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So 1♣-1♥; 1♠-1N; 2♦ = 18-19 bal. / very strong hand 1♣-1♥; 1♠-1N; 2N = D reverse I've often wondered if/when the 2♦/2N "switch" where 2♦ = 18-19 bal. (possibly including 2245 and even 3M4D5C) 2N = D reverse (possibly excluding 2245 and even 3M4D5C), something that many T-Walsh pairs already play over 1♣-1♠ or 1♣-1N, will become standard over 1♣-1M(=4+ M), among pairs unwilling to play T-Walsh, or over 1♣-[1M-1](=4+ M) among pairs unwilling to play the kind of T-Walsh where 1♣-[1M-1]; 1N = 18-19 bal. or similar.

Yes. So 3♥ guarantees a fit unless Responder is strong enough to deal with the consequences of not having one. I also think that a regular partnership playing standard should be able be able to bid the slam with some confidence. My system is less ambitious than these on slam deals.

If you * treat 2245 as balanced * rebid 2♣ with 11-15, 4OM4D5C * rebid 1♠ with 4+D6+C then the only shape where you would rebid 2♦ to show 11-15, 4+D5+C is (31)45. Then it might be better to play 1♣-[1M-1]; 2♣ = 11-15, either 4OM5+C or 3OM4D5C 1♣-[1M-1]; 2♦ = 11-15, 3M4D5C

I play 1♣-[1M-1]; 1N as a kind of NF, limited Gazzilli, although "NF lebensohl" might be just as descriptive. Specifically, 1♣-[1M-1]; 1N = Rules of 19-21, 3-S4-H2-M6+C / Rules of 25-27, 3-S4-H2-M5+C, unbal. / 17-19 bal., 2-3 M Not intended as a stand-alone gadget, though.

2♦-2♥ (canapé preempt in S / Rule of 31+, 5+ S, unbal. // P/C) 2N-3♣ (Rule of 31+, either 5S4H, 6+S4+H, or 1-suited w/ 6+ S // relay) 3♦-3♥ (1-suited // relay, usually 2+ S) 3♠-3N (short C or 6(32)2 // relay) 4♣-4♦ (6(32)2 // relay) 4♠-4N (6322 // RKC(S)) 5♣-5♦ (1 or 4 key cards // SQ ask) 5♠-5N (SQ, no HK // CK ask (now hoping for AKQxxx-Axx-KQ-AK)) 6♣-6♠ (no CK // to play) P Not particularly elegant, but it helps to know that Opener has 6+ S and that he could not have a weaker hand, point-wise, for his bidding up to and including 4♠. (I've stolen the '(--- // ---)' notation from Fluffy.)

http://info.ecatsbridge.com/Systems/2015WorldTeamsChampionships-Chennai/Bermuda%20Bowl%20PDF/Sweden/Warne-Bergdahl.pdf

Just an observation: Consider a hypothetical "7A RKCB" that is very much like 6A RKCB except that * the combination of both non-key (suit) kings is treated as 1 key (suit) queen * the combination of both key queens is treated 1 one key card * the Roman-style responses to 4N are, more or less as a result, 5C = 0, 3 or 6 key cards (=> relay => (e.g. step 1 = 0 extra key queens; step 2+ = 1 extra key queen)) 5D = 1, 4 or 7 key cards (=> relay => (e.g. step 1 = 0 extra key queens; step 2+ = 1 extra key queen)) 5H = 2 or 5 key cards, 0 extra key queens 5S = 2 or 5 key cards, 1 extra key queen Then, by declaring that * each key card is worth 4 points * each key queen is worth 2 points, the responses can be rewritten as 5C = 0,2,12,14,24 or 26 points (=> relay => (e.g. step 1 = 0, 12 or 24 points; step 2+ = 2, 14 or 26 points)) 5D = 4,6,16,18,28 or 30 points (=> relay => (e.g. step 1 = 4, 16 or 28 points; step 2+ = 6, 18 or 30 points)) 5H = 8 or 20 5S = 10 or 22 points or, alternatively, as 5C = 0-2, 12-14 or 24-26 points (=> relay => (e.g. step 1 = 0, 12 or 24 points; step 2+ = 2, 14 or 26 points)) 5D = 4-6, 16-18 or 28-30 points (=> relay => (e.g. step 1 = 4, 16 or 28 points; step 2+ = 6, 18 or 30 points)) 5H = 8 or 20 points 5S = 10 or 22 points, where each interval is to be understood as consisting of even integers only. To make the above scheme look even more like in the OP, we also declare that * each non-key king is worth 1 point and change the scheme to 5C = 0-3, 12-15 or 24-27 points (=> relay => (e.g. step 1 = 0-1, 12-13 or 24-25 points; step 2+ = 2-3, 14-15 or 26-27 points)) 5D = 4-7, 16-19 or 28-31 points (=> relay => (e.g. step 1 = 4-5, 16-17 or 28-29 points; step 2+ = 6-7, 18-19 or 30-31 points)) 5H = 8-9 or 20-21 points 5S = 10-11 or 22-23 points, where each interval is now a standard integer interval. But despite appearances, this is really the scheme we started with, because each interval n-(n+1) just informs us that in addition to the number of key cards and key queens determined by the even integer n, Teller can also have 1 extra non-key king. (Duh!) So here we have a Blackwood-like gadget that looks a lot like the one in the OP, the main difference being just that aces are now worth 1 point less.

