Lovera

This is one position of guard squeeze. With Cicchelli notation it may be written ♠ C(=sq.card)/N(=idle diamond card) 2dr/3drimp(double menace with impasse situation) in diamond, -/1d(=8 of heart), 2f(=K9 of club)/-. (See "Double squeeze tecnique ..)(Lovera)

After 5♥ bidding will end in 6♠. I (avoiding x) perhaps may have better chance. Infact with opening lead of Ace of heart or J of club by E 6♠ is factible.(Lovera)

There is a convention called "Reese" that is used on 2♣ bidding :2♣-2♦, 4♦ (a jump show that this suit is compact) query for Aces -4♠(=second step 1A-spade), 5♣(=? for K)-5♦(=0Kings 1.st step with 3 diamonds almost) than with 5♠ or 6♣ is known the Queen presence in the same way for 13 tricks.(Lovera)

In this case you don't know that spade is unilateral because suit is divided 4-2 and justly consider 10 both controlled by opp otherwise at 9th trick you can lead little spade for K, Q to King of heart for a ruffing to realize a simple squeeze (E2) against E on last diamond. This squeeze positions have been analyzed very clearly by Bertrand Romanet in his book "Le squeeze au Bridge".(Lovera)

100 of honors.

It strange but how it is posted this one seemed an open problem (i.e. we already know about Q+K before bidding for Q). Than investigate under 5 doesn't cost anything (fourthemore we can have some datum) and we can figure QJxxxx of trump+A of heart (instead A of spade - ambigous situation). I don't exclude 5♥(=interrogative) to clear.

You can get grand in more ways :here ruffing two hearts+tops winners or lacking club Q but only if diamond break ruffing a spade. Althoug 4♥ could/should be considered as Rkcb(14d) than 5♣=2keys w/o Q and you may query 5♠=?Q/doubleton so:5NT=yes(alone), 6♣=yes+Q/doubleton in club, 6♦= no Q/d. Eventally bidding problem could arise when there is a tripleton in spade with Q in club because a grand is only if you query directly 6♣ (bypassing spade - but why?)=6NT(i have Q/doubleton)-7♦, bye.

I think that establish diamonds is easier than hearts (8 vs 7) ruffing one (dia 4-2) with 10 of spade (J onside or toghether longness).

I bidded 4♠ and ended with 2 more. A thing that confuse me is that if i bid 3♠ over 1♠ for me is 14-18 but i know that other players use this bidding as limitative (with minus force). Infact my intention was to bid six but having to chose between 4♦(=splinter? that i don't use) i bidded 4♠ for pass.

I find this topic much interesting yet for different ways to manage this convention. Has i have already said i use Stayman system that rules interfence vs 1NT(16-18) so:(31) ♠ QJ10874 ♥ QJ76 ♦ K9 ♣ 2 1NT-2♣ ;(32) ♠ K103 ♥ AJ7 ♦ QJ43 ♣ AQ6 1NT-X ;(33) ♠ KJ85 ♥ Q1096 ♦ 7 ♣ AQ84 1NT-2♣ ;(34) ♠ KQ83 ♥ QJ10973 ♦ 7 ♣ A4 p-p-1NT-2♣. As you may see X occurs when hand is "pair" of force to that one of 1NT and is balanced whilest 2♣ when unbalanced than, to integrate, 2♦(=heart+spade), 2♥ and 2♠(=Major+minor) should be weak.

I have a light impression (but perhaps i'm wrong) that delayed duck, a secondary squeeze, it is less known on refering to the other types.

Perhaps 4NT bidding could be seen how a "thin thinking" but save all what is said untill 4♠ (with optimal result as "last train for bidding"). The concept of bidding is don't last any possibility (here 4NT) in critical situations (that doesn't do damn yet if slam is not=5♣).

When will be possible to see the four hands ?

It' d be suggest for the case of 0 keys to save 5♣ bidding. For the other situations should be ininfluent if 30 or 14.

It should altough work a delayed (three losers with CLE in two suits) duck diamond/club squeeze yet.(Lovera)

I don't know how much is it good but i have quickly thinked that if spade are 3-3 with Q offside (2 points) ruffing third round and after running club (K offside 3 points than 5 in E) with A of diamond we could have 12 tricks.

Here ducking twice you are sure to fullfil the contract because need you only two tricks having 2 wiinners in spade, heart and club and one in diamond. In this situation when an ace is in a hand of any opp and (more) there is an impasse you must first eliminate ace (seen as reentry for longness) and when this suit is sectioned try for impasse. It has to consider that S has no entry when A and K of club when lead and if suit is blocked you can move suits only from dummy.

..It are examples that justify this opening : ♠ - ♥ 2 ♦ KJ98 ♣ KQJ98742 , ♠ Q83 ♥ 4 ♦ - ♣ AK10987654 , ♠ - ♥ - ♦ KQJ87542 ♣ QJ1032 , ♠ Q763 ♥ - ♦ 2 ♣ AQJ108642.. [9 or 10 tricks principally from longness] (From "Il libro completo del bridge di gara by Guido Barbone pagg. 53-54 Open bidding of five in a minor suit).

http://www.bridgebase.com/vugraph/schedule.php

I too think 4♣ should be a better bidding partner following with 4NT(=RKCB 3014-club trump), 5NT(=2 keys+void). It has to consider that with 5 (presumibly) entries in dummy we can get grand easyly.(Lovera)

Althoug i instead open and visualize directly with indication on post 1 whilest with this one (almost in first page) i don't see schedule.

About the "rule of 2 and 3" i suggest "to convert" in rule of 1-2-3: when you must value level of pre-empt (preferibly using LTC) consider to lose a trick if vul vs not vul, two tricks if pair vul and three tricks when you are not vul vs vul.

The other squeeze solution is clash requiring heart controll (9 or J) in W (because automatic position doesn't work): win diamond in dummy and let you cash A ♠ then run diamond pitching a club on 10 ♦. W if retain heart should descart club on last diamond otherwise you can cash separatly J and K of spade (Q of spade in W) and you descart 3 ♥ in dummy and now last heart squeezes E in spade and club.

I think that this five-losers hand is not just for the2♣ bidding (requeiring almost 26 points). Perhaps 25+1(=5th spade)+1(doubleton in heart)=27.(Lovera )

Than S should be max 8 points ( is it correct ?). However i pass.