straube Posted October 20, 2017 Report Share Posted October 20, 2017 AQT8 J8 K862 K93Void AKQ732 QJT AQJT South is captain. Let's say North shows pattern and 8 QPs at 4C. A good one for reverse relays no doubt. 1) With rules of short to long (and stop with odd for singleton or doubleton), king parity last 4D-4N even heart, odd club, even spade (KQT8 J8 A62x K93 still possible)5C-5S odd diamond, DKCards placed 2) With rules of short to long (and stop with odd for singleton or doubleton), king parity first 4D-5C even kings, even heart, odd club5D-5H even spade (KQT8 J8 Axxx K93 still possible)5S-6C odd diamond (KQT8 J8 Axxx K93 still possible) 3) With rules of long to short (and stop for singleton only), K parity last 4D-4H even spade4S-5N odd diamond, odd club, even heart (rule change), DKCards placed 4) With rules of long to short (and stop for singleton only), K parity first 4D-4S even K, even spade4N-5S odd diamond, odd club, even heart (rule change) (KQT8 J8 A862 K93 still possible) Quote Link to comment Share on other sites More sharing options...
foobar Posted October 20, 2017 Report Share Posted October 20, 2017 AQT8 J8 K862 K93Void AKQ732 QJT AQJT 4D....4H (relay; even K parity...could be S(AKQ)...K(D|C) among others)4S....4N (relay; nothing in hearts; even spade...S(AKQ) is out the window. S(KQ)...D(A)...C(K) or the above? Can relay, but won't be resolved: 5C....5S (relay; odds in ♦+[♣) Quote Link to comment Share on other sites More sharing options...
awm Posted October 20, 2017 Report Share Posted October 20, 2017 Several things about this hand: 1. We would reverse relay 2. If we can't reverse relay, we would use a shortness-showing sequence. Ignoring all of that: 4♦ - 4♥ (even spades)4♠ - 5♦ (odd diamonds, odd clubs, even hearts). At this point it's either: ♠AQ + ♦K + ♣K♠KQ + ♦A + ♣K But it does not really matter. Partner's heart jack is not something we can find out for sure, so I'd just blast six in either case. I suppose we can resolve via: 5♥ - 5NT (on the actual hand, we pair ♠+♦ since one has king and other other queen, and we show that lower suit has the king; on the other option we would pair ♠+♥ since one has KQ and the other neither honor, and bid 5♠ since higher suit has king). Quote Link to comment Share on other sites More sharing options...
yunling Posted October 21, 2017 Report Share Posted October 21, 2017 4♦-4♠(ask 1&2, 0/3/6)the ask is not effective with shortage in suit 1, but a next move is okay here since if partner is 1/4 in suit 2&3 5H signoff is still possible. 4NT-5NT(ask 2&3, 2/5 no honor in H even Kings)♠KQ ♦A ♣K is still possible but won't matter Quote Link to comment Share on other sites More sharing options...
sieong Posted October 22, 2017 Report Share Posted October 22, 2017 EPCB.4♦ - 4♥ (even ♠)4♠ - 5♥ (even ♥ odd ♦ odd ♣) At this point this converges to classic PCB, but one step worse. Quote Link to comment Share on other sites More sharing options...
nullve Posted November 1, 2017 Report Share Posted November 1, 2017 (edited) AQT8 J8 K862 K93Void AKQ732 QJT AQJTAuction, with some ambiguity left out: (...)4♣-4♦ (11-13 hcp (say); relay)4♥-4♠ (even # of kings; relay)5♣-5♦ (odd # of queens, no ♠K; relay)6♣-? (♦K, ♣K, ♠Q, no ♠J). So Teller could still have Qxxx xxx AKxx Kxx. :( Edited November 2, 2017 by nullve Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.