Puzzle #10

The Case of the Delayed Entry

by Jeff Rubens

Although some of the earliest inferential problems published were from a defender's point of view, that format is less common than layouts as seen by declarer.

 NORTH ♠ — ♥ A 5 ♦ K Q 6 5 3 2 ♣ A J 8 5 4 WEST ♠ 7 5 ♥ K Q J 7 4 2 ♦ J 9 ♣ K 7 6

South is declarer at six spades. If West leads any heart or the club king, the contract can be defeated; otherwise, it can be made. Despite a combined holding of exactly 26 highcard points, against best defense North-South cannot make game in any strain other than spades. Defining spot cards as deuce through ten, the sum of South's spot cards in the red suits is one less than the sum of South's spot cards in the black suits.

(1) Find a South hand that meets the given conditions.

(2) (harder) Show there is no other such South hand.

#### Solutions

South's hand is: A K Q J 9 8   9 8   10 8   Q 10 9

It is not too hard to verify that this South hand meets all the given conditions. (For example, after a low-club lead, declarer wins the queen, cashes, say, three top trumps, and attacks diamonds. The defense wins the second diamond and leads, say, a heart. Declarer wins, ruffs a diamond, returns to dummy with a club finesse, and pushes diamond winners through East.)
It is more tedious to show that the hand is unique. One possibility is to deduce the following in the order given: (a) South lacks the ace of diamonds. (b) South holds exactly six spades. (c) South has exactly 6=2=2=3 distribution. (d) South's black-suit spots must be 9-8 of spades and 10-9 of clubs. (Other orders of deduction are possible.)