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Puzzle 7E75 by Richard Pavlicek
You may not approve of the bidding, but the presence of 12 top tricks will usually produce 13 — or so say the optimists of the bridge world. Alas, not this time.
7 NT South
None Vul |  A 4 2 A 7 5 2 A K Q J 10 6 |
| Lead: 9 |  |
| K 8 7 5 K 4 3 4 2 A K Q 8 |
| West Pass Pass Pass Pass All Pass | North 1 2 3 5 7 | East Pass Pass Pass Pass Pass | South 1 3 4 NT 5 NT 7 NT |
After West leads the 9, the contract can always be defeated with best defense. Further, the following conditions exist:
| 1. At least one defender has a singleton. 2. West has exactly two even spot cards. 3. The sum of West’s club spots equals the sum of East’s spade spots. |
What are the exact East-West hands?
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SolutionFor starters, both majors must split 3-3. If not, declarer would have a positional major-suit threat behind at least one defender, allowing 7 NT to be made on a compound squeeze. (To understand this, see my article on “Pure Squeezes” under Odds and Theory.) For the same reason, East cannot hold five or more clubs. So West cannot have a club singleton; nor East a diamond singleton (East will have too many clubs); nor East a club singleton (he’ll have at most 12 cards). Therefore, West’s lead must be the singleton, giving West 3=3=1=6 shape, and East 3=3=5=2.
If West has only one even club spot, his minimum club-spot count is 26 (J-9-7-5-3-2); whereas, East’s maximum spade-spot count is 25 (10-9-6). Therefore, West has two even club spots; and East has all the even major-suit spots ( 10-6 10-8-6). East’s third spade must be the nine, else his maximum spade-spot count is 19 (too low to match any possible West club-spot count). The only West club holding with a spot total of 25 is J-9-7-4-3-2. So the full deal must be:
| A 4 2 A 7 5 2 A K Q J 10 6 | |
| Q J 3 Q J 9 9 J 9 7 4 3 2 |  | 10 9 6 10 8 6 8 7 6 5 3 10 5 |
| K 8 7 5 K 4 3 4 2 A K Q 8 |
And as most people would argue: Who really cares!
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© 2013 Richard Pavlicek