Would you also have stopped reading about RKCB if this pair of hands were used to promote it?

If the 1N range is narrow enough (e.g. 15-17 using a good evaluator), there's no need to be able to stop in 2N, so you can play 1N-2N = Puppet Stayman, like Meckwell do.

A slightly different scheme: 2N-3♣: now also with 5+S3+H 2N-3♣; ?: 3♦ = 4 M or 3 S 3♥ = 5 M 3♠ = 2S2H 3N = 2S3H Not that it solves the information leakage problem.

Corrected bad wording in OP.

The obvious choice would be to offload the 3N rebid. But then there would also be information leakage to worry about, so I don't know.

I'm not sure which hand types you have in mind, or what you regard as normal, but it would certainly be possible to change 2N-3♣; 3♥-3♠ = 3+ H to 2N-3♣; 3♥-3♠ = 3+ H or strong options and then use more of the available space after that. (I.e. not let Opener just bid 4♥ with 5 H, which is kind of preemptive.)

Suppose Responder to a 2N opening would never use Puppet Stayman without a 3-card or longer major. Then it should be possible to play 2N-3♣; ?: 3♥ = 5 M 3♠ = freed up! 2N-3♣; 3♥-?: E.g.: 3♠ = 3+ H ...3N = 5 S ......4♣ = 2-S3+H5+C, F4N .........4♦ = 5S3(+)C, RKC(♣) .........4N = 5S2C ......4♦ = 2-S3+H5+D, F4N .........4♥ = 5S3(+)D, RKC(♦) .........4N = 5S2D ......4♥ = slam try in S ......4♠ = to play ......4N = 2344 or 1444, quantitative ...4♥ = 5 H 3N = 3-4S2-H, NF ...P = 5 H ...4♠ = 5 S 4♣ = 3+S2-H5+C, F4N ...4♦ = 5H3(+)C, RKC(♣) ...4♠ = 5 S ...4N = 5H2C 4♦ = 3+S2-H5+D, F4N ...4♥ = 5H3(+)D, RKC(♦) ...4♠ = 5 S ...4N = 5H2D 4♥ = 3244 or 4144, F4N ...4♠ = 5 S ...4N = 5 H Thoughts? Edit: Changed 'at least a 3-card major' (ambiguous at best) to 'a 3-card or longer major'. Sorry about the confusion. I knew something had gone seriously wrong when I read post #11.

I'm morally with you, but it reminds me of Samuel Johnson's refutation of Bishop Berkeley's immaterialism.

Of course.

1N. I'll pass if partner tries to transfer to hearts.

Just an idea: 2N-3N = 4S5H, NF, because it allows 2N-3♦ = 5+ H or bal.(?) slam try with similar continuations as I suggested after 1N-2♦(5+ H or bal.(?) slam try*) in post #10 here: http://www.bridgebase.com/forums/topic/72920-inverted-stayman/page__pid__869627#entry869627, i.e. 2N-3♦; ?: 3♥ = 2-3 H or 3433 (=> 3♠ = relay, cancelling the transfer) 3♠ = 4-5 H, not 3433 (=> 3N = relay, cancelling the transfer) * similar to, and quite possibly inspired by, glen's Jacoby Stayman

A very similar hand (6-K65-Q987-AKJ98) was recently discussed here: http://www.bridgebase.com/forums/topic/72910-its-easy-for-opener-to-misrepresent-5-4-minors-distribution/

My somewhat Mulberry-like structure when shape resolution is almost complete and the range shown is 2 (Bergen) points wide: http://www.bridgebase.com/forums/topic/72952-2c-gf-relay-response-in-4cm/page__pid__870516#entry870516 (post #8)

Maybe they take for granted, or are willing to gamble, that they have similar views on global warming and therefore play the same range.

But there doesn't have to be a simpler underlying meaning. A pair could agree to play e.g. 1N = 15-17 if global warming is real 12-14 otherwise, couldn't they? Then if the players disagree on whether global warming is real, the agreed meaning couldn't be simplified to just '15-17' or just '12-14'.

From the 'variable NT range' thread: In a somewhat different vein, suppose M and M' are two different legal meanings of a call c and that c = M if P is true M' otherwise, where P is a proposition not necessarily about bridge, e.g. one of * 'Holocaust took place.' (almost universally accepted as true) * 'Global warming is real.' (slightly more controversial, perhaps) * 'God exists.' (many believe it, many don't) * 'Peanuts are nuts.' (I was recently shocked to find out they're not) * 'Justin Bieber just broke up with Selena Gomez.' (an annoying bit of trivia if true, but according to Wikipedia they had already broken up in 2013) * some lofty mathematical result (not open to debate, but not exactly common knowledge either) Would someone playing this be required to state whether they believed P to be true? (c could also be a defensive signal, of course)

I got 100% in #6454 Express - Free TCR 90 Automated Fun (6 boards) in November. Just normal, bad bridge.

Pass. Easy LoTT decision.