Competition Problem 8
West to lead and East-West to defeat South's three no-trumps.
Successful solvers: Robin Adey, Bob Bignall, Jean-Marc Bihl, Steve Bloom, Clint Fyke, Andy Prothero, DaniŽl de Lind van Wijngaarden, Wim van der Zijden
West must lead ♦10, which East must duck. If South ducks too, West must follow with a low diamond to East's ♦Q and South's ♦K, in which case South gives up a club and East returns a low diamond to South's ♦J. If instead South wins the first trick with ♦J and gives up a club to East's ♣K, East must return ♦Q, which South ducks, and follow with a low diamond to South's ♦K. In both cases we have had three rounds of diamonds, East retaining ♦A, and one of clubs. Now, if South leads:
A. a club, West wins with ♣A and East discards the ♦A so that West can score two diamond tricks to defeat the contract.
B. a diamond, East must return a heart, for a spade return would allow West to be squeezed, with or without the count, by three or four rounds of spades, South being able to discard a club on the thirdónote that North needs a heart entry for the squeeze to work and so does South.
C. ♠Q and another spade, North playing low, East must again return a heart, for much the same reason as in B.
If East had not deliberately blocked the diamonds by hanging on to the ♦A, declarer would have been able to squeeze West by playing three rounds of spades. A club can be discarded on the second, but on the third West must either give the ninth trick immediately by unguarding hearts (or clubs), or must discard a diamond winner, allowing the ninth trick to be established in clubs.
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: 11 March, 2017