Lovera

Instead, because if trump break is easy, needs to think to sure contract in the other eventuality. When partner bids 5NT usually suggest King only in this case. But 5NT has other meaning about any value (i.e. in a suit below query) that cannot bid at next level or upper trump agree. Here with J10 more in trump (Instead of xx) and K in club is better to bid 6♣ meaning Q in spade plus K in suit indicated to difference with 5NT=(yes King).

The problem is that if we had Q we'd known King's location: with 5♠ meaning yes Q in trump plus K in spade or conversely with 6♣ yes Q in trump plus K in club and not K in spade and we stay at six level but we have the Q also for referring about K. When Q is not we can probably avoid to get six if,as in this case, the partner hand is poor not having any extra. Than my choising to indicate with 5[NT] a top honor (between K or Q -let's suppose Q in spade and K in club), bye.

I also think that situation in heart, when Q is denied, is clear (as already aguahombre said): in the suit there is not 10 card fit that could be helpfully searching if to play 6♥ or 6NT. The hand of N is really in minimum and as we can see there is probably to lose a trick in trump lacking QJ10. Than the answer to 5♠ being 5NT(=King) and not passing at next level should sure an high honor (K or Q) in query suit(=spade) and not ultherior extra in the hand avoiding to get at 6♥ only how a remaing (and forced) choising (in this way recupering sign off).(Lovera)

Let us see, detailing more, the hands we talk about to explain. I call the hand of W in post #1 "A", the hand of E "0" and the hands of gnasher in post #13 in order to present "1", "2" and "3" so having four cases. After a bidding that has discoverd from every side 14 and more points to think slam after RKCB and relative answer of 5♥(=2k-Q), 5♠ query for Q in all our cases. Than : with A0 hand the answer is 6♦. Now, not indicating Q in club (otherwise 6♣) is likely Q in spade more and, as i have already said, bidding ends in 6♥. In case A1 after 5♠ instead we have the negative case recourring in trump 6♥. And then the two similar cases A2 and A3: the answer to 5♠ query for Q is 6♣(=Q in club is) and 6♦ is for controlling that in diamond suit Q is with 6NT (an eventually 6♠ signal a third Q). West considering doubleton in spade usefull can bid 7♥. Fourthemore we can consider, as undermeaning or support, that when two balanced hands are on the line searching for grand we need: four Aces obviuosly and more (considering also trump agree) other almost six high honors between K and Q i.e. four K and two Q or three K and three Q it being a positif condition to get on.

This interesting topic it seems to me had indicated a point that may be considered such a" little nice problem ". Complication are becoming to balanced and managed answers to apply on a generality of cases. I have done mathematician studies and about this type -i say - of algorithmic problem that i like very much too and also me as you we are working at this (RKCB) one to estabilish level of compatibility for vary elements to insert. Than i ask you to be patient because anytime answeres are not quickly to tell.

So when you switch on a suit (5♠, 6x) you query for Q, when you bid 5NT you want to know about K. Either first case or the other you sure to have all 5 keys plus Q and are looking for a grand.

Vampyr says :"I'm afraid I didn't understand most of your post" than i try to explain. When partner bids 2♦ with this convention indicates this type of hand and relative range of force requiring stopper(s). Opener has a stopper in spade (AK) but not in heart and bid 2♠. If had have K♠ and A ♥ must bid 2♥(=only sure stopper in heart). With two stoppers in heart and spade instead 2NT. This information at second level, If seven tricks are available, allow to play 3NT otherwise starts searching for a minor suit to a game o slam. This Stayman conv. is not much known and that's why i have crossed it with transfert making 2♦ ambigueous, bye.

It should be considered that if is insert doubleton togheter Q (because are third round controlls) in RKCB answers the same has to be done for singleton and K (second round controlls).

About first question i can agree and, if it is, you must read "if you use 2♣-2x, 3♣-to pass". Infact this using combined is particolar. 2♦ ambigoeus is a my idea explained in 2♦ Stayman where i remind you (see in BBO Forum event or clicking in this box on Lovera and visualizing my profile via "Find My Content"), bye.

I also talk about in "How not to get to 7" in Interesting Bridge Hands.

