one. Is this a sufficient reason? Not at all. He could lead an ace to look at dummy and at his partner signal. Leading an ace is less dangerous than underleading a king. His trump lead makes sense only if he thinks that he can stop diamonds. So West holds the K, eventually four to the K.

Does this reasoning make sense in retrospect only?
July 29, 2017
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@Stefan: I don't know ;-)
I've found that hand in an italian bridge forum and the bidding was not explained. I can just guess that 3 was third suit forcing - and can't understand why South did not bid 3NT.
As to the diamonds, according to W.S.Root, after 1-2;2-3;4, West shoud lead the 2 from 73 A953 1098 K1062 to get tricks in the side suits before the declarer can discard them.
So, as David Wetzel said, the declarer can conclude “that West has two black aces and is well trained enough to not blow tricks by leading one”. I.e., if holding 73 A953 1098 A1062 a “well trained” West could lead a trump.
July 25, 2017
Sergio Polini edited this comment July 25, 2017
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This was my conclusion, but it was wrong…
If West has the K a trump lead is a suicide.
But if West has two black aces, can hardly have the K. From his point of view the K can be in dummy, and a passive trump lead looks very dangerous.
But what if West has four to the K?
West's hand: AJxx xx Kxxx Jxx.
July 25, 2017
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Thanks! Edited.
July 24, 2017
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Double Dummy Solver: http://www.bridgecaptain.com/downloadDD.html
You can:
- load a lin/pbn file
- select a deal and play it
- press F5 (Input custom deal) and edit the current deal
- add the new deal to the current file or save it as a new file
- play the new deal.
May 26, 2017
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Yes, I think that Precision can help a lot here. But what if we swap minors?
May 4, 2017
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Kurt Häggblom: The assumption is that you play a card because you have to, not because you choose to do so.
Michael Kawalec: With “repulsion of equals” you don't have to make any assumptions.

However, such an assumption could be useful.

First example: AJ10 / xxx
Declarer's goal is two tricks. Sitting East, you have to play an honor if you hold one. Otherwise you gift declarer his/her goal. Your choice is “restricted”.

Second example (Reese): KJ974 / A6
Declarer's goal is four tricks. Sitting East, you would not drop an honor under the Ace, unless “restricted” (and if you drop the Q, declarer must finess the nine).
BTW, Q and 10 are not “equal”, they are somewhat “equivalent”.

Third example (Pavlicek): AQ9xxx / Kx
Declarer's goal is five tricks. Sitting East, you have not to play an honor under the K if you hold J10x. You gift declarer nothing if you don't, but you can choose to falsecard (but if you always falsecard, declarer will play the odds.)

So I would not say: “The play of a card which may have been selected as a choice of equal plays increases the chance that the player started with a holding in which his choice was restricted”, but something like: "When two equivalent cards are missing and a player is forced to play one of them if he/she holds it (or both), he/she should be assumed not to have had a choice."
March 22, 2017
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You can also use Google.
For example, there was an article “Suggestions on how to make a good bridge class on laptop without internet” by by Adriano Rodrigues, Sept. 11, 2015.
Getting more and more content would be a long way…
If you look for ‘bridgewinners “bridge class”’, you get it very quickly.
March 19, 2017
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“they contrived a calculation that added the requirement that East and West must play randomly from equals”

Actually, I think that Borel & Chéron explain better: the proper play by South does not depend on East's choice, South can and should ignore East's strategy (because he/she could misguess!)
Then, when you calculate the posterior probability, you assume that J and Q are just… two balls in an urn: “who knows” assumption -> uniform distribution.

This is roughly what they say in their first proof (see my comments above). Kelsey & Glauert also warn against trying to guess an opponent's strategy, but they do it after their arguments.
March 13, 2017
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Yes, I assume 50%, and finesse the 9, because I don't know.
The third proof by Borel & Chévron (pp.389-390) is:
A = East held J, P(A) prop. to 11
B = East held Q, P(B) prop. to 11
C = East held QJ, P(C) prop. to 12
P(J|A) = 1
P(J|B) = 0
P(J|C) = P(Q|C) = 1/2 (“who knows” assumption -> uniform distribution)
By Bayes' theorem, since P(J|B)=0,
P(E)=P(J|A)P(A)+P(J|C)P(C) prop.to 11+12/2=17
P(A|J)=P(J|A)P(A)/P(E)=11/17
P(C|J)=P(J|C)P(C)/P(E)=6/17

It is an expert East who chooses 50% because then T=12+5j+5q, i.e. he/she wins on average if I do not consistently finesse the 9 (T<max if j<1 or q<1).
March 13, 2017
Sergio Polini edited this comment March 13, 2017
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“If East would NEVER play the J from QJ” etc.
Sure, but how could I know?
Borel & Chéron say that South can know wether East is a poor player, but cannot know if he/she is naive (never the Q from QJ) or cunning (never the J from QJ).
BTW, I didn't calculate anything, I just quoted from Borel & Chéron.
March 13, 2017
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