My friend David Berkowitz played this deal at Jourdans Bridge Club in Boca Raton, Florida:
K Q 9 4
10 6 4 2
A 10 2
A 3 2
Q J 9 8 7 5
|K J 10 9 6 5
Q 9 8 7
|A Q 2
J 10 7 6
K J 4 3
West opened 2, East responded 2 (not forcing) and David as South ended up in 3NT.
West led a spade, won by declarer.
He knocked out the heart ace and that meant he had the following sure winners:
2 spades, 3 hearts, 2 diamonds, 2 clubs. That's 9 tricks, so it was just a matter of overtricks.
What do you think of the club suit?
Assuming declarer plays East for the queen, he can take 3 tricks (lead the ace, then finesse). Can he get 4 club tricks and make many overtricks?
It looks impossible. When declarer wins the A and leads the 10, East will cover, of course. If declarer doesn't lead dummy's 10, East still gets a club trick.
But, look what happened. When West won his A, he continued spades. Declarer won and cashed the red suits to leave:
|What a curious ending. East had to keep all four clubs (else the suit would run). So, David threw East in with the spade. East had to lead a club and voila! -- 4 club tricks for declarer.
Plus 660 was a top board.
Note -- West could have broken up this ending by playing a club when he was in.