Infact this Stayman convention search for minor suit fit and complete scenario together its sister 2♣ for major suit fit and allow to play NT also. How ? You have this information enter second level when partner bids 2NT meaning that (s)he has sure stoppers in both major suit (i.e. two Aces in heart and spade suit). If it is not starts minor fit search. The 2♣ Stayman, you surely know, allow to play in club when bidder repeat club -3♣(=weak)- than using 2♣ and together 2♦(=strong 11/+ p.) 2♠ is weak transfert and may be used how direct transfert for diamond suit. I anew suggest to see my topic on 2♦ Stayman in Forum event (for combined useing with transfert for heart suit fit-2♦ ambigous), bye.

This hand is just for 2♦ Stayman (see my topic 2♦ Stayman in Forum event to know more) : Opener 1NT has ♠ AKxx ♥ Jxx ♦ KQxx ♣ Ax - 2♦(=? for stoppers), 2♠(=AK), 3♦(longness searching fit)-4♦ ending in 6♦.

About first hand i've already said, the second and third are similar but 5th diamond/3th club. Than with 5♠(=?)-6♣(=Q in club), 6♦(=?)-6♠(=Q in spade too)/6NT(=xx in spade). The concept is to remain at 6 level. If partner bid before 6♦ (than not 5♠) and in spade is not Q the answer is 6NT that is not excluding that in club there is Q. If partner bid 7♣: If Q is not=7♥, If Q is in club=7♦..

This is another hand, for the purpoise of this topic (two suits exchange), that i've seen in other one: ♠ Qx ♥ Axxx ♦ Qx ♣ KQxx, heart is trump suit. Tell me if you are agree untill now.

Well, infact there is not conflict because art. and nat. convention are different and the using frequence it'd depend about its complexity (i.e. art. conv. allow to cover more possibilty) and about method this is not a mine convention. You can see, re-reading what i has told, that combined use already cover many situations and also query again help to solve with adjuntive information. I try to explain it step by step than slowly, bye.

Hi Zelandakh, you know that are two ways -artificial and natural one - to approach a problem each other with any difficult/complexity and i like more natural method(s) and relative logic. I hope don't make a conflict when i started to talk about 3rd round control and relative position(s), bye.

To be clear: i don't discuss SSA (that can be a convention that you probably know well) but i don't. My knowledge are of "old school" and all on natural biddings.Fourthemore i stay on RKCB before passing (eventually) on other slam bidding because the solution must be find here. And infact what' s happen if anyone don't know SSA or Kickback ?

When you tells about "slow spade loser" it is better if explain with an example, thanks. If you query with 5♠(=? Q) instead you know there is not Q/xx (and than there is other one) because answer is 6♥(=neg.ve or trump), bye.

Via 5NT could be 6♥ meaning 0 or 4 Kings (and in this case 0 it's impossibile)

It seems to me that at this point we are in the centre of musical labyrinth. Minotauro probably is sleeping and we can go around freely to discover its many rooms. Have fun : http://www.google.it/search?hl=it&redir_esc=&client=tablet-android-samsung&source=android-unknown&v=141278776&qsubts=1434381754422&action=devloc&q=ABBA+-+Interactive+Timeline

In RKCB King search is based on cue bidding (probably it's sufficent because other cases -4 Kings- are rare) and structurarly there can be a solution (resolved in other way).

So standing the things neither W nor E can contribute for grand but what's happen if 4♠ and K♣ are traded ? Now we have KQ in spade and A4 in club and after 6♥ E can count 13 tricks (1 in club, 5 in diamond, 4 in heart and 2+1A of partner in spade) and bids 7♥. Another grand is if in spade we have xx and in club AKQ because query in club for third round controll starts if is Q or xx (but this one only in query interested - xx aside is not indicated).

I'd use 2♦ Stayman ( i suggest to see my topic in Forum event) for Question #2 : 1NT(=strong 15-17)-2♦(=i'm not interested to major but minor suit tell me yours sure stoppers), 2x(=only stopper in suit indicated or two stoppers if 2NT)-4♦(=longness in diamond estabilished as trump suit)..

We need data because looking for a grand possibility in a suit upper trump than suring all keys and authorizing partner to bid 7 if has extra. There is not ambiguity for answering. In this case after 6♦ the 6♥ by RKCB bidder tells that (s)he counts only 12 tricks and partner confirm passing.

Althoug in RKCB query for Q in spade should be 5♠ with answers : 5NT=Q only in spade, 6♣=Queens in spade and club, 6♦=Q in spade and diamond, 6♥= no Q in spade. In this case we know with 6♦ that are two Queens (in spade and diamond suits).