Looking for some inspiration for this edition’s bridge article, I asked around at a recent family gathering if anyone had an interesting problem they could share with me.  Lior, the husband of our niece Sheli, presented me with one such from a social game in which had played opposite Sheli. In the hand in question, she dealt and opened 1NT. He held a strong hand with 15 high-card points, a strong six-card spade suit and a void in diamonds:  A K J 9 5 4, © K 4 2, ¨ -, § A 9 7 5.  Slam bells immediately rang in his ears but he was not exactly sure how to go about the bidding. The actual result in the social game was irrelevant and he could not remember Sheli’s exact hand but his question was, faced with the same situation in a pairs tournament, how to bid the hand to its maximum potential.


Firstly, let me say, that slam is by no means certain with a combined partnership holding of between 30 and 32 HCP.  The partnership could be missing both red Aces, making 6NT impossible. That is not to say that 6 might not still be on because of the diamond void. The art here is to treat the hands like two parts of a jigsaw puzzle and, through judicious bidding, explore how well they fit together.


The first step would be for Lior to use the pretty standard Jacoby transfer bid of 2 H.  While this does not fully describe his hand - it simply shows a hand containing 5 or more cards in spades but an indeterminate number of points – it does set the base for more descriptive continuation bids.  Unless Sheli has both a four-card spade fit and the maximum 17 HCP, in which case she would “super accept” the transfer by bidding 3, she would now simply respond 2.  Assuming the latter, Lior is no wiser than he was at the outset - Sheli’s 1NT bid promised at least 2 spades and15-17 HCP – so it’s now up to him to describe his hand in such a way as to solicit additional information from his partner as to her holdings.


There are two ways to do this:  The first is to bid a suit below the 3NT level to show a second suit of 4 cards or more – in this case clubs. His 3♣ bid says to partner: “Give full value all your points in clubs and downgrade queens and knaves in the other two suits, hearts and diamonds. It’s then up to you to bid 3, 3NT or 4, depending on whether you hold 3 or more spades and whether you can or cannot help in the club suit.”


The second way, my preference in the given hand, is to make a short-suit slam try, by jumping to the 4 level in a suit in which you are void or have a singleton – diamonds in this case – promising a good, self-sufficient spade suit with at least 6 cards. A bid of 4¨ would allow Sheli to evaluate her hand, downgrading points in diamonds other than the Ace, and giving full value to all honor cards in hearts and clubs.  With wasted values in diamonds and no spade support, she would simply bid 4, but with spade support and good values outside of diamonds, she would make a constructive bid towards slam.


To illustrate further, let us assign Sheli this seemingly innocuous hand:  Q 8 3, © A Q 9 5,  ¨ A 8 6 2,  ♣ K 3.


Partner’s 4¨ bid transforms her hand into a giant and she indicates this by bidding 4©. Emboldened by her constructive bid, Lior would now check for Aces by bidding 4NT.  The couple play Roman Key-Card Blackwood (RKCB), so she would reply 5ª, showing two key cards, the red Aces, and the Queen of spades.  A bid of 5NT would now elicit a 6¨ response from her, showing one King.


Thus far Sheli’s bidding has shown 13 of her minimum 15 points – 2 Aces, a King and the spade Queen and, from the way the pieces of the puzzle have fallen into place, it’s pretty sure that the additional points for her 1NT opening are by way of the ©Q or ♣Q.  Thus Lior can count 12 tricks on top in either 6 or 6NT. In a matchpoint tournament, where the 10-point premium of making 12 tricks in NT over 12 tricks in spades can mean the difference between a good and a poor score, should he go for the NT slam? 


The bidding so far:



























I strongly suggest not going for 6NT.  The majority of pairs with a void in the one hand and such a compelling spade suit will end up in the trump contract, so at worst one can expect a slightly below average score playing in 6, making 12 tricks.  True, 6NT may give you a near top but there is a good possibility that the hand will yield 13 tricks played in spades, so 6 will outscore 6NT. Indeed, if you are truly desperate for a top, bid the grand slam, 7.  Consider the following possible full layout:






 A K J 9 5 4



© K 4 2



¨ -



♣ A 9 7 5

  7 6 2

© J 10 8 3


© 7 6 

¨ K Q J 9


¨ 10 7 5 4 3

♣10 6 4 2

 Q 8 3

♣Q J 8


© A Q 9 5



¨ A 8 6 2



♣ K 3



Only 12 tricks are available in 6NT but 13 are available in spades.  Where does the 13th trick come from? How should Sheli play to make 13 tricks on the lead of the ¨K?  The answers can be found …..


In the meantime, dear reader, happy bidding the pieces together.



The 13th trick comes from ruffing a club in declarer’s hand.  Sheli should win the opening lead with the ¨A in dummy, discarding a club from the closed hand. Next she should draw two round of trumps with dummy’s high cards; win the next two tricks with the King and Ace of clubs, then ruff a club with her Q; re-enter dummy with the ©K to draw the last trump; and cash her high hearts to claim 13 tricks – 6 spades, three hearts, one diamond, two clubs and a club ruff.

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Alan Caplan

Alan Caplan was born and raised in Johannesburg, South Africa. He was an active member of Bnei Zion and, subsequently, Habonim following the merger of the two movements. The year after high school ...

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