Zelandakh Posted March 23, 2015 Report Share Posted March 23, 2015 (Opener) ♠ AK5 ♥ AQ10 ♦ AKJ7 ♣ AQ7 (Partner) ♠ Q8762 ♥ 954 ♦ Q6 ♣ 832 3NT - 5 NT, 6NTHow do you think he would have bid this pair of hands?(Opener) ♠ A35 ♥ AQ10 ♦ AKJ7 ♣ AKQ (Partner) ♠ Q8762 ♥ 954 ♦ Q6 ♣ 832 3NT - 5 NT, 6NT Quote Link to comment Share on other sites More sharing options...
Lovera Posted March 24, 2015 Report Share Posted March 24, 2015 How do you think he would have bid this pair of hands?(Opener) ♠ A35 ♥ AQ10 ♦ AKJ7 ♣ AKQ (Partner) ♠ Q8762 ♥ 954 ♦ Q6 ♣ 832 3NT - 5 NT, 6NTIn book indicated, under ex. 35) was written this :"the two Queen of partner can consider sure tricks and yet the 4th and 5th card in spade suit sound be strongly precious. Because opener sure 27 points partnership has a combined force of 31 points in high cards and also a suit with five cards; generally it is enough to realize a little slam. Bidding of 5 NT by partner asks for 6 excluding grand slam". If you move King of spade in club we cannot say that spade Queen is a sure trick in the same way how in ex. 35) was said and than 3NT - 5NT, p Quote Link to comment Share on other sites More sharing options...
Zelandakh Posted March 24, 2015 Report Share Posted March 24, 2015 How did Opener know to discount the Q in this example but count it on #37 from your earlier post where ♦A95 was also held? Quote Link to comment Share on other sites More sharing options...
Lovera Posted March 25, 2015 Report Share Posted March 25, 2015 How did Opener know to discount the Q in this example but count it on #37 from your earlier post where ♦A95 was also held?I think i have understood (in traslation) what have you asked for: this is the explanation by Stayman about ex. 36): "bidding 4♥ or 4♠ answering 3NT by opener, partner simply tells that there is major productivity in suit[=because unbalanced- 6/+ cards] vs NT[=balanced hand]. Opener can single raise to show an maximum hand, highly efficent, with all points make by Aces and Kings how in the example 37): (see) [en passant the same points but with 4 Aces +1point, different plus values]. Quote Link to comment Share on other sites More sharing options...
Lovera Posted March 28, 2015 Report Share Posted March 28, 2015 Obviously in ex. 37) without Q in spade 3NT - 4♠, 5♠ - p and, for the hand of this topic 3NT - 5♠ (or 5♥ transf. here, two winners Q), 6♠ counting 12 tricks. But i suggest with hands with no queen and shape 6-3-3-1 to use South African Texas(=double transfert in major suit) meaning 4♣ transfert for heart and 4♦ for spade and opener decide to end in slam or not (i.e. AKQ in spade and Axx in club in ex. 36) knowing in hand of partner there is a singleton, bye. Quote Link to comment Share on other sites More sharing options...
PhantomSac Posted March 29, 2015 Report Share Posted March 29, 2015 Surely north can just keycard and ask for the queen if his partner shows six spades. You have 12 tricks if you have 6 spade tricks, and 11 tricks if you have 5. I would assume when my partner pulls 3N to 4S they have 6, they would just pass with 5 (unless they are going to bid or try for slam in which case we are gonna make 7N). I also think you should not jump to 3N after kokish to show a different range, it is more important to be at 2n and let partner stayman or transfer especially when your hand is that strong rather than try to show 27-28 or whatever. In this case partner will texas or bid 3H planning to slam try, in either case you can get to slam (even over texas, since as described earlier it is safe to keycard and ask for the queen if partner has shown 6 spades). Quote Link to comment Share on other sites More sharing options...
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