Double Dummy Corner

 

Competition Problem 165a

composed by Steve Bloom
presented for solving in December 2018

DR7

♠ A82

 Q5432

 Q432

♣ 6

♠ J9

 10

 J10987

♣ KQJ104

♠ Q7654

 J876

 5

♣ 987

♠ K103

 AK9

 AK6

♣ A532

West to lead and East-West to defeat South's contract of six hearts.
What happens if the
♠10 and ♠8 are swapped?

Successful solvers:  Sebastian Nowacki, Zoran Sibinović, Rajeswar Tewari

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Solution

Declarer threatens to give up a club to rectify the count and squeeze West in the minor suits, North discarding a spade on the second round of clubs.  North has entries on a spade and a club ruff to enable trumps to be drawn via a finesse of the 9 and this squeeze to be put into effect.

So West leads a diamond, threatening to give East a ruff if declarer gives up a club.  Declarer’s best try is to win in hand and follow with the A, A, club ruff, trump finesse, third trump.  In the seven-card position West must keep three diamonds to guard North’s holding and two spades to guard against the finesse of the 10.  Suppose this is then the position:

♠ A82

 Q

 Q43

♣ none

♠ J9

 none

 1098

♣ KQ

♠ Q7654

 J

 none

♣ 9

♠ K103

 none

 A6

♣ 53

Declarer plays a spade to the A and discards a spade on the Q.  West safely discards a spade but is squeezed without the count on a spade to the K.

To defeat the contract West keeps K4 instead of KQ.  The seven-card ending is then:

♠ A82

 Q

 Q43

♣ none

♠ J9

 none

 1098

♣ K4

♠ Q7654

 J

 none

♣ 9

♠ K103

 none

 A6

♣ 53

A.      If declarer plays a spade to the A as before, and comes to hand on a second spade after drawing the last trump, then West discards the K.  East has spade winners if allowed in on the 9.

B.      If instead declarer comes to hand on a diamond, East discards the 9!  West’s K4 are now a major tenace over South’s 53.

C.      With West’s club holding now weakened the defence cannot easily make two club tricks, so declarer might instead try exiting on the 3 to rectify the count for a minor suit squeeze against West.  To foil that plan West completes the club sacrifice by rising with the K!  If that wins, East will make the J.  If North ruffs it, East will make the J and a spade trick.

If the 10 and 8 are swapped and play follows as above to the first six tricks, then the seven-card ending is now:

♠ A102

 Q

 Q43

♣ none

♠ J9

 none

 1098

♣ K4

♠ Q7654

 J

 none

♣ 9

♠ K83

 none

 A6

♣ 53

South advances the 3.

D.      If West plays the 4, the minor suit squeeze materialises as in C.

E.      If West rises with the K, North ruffs.  The minor suit winners are now played until East ruffs and then has only spades left while North and South still have three each.  A spade to the J and K lets North lead the 10 to pin the 9.

However, an alternative solution has come to light:

South wins the A and A, ruffs a club, comes to hand on the A, ruffs a club, finesses the 9, and ruffs a third club with the Q, leaving:

♠ A102

 none

 Q43

♣ none

♠ J9

 none

 1098

♣ K

♠ Q765

 J8

 none

♣ none

♠ K83

 K

 K6

♣ none

The 3 to South’s Kif East ruffs declarer easily makes the restis followed by the K and a diamond to the Q.  East must ruff and lead a spade, giving three spade tricks as in the intended solution.

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, 2018
Date last modified: 03 June, 2019