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THE BRIDGE WORLD

Puzzle #3

Little Things Mean a Lot

by John Lowenthal

   Inferential puzzles, in which the solver is required to deduce aspects of the unshown hands from any of a wide variety of clue forms, run the gamut from straightforward to intricate. For the initial instance of that genre in this series, we present one of the less complicated kind. It gets its play point across with minimum trappings.

NORTH
K J 8 5 2
K 8 6
9 7 4
K 3
SOUTH
A Q 10 7 3
A 5 2
A 6 3
A 4

   If West leads the jack of clubs against South's contract of four spades, perfect defense can defeat the contract. However, if West leads the deuce of clubs, South can make the contract.

   What are the suit distributions of the East-West hands?

Solution

   Four spades can be defeated only if East can ruff the opening lead; therefore, clubs are 9-0.

   The size of the club led can make a difference only if declarer can gain by playing the king and ace of clubs at trick one, later stripping West's red-suit cards, and then throwing West in with a club. (The club-deuce lead will prevent West from ducking the second round of clubs.) Thrown in, West will have to lead a club, letting declarer throw a heart from one hand and a diamond from the other; then, another ruff-sluff will let declarer get rid of his last red-suit loser.

   What can West's distribution be? He cannot be void of spades, for with four red cards he could retain a safe exit in a red suit. Nor can he have three spades, since East would not have a trump to ruff with at trick one. If West has two spades, his distribution would have to be exactly 2=1=1=9, else a red-suit ruff at trick two and a club continuation would lead to a successful defense. But if West has 2=1=1=9, he cannot defeat he contract even with the club-jack lead; declarer does not unblock in clubs, strips West's red singletons, and throws West in with the second round of spades for the same successive ruff-sluff endplay.

   Therefore, West has one spade. For the club endplay to be possible, his three red cards must be two hearts and one diamond. Therefore, West's suit distribution 1=2=1=9 and East's is 2=5=6=0.

(Adapted from The Bridge Journal.)

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