Puzzle 7F20 (Sep 94) by Richard Pavlicek

, as even if it failed, it would surely be a great sacrifice at favorable vulnerability.
![]() | A 6 5 4 3 J 9 5 4 3 3 2 2 |
![]() | |
![]() | Q 10 7 2 K 8 7 2 5 4 5 4 3 |
| West 1 ![]() Dbl | North 2 ![]() All Pass | East 2 ![]() | South 4 ![]() |
Alas. The postmortem revealed that East-West could not make any game and there were no void suits. What are the exact East-West hands?

Solution
is defeated with a trump lead; 5
is defeated with the
K lead; and 3 NT is defeated with the
A lead (South must unblock the
7).
![]() | A 6 5 4 3 J 9 5 4 3 3 2 2 | |
K A 10 6 A K Q J 10 9 8 7 6 | ![]() | J 9 8 Q K Q J 10 9 8 7 6 A |
![]() | Q 10 7 2 K 8 7 2 5 4 5 4 3 |
As evidence that this is the only layout, consider that that East-West must hold at least 11 top tricks (five in each minor plus the
A) so some kind of blocked position is required to defeat five of a minor. If West, instead, had
K
A-10-6
A-K
Q-J-10-9-8-7-6, it would be easy to make 5
(ruff a heart).
The spades must be divided exactly as shown to allow N-S to run that suit in notrump. Any other layout would give E-W a stopper.
If the hearts were divided in any other way, 5
could be made either by a simple heart finesse; or if East has the
A, a side entry to the East hand would give the defenders an impossible task.
[Corrected 9-30-99. In the original puzzle I did not specify no void suits and Scott Cardell noted there were alternate solutions based on East having all nine diamonds. Thanks, Scott, for the eagle eye.]

Copyright © 1994 Richard Pavlicek. All rights reserved.