Main     Puzzle 7E75 by Richard Pavlicek    

No Squeeze

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 by South

None Vul
S A 4 2
H A 7 5 2
D A K Q J 10
C 6
Lead: D 9Table
S K 8 7 5
H K 4 3
D 4 2
C A K Q 8

West

Pass
Pass
Pass
Pass
All Pass
North
1 D
2 H
3 S
5 S
7 D
East
Pass
Pass
Pass
Pass
Pass
South
1 S
3 C
4 NT
5 NT
7 NT

After West leads the D 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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Solution

For 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 (S 10-6 H 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:

S A 4 2
H A 7 5 2
D A K Q J 10
C 6
S Q J 3
H Q J 9
D 9
C J 9 7 4 3 2
TableS 10 9 6
H 10 8 6
D 8 7 6 5 3
C 10 5
S K 8 7 5
H K 4 3
D 4 2
C A K Q 8

And as most people would argue: Who really cares!

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© 2013 Richard Pavlicek