Competition Problem 165a
to lead and East-West to defeat South's contract of six hearts.
Successful solvers: Sebastian Nowacki, Zoran Sibinović, Rajeswar Tewari
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:
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:
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:
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:
The ♦3 to South’s ♦K—if East ruffs declarer easily makes the rest—is 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.
Hugh Darwen, 2018