Bridge Deal of the Week (February 15 2017)

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Problem

The Auction:

West North East South
  1 Pass 1♠
Pass 3♣ Pass 4NT
Pass 5 Pass 5NT
Pass 6 Pass 7NT
all pass      

West has led the ♠5. You can count on four (hopefully one more) tricks in spades, two in hearts and diamonds, three in clubs. You miss the Q and Q, and although a chance of winning a single finesse is 50 %, the probability of two finesses both being successful is only 25%. Besides – you cannot afford to lose a trick as you have just declared 7 NT.

How are you going to make it?

 

Solution

West has led the ♠5 to dummy’s Ace (trick 1).

You lead a small club to the ♣A (trick 2) and take the third trick with the ♠K, ditching the 3 from dummy’s hand. Then you lead the ♣Q, a small club is played by West while East plays the ♣10 (trick 4).

At this point, the opponents have three spades (10xx) and three clubs (Jxx) left. If the remaining spades would be distributed 2-1, you could hope for an extra trick in spades, but somehow you doubt it – there must be a reason why West chose spades as the opening lead. And since East dropped the ♣10, it seems probable that West has the rest of clubs.

Next you lead the 9 to dummy’s ace (trick 5), then the ♣K – East discards a small diamond and so do you (trick 6). West does indeed hold the remaining two clubs.

After that you lead a small diamond to your king (trick 7) and lead the ♠Q. West discards the 7, you ditch the 2 from dummy (trick 8). Now that the opponents have only two hearts left, so you lead the 10, West plays the 8, you play the K from dummy and West’s Q drops under the king (trick 9).

You lead hearts from dummy to your J, the opponents have no hearts left, so East discards the 9 and West the ♣8 (trick 10). Now you can take a trick with the ♠J, West ditches the 10 and you discard dummy’s ♣9 (trick 11).

The opponents have two diamonds left and as West still holds one club and East one spade, you can safely lead the 6. West plays the Q, you win the trick with the A (trick 12) and take the last trick with the J.

 

   A  
   AK632  
   AJ5  
   K953  
 65 Deal  107432
 875  Q4
 Q107  9843
 J8642  107
   KQJ98  
   J109  
   K62  
   AQ  

Smoke and mirrors (and a successful grand slam!). It appears West chose his shortest suit as the opening lead. If you had known the overall layout of cards, you could have led the AK and won five tricks in hearts. Instead of focusing on your longest suit, you discarded a small heart early in the game. East followed suit and pitched a heart too, making it easier for you to lead the AK – if the queen had not dropped, you could still have tried to guess and finesse the diamond queen.

During the last tricks both East and West were squeezed – they had no hearts left and chose to discard diamonds and preserve their (potential) winners in clubs and spades.

7♠ or 7 could have been successful too, although 7NT vulnerable of course yields the most points (2200).

To make 7♠, you would have to use trump coup, which would have been easy as West would probably have led his longest suit – clubs. After winning tricks with the ♣AQ, ♠A, and AK, you could have led the ♣K and East would probably have ruffed. After overruffing you could have pulled the trumps and cashed in the J. West would be squeezed and to preserve the ♣J would have discarded a diamond, so you could have taken the last tricks with the AKJ.

If 7 would have been played, North would have become the declarer. East would have led spades (longest suit), so after taking the first trick with the ♠A, North would have needed the nerve to lead the A and K successively.

Par Contract Analysis

The par contract on this deal is 7NT by North-South.

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