Competition Problem 130a
is in four no-trumps.
Successful solver: Wing-Kai Hon gave the only correct solution, being the only one to notice that the play in (a) depends on which spade is led. Tables
(a) The play depends on which spade is led.
A. If West leads the ♠K, North wins and South comes to hand on a heart to lead the ♣9, optionally cashing the other heart winner first. West might as well go up with the ♣A and lead the ♣Q, but when the other heart winner and the remaining club winners have been played, West must either discard a spade, allowing declarer to cash the spade winners and endplay West in diamonds, or bare the ♦AK, allowing South to establish diamonds by losing two with an entry in spades.
B. If West leads a low spade, South wins in hand and leads the ♣9. West does best to take the ♣A and lead another low spade to South, who now leads the ♦Q. West does best to win this and lead a third spade to Northís ♠Q. On this or the previous spade East must discard a diamond to prevent a safe duck by declarer in that suit. Now, in some order, the ♥A, ♠Q and two club winners leave this 4-card ending with North on lead:
North plays the♣10, then the ♥10. East ducks but then either the ♣7 wins the trick or Eastís ♣8 is the stepping stone to the ♥A.
(b) To defeat the contract West must lead the ♥9 and East must allow Northís ♥10 to hold! The gambit pays off because South cannot now make two more heart tricks and also play both black suits through West. The best shot is to come to hand on a heart and lead a club, but West simply cashes the minor suit winners and puts North in with a club. The ♠K will be the defendersí fourth trick.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2015