Competition Problem 14
South to make six spades. West leads the ♣5.
Successful solvers: Steve Bloom, Ian Budden, Tim Chanter, Hess Cheng, Leigh Matheson, Bircan Öztürk, Ertoz Suiçmez, Daniël de Lind van Wijngaarden, Wim van der Zijden. Many other solutions were submitted but these failed to distinguish between lines B. and C.
South captures East's ♣Q and crosses to the ♠A to lead a diamond, which East covers. After two rounds of diamonds South comes to hand on a spade, North preserving the ♠4, to take the ruffing finesse against West's ♦10. Assuming West covers, North ruffs and leads the ♠4 to South's ♠5. North discards the club loser on South's good diamond, and South leads the ♥Q.
A. If neither defender has kept three hearts, North overtakes with the ♥K. East wins the trick but the North hand is now high.
B. If East has kept three hearts and two clubs, North plays low. If East wins, North is high again; so East ducks. Now South leads the ♣10. Assume that East's clubs are ♣K-6 and that West now has two clubs, a diamond and a heart, though the following play works whatever East's clubs are and whatever West has kept. If West plays low, North discards a heart and East is endplayeda club return sets up South's last club and a heart return sets up a heart for North. If instead West plays the ♣J, North ruffs and leads a high heart, covered by East and ruffed by South. At trick 12 South puts East in on the ♣K and North's heart winner takes the last trick.
C. If East has kept two hearts (opposite West's three) and three clubs, North again plays low and again East should duck. This time, however, South leads the ♣2 (not the ♣10, which West would cover) and North ruffs. South ruffs a heart and exits on a club. If West wins, North makes a heart; if East wins, South makes a club.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
© Hugh Darwen, 2014
Date last modified: 03 June, 2019