Puzzle 7F15 (1993) by Richard Pavlicek

The Old Insurance Play

South’s bid of 7 NT is surely a desperate move (7

is more sensible), but we have all done worse things at the bridge table. East doubles for a club lead (ha), and West leads the

7 NT x by South

—
K 2
A Q J 8
K Q J 10 9 8 7
2
Q J 9 8 7 6 5 4
K 9 7 6
— [W - E] 10 7 5 4
10
3 2
A 6 5 4 3 2
Lead: Q A K Q J 9 8 6 3
A 3
10 5 4
—

West
4
Pass North
5
Pass East
Pass
Dbl South
7 NT
All Pass

Which heart honor do you win at trick one, and how do you make this contract. Warning! It is harder than it looks.

The contract looks easy at first sight. Declarer has two top hearts, eight spades and at least three diamonds with the finesse. But this is an illusion. Declarer cannot win three diamond tricks due to entry problems — the finesse must be postponed until the end (else South will be locked out of his hand) and it cannot be repeated. If you played out the hand in typical fashion, you would discover the annoying ending in the diamond suit.

Return to Main

Solution

The solution is to win the opening lead twice — an “insurance play” to guard against losing the first trick. That’s right; North’s king and South’s ace are both played at trick one. This sets the stage to put pressure on West as the spades are run. The diagram shows the ending before the last spade is led.

—
2
A Q J 8
—
—
J
K 9 7 6
— [W - E] —
—
3 2
A 6 5
3
3
10 5 4
—

When South leads the 3, West is squeezed. If he throws a diamond, the 2 is discarded from dummy; then the lead of the 10 wraps up four diamond tricks. If West instead discards his last heart, dummy throws a diamond; then the diamond finesse can be repeated using the 3 as a reentry to the South hand.

Return to Main

	— K 2 A Q J 8 K Q J 10 9 8 7
2 Q J 9 8 7 6 5 4 K 9 7 6 —		10 7 5 4 10 3 2 A 6 5 4 3 2
Lead: Q	A K Q J 9 8 6 3 A 3 10 5 4 —

	— 2 A Q J 8 —
— J K 9 7 6 —		— — 3 2 A 6 5
	3 3 10 5 4 —