Bridge World Extra! Newsletter

BEFORE BIDDING CAUGHT UP

When contract bridge began, in 1925, players of whist-type games had accumulated hundreds of years of experience in card-play techniques. Contract bidding techniques had to start from scratch. During the 1930's, when bidding theory was taking what in retrospect seem baby steps, the theory of play was easily seen to be much further advanced. Consider this simple-looking deal from an instructional article by Alfred P. Sheinwold:

NORTH
S 9 6 3
H 8 7 6 4
D 8 5 2
C Q 10 3
WEST
S 8 7 4
H 5 3
D 10 9 7 6 4 3
C 7 4
EAST
S 5 2
H Q J 10 9 2
D Q J
C A K J 9
SOUTH
S A K Q J 10
H A K
D A K
C 8 6 5 2

After East opened one heart, South, perhaps greedy for his honors, wound up in four spades instead of three notrump. West led the five of hearts, and South recognized the potential need to ruff his fourth club in dummy. So, he led a club at trick two. East won, and shifted to a trump. South, seeing no reason to change his plan, continued clubs. He hoped that if East had started with four clubs he would eventually be unable to lead a third round of trumps.

The main idea was good, but the execution was poor, since South was now exposed to a heart ruff. Declarer should have cashed his remaining heart honor at trick two to avoid this danger. If hearts were six-one, the actual line of play would expose a heart ruff anyway.

SIX THE HARD WAY


TO THE EDITOR:

In a previous edition, Charles Blair extended a previously published puzzle by showing how this hand:

S 4 3 2 H 4 3 2 D 4 3 2 C 5 4 3 2

can win six tricks with clubs trumps and no irregularities. Construction fans may find it an even greater challenge to create a deal on which this hand can take six tricks with no irregularities and clubs not trumps.

Solution to SIX THE HARD WAY
(letter to the Editor continues)

NORTH
S A K Q J 10
H --
D A Q J
C A K Q J 10
WEST
S --
H A K Q J 10 9 8 7 6
D K 7 6 5
C --
EAST
S 9 8 7 6 5
H 5
D 10 9 8
C 9 8 7 6
SOUTH
S 4 3 2
H 4 3 2
D 4 3 2
C 5 4 3 2

With diamonds trumps, West leads four rounds of hearts. North and East throw clubs; South ruffs (one trick). South leads a spade. West ruffs, and leads a fifth round of hearts; North and East throw their last clubs. South ruffs (second trick), then runs clubs (four more tricks). [Note that all three "opponents" have only trumps remaining thereafter.--Editor]

For this problem, it should be possible to show that six tricks is the maximum. Someone has at least four trumps, which play on tricks South loses, and South must lose at least three side-suit tricks before gaining the lead. These requirements significantly curtail the possibilities; it is probably straightforward, though tedious, to give an impossibility proof for seven tricks.

submitted by Chip Martel

Editor's comment: A similar solution to the unposed version was submitted by Bob Munson. One of his three "opponents" has 10-3-0-0 shape, instead of 9-4-0-0, so his deal lacks the all-trump ending. Please don't ask us if that makes it lesser or greater! Definitely on the greater side, Munson provided proofs of maximum for both cases--clubs trumps and clubs not trumps. We wouldn't call his proofs rigorous, but neither would we call them tedious--and they were convincing.








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