Competition Problem 167a
West leads to South’s four no-trumps. East-West to defeat the contract.
Successful solvers: Steve Bloom, Ed Lawhon, Steve McVea, Andrew Prothero, Sebastian Nowacki, Rajeswar Tewari, Andries van der Vegt, Wim van der Zijden. Suggested DRs ranged from 4 to 6, averaging at 4.8.
West must cash the ♣A and then switch to the ♦10! Declarer’s best try is to win with the ♦J and then use the red suit entries to hand in order to take the major suit finesses, West covering the ♠J when it is led. This is then the position with North on lead:
North tries the ♥4, on which East must throw the ♣5 and West a spade! North can cash the ♦K and throw West in with a spade but that player merely exits on a low club to win the last two tricks.
Trap: If West exits on another card at trick two—say the ♠K—then declarer can play three rounds of hearts immediately. This is the position with North on lead:
North advances the ♥4.
A. If East discards a diamond, then so does West and three rounds of diamonds follow. If West now discards
1. a club, then North plays the ♠Q and another spade to throw West in for a club lead.
2. a spade, then North exits on a low spade. West exits on the ♣2 but it will not have escaped the reader’s notice that South still has the ♣3, so East wins with the ♣5 and has to lead into North’s spade tenace.
B. If East discards a spade or the ♣5, then North can exit with a low spade. Defenders can score either the ♣Q or a second spade trick, but not both.
An additional line arises if West leads the ♦Q instead of the ♦10 at trick two. Declarer takes the spade finesses and four rounds of hearts in some order. To avoid lines A. and B. East must discard the ♣5. Now South must discard a club, West a spade. In that case North can simply establish a long spade, losing two trick to East and having the ♦K and ♦J for entries.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2019