Double Dummy Corner

 

Competition Problem 142a

composed by Jean-Marc Bihl
presented for solving in December 2016

DR7

♠ 1092

 A4

 543

♣ AK1072

♠ 76

 KQ63

 Q976

♣ Q84

♠ AQ843

 72

 AJ2

♣ J93

♠ KJ5

J10985

 K108

♣ 65

South to make two no-trumps.  West leads the 6.

Successful solvers: Steve Bloom, Radu Mihai, Sebastian Nowacki, Rajeswar Tewari, Wim van der Zijden  Tables

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Solution

The favourable high card placements in spades, hearts and diamonds, combined with the 3-3 split in clubs, gives declarer an apparently easy ride to eight tricks, perhaps even nine.  The defence might aim for six tricks via three in diamonds and one in each of the other suits, but then West will need an entry in hearts and if West covers the J on the first round of hearts from South, then declarer can get four heart tricks, two in club and one each in spades and diamonds.  So a passive defence is best, avoiding leads that would help declarer to make three tricks in spades and diamonds.

East’s best play at trick one is the J, which is allowed to hold.  Now if East clears the diamonds declarer gets home easily by playing on hearts, switching to clubs if West fails to cover the J.  A similar observation applies if East exits with a low spade (to the J) or a heart.  East’s best return at trick two, therefore, is the 3 to West’s Q—which North must allow to hold.  Setting up the diamonds still doesn’t help the defence as East comes under pressure on North’s clubs and will have to let South make two tricks in spades, one way or another.  West’s best play now, therefore, is to advance the K—but again North ducks!  West continues with a low heart to the A but now North leads the 10.

A.      If the 10 holds, North follows with a diamond, guaranteeing three tricks in those two suits to go with four in clubs and the A.

B.      If East covers, declarer wins with the K and runs the clubs, discarding hearts.  Whether East keeps two spades or three, a diamond from North now guarantees two more tricks.

C.      So East does better to win with the A and return a club.  North’s clubs are run and East comes down to Q8 and A2.  Now South cannot make the K as well as the KJ, so discards a heart and the 10.  

Here is the four-card ending with North on lead:

♠ 102

 none

 54

♣ none

♠ none

 Q6

 Q9

♣ none

♠ Q8

 none

 A2

♣ none

♠ KJ

J

 K

♣ none

The two spade winners now catch West in a vice: a heart discard sets up South J whereas a diamond discard makes North 54 equals against East A3 when South leads the K.

Trap: If declarer plays the A at trick three, with the idea of setting up the hearts, West wins the second heart and returns a club.  If North cashes the clubs South is forced to discard hearts and then East can always come to three trick in spades and diamonds.  If instead North leads the 10, East can either win with the A and play A and another diamond, scoring a second spade at the end, or duck and then play ace an another when North next leads a spade or diamond.

See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.

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© Hugh Darwen, 2016
Date last modified: 11 March, 2017