Shape-oriented conditional key card asks

[nullve]

Codo, by this do you mean the convention I suggested earlier (first step = bad; steps 2-5 = good + key card response) or that you should bid 5m with the bad hand and the normal Minorwood response with a good hand. If you play the second of these then I strongly recommend trying out the former which I think is several orders of magnitude better.

Yes. I play it the same way as you do, anything else looks horrible to me. Is there a name for it? In German it is "bedingte Asfrage", but I have no idea what it is called elsewhere..

It was in one of Rosenkranz' books about 20 years ago - I think he called it "RKCB min/max zoom" - not very catchy, but I still use that description. He also had some relay situations where, after showing shortage, you bid step 1 with a void and make a "zoom" RKCB response with singleton.

Funnily enough I am doing this too in certain auctions - using the first step to show a void. I have never read any books by Rosenkranz but it does rather sound like I should!

Ok, so I didn't invent "shape-oriented conditional key card asks", either. (They are "shape-oriented" in the sense that "min/max" is interpreted as having more to do with shape (e.g. 5143 is "worse" than 5044) than general values.)

 

In my system, these conditional key card asks are used in situations where

 

• Teller's shape is only nearly resolved, in the sense that he can still have different but qualitatively similar shapes, at the level of 3♠ or higher;
  
• Teller's hcp will be known to lie within a certain narrow range once shape is fully resolved;
  
• bidding space is so scarce that key card investigations cannot always wait until Teller's shape is fully resolved.
  

If 4T+1 is available as a conditional key card ask with trump suit T, then the continuations are

 

4T+2 = best shape

...Now steps as if responding to Parity Key Card Blackwood, i.e.

...4T+3 = even # of working* key cards

......4T+4 = TQ ask

......(...)

...4T+4 = odd # of working* key cards, no TQ

...5T = odd # of working* key cards, TQ

others = worst shape, steps as if responding to Parity Key Card Blackwood

In detail:

4T+3 = worst shape, even # of key cards

...4T+4 = TQ ask

...(...)

4T+4 = worst shape, odd # of key cards, no TQ

5T = worst shape, odd # of keycards, TQ.

 

* i.e. outside any void (if any) in Teller's hand

 

Note that switching 'best' and 'worst' doesn't work as smoothly, because over 4T+2 (now: worst shape), there must be a reliable way to get out in 5T if Asker was only interested in slam opposite the best shape.

 

If 4M-1 is available as a conditional key card ask with the major suit M as trumps, then the responses are usually

 

4M = worst shape(s)

others = best shape, steps as if responding to Parity Key Card Blackwood (but steps as in regular RKC would obviously work as well)

 

although in at least one case I currently play

 

4M = worst shape(s)

4M+1 = best shape (steps as if Responding to Parity Key Card Blackwood over this)

others = next best (next worst) shape, steps as if resonding to Parity Key Card Blackwood.

 

---

 

I'd like to know more about similar (shape-oriented) structures, including Rosenkranz's (the original?) and Zelandakh's.

[Zelandakh]

It is usually best for the first step to be negative. Then less information is leaked if Asker only had a slam try and no longer needs the key card information. For your case (4X+1 as the ask) you could, for example, use:-

 

4♦ = CKCB for ♣

... - 4♥ = decline slam try

... - ... - 4♠ = RKCB

... - ... - ... - 4NT = 1 or 4 key cards

... - ... - ... - 5♣ = 0 or 3 key cards

... - ... - ... - 5♦ = 2 or 5 key cards without ♣Q

... - ... - ... - 5♥ = 2 or 5 key cards with ♣Q

... - ... - 4NT = MKCB (use when 2 key cards are not enough for slam)

... - ... - ... - 5♣ = 0-2 key cards

... - ... - ... - 5♦ = 3 key cards

... - ... - ... - 5♥+ = 4-5 key cards

... - ... - 5♣ = sign off

... - 4♠ = accept slam try, 1 or 4 key cards

... - 4NT = accept slam try, 0 or 3 key cards

... - 5♣ = 2 or 5 key cards without ♣Q

... - 5♦ = 2 or 5 key cards with ♣Q

--

 

Obviously it is possible to rearrange the responses as desired, including moving over to parity responses if that fits better with the methods. It might even be necessary for the X+1 ask as you can see there is a hole here when max+2kc+♣Q is not enough for slam (it works better if 4X is the ask). Similarly, what accepting or declining the slam try is based on can vary by auction. In the general case it might be overall strength but in specific auctions with unresolved shape and known strength it is instead based on the shape being held.

 

The same structure also works quite well for zooming in natural bidding sequences. Take, for example, the splinter auction 1♠ - 4♦. Clearly much more is known about Responder's hand than Opener's here so it would be nice if Opener could do the asking but after the popular 4♥ as a general slam try, it seems that that ship has sailed. In fact, it is possible not only to arrange for Opener to be the Asker but also to resolve whether the splinter is based off of a singleton or a void:-

 

1♠ - 4♦; 4♥

==

4♠ = decline slam try

4NT = accept slam try, ♦ void

5♣♦♥♠ = accept slam try, singleton ♦, normal RKCB response

--

 

This trick works essentially any time we are in the slam zone and the known hand is bidding beyond 3NT (for a minor) or 4M (for a major). Since it can be used so consistently across a fairly wide range of auctions and similar logic, it is easy to incorporate into any bidding system. But it works best in systems with many auctions with clearly-defined known and unknown hands if you want to use it for just the basic situations.

[nullve]

... - ... - 4NT = MKCB (use when 2 key cards are not enough for slam)

... - ... - ... - 5♣ = 0-2 key cards

... - ... - ... - 5♦ = 3 key cards

... - ... - ... - 5♥+ = 4-5 key cards

Nice idea!

 

Is this MKCB (a variant of?) kenrexford's Middle Key Card Blackwood?

[Zelandakh]

Is this MKCB (a variant of?) kenrexford's Middle Key Card Blackwood?

I just used M for Modified. I am fairly sure that Ken's suggested method is considerably more complicated.