Puzzle 7E75 (Jan 79) by Richard Pavlicek

7 NT by 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 Wests club spots equals the sum of Easts spade spots. |
What are the exact East-West hands?

Solution
If West has only one even club spot, his minimum club-spot count is 26 (J-9-7-5-3-2); whereas, Easts 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). Easts 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!

Copyright © 1979 Richard Pavlicek. All rights reserved.