TO THE EDITOR:
In a previous edition, Charles Blair extended a previously published puzzle by showing how this hand:
4 3 2
4 3 2
4 3 2
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
A K Q J 10
--
A Q J
A K Q J 10 |
WEST
--
A K Q J 10 9 8 7 6
K 7 6 5
-- | | EAST
9 8 7 6 5
5
10 9 8
9 8 7 6 |
| SOUTH
4 3 2
4 3 2
4 3 2
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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