June Problem

	9 A 10 8 3 10 8 Q J 9 6 5 3
6 Q 9 6 A K 9 6 2 K 10 8 2
N-S Vul

West 2 All Pass

North Dbl

East 5

South 1 5

Trick 1. W

Lead A

2nd 8

3rd J

4th 3

You lead the ace of diamonds to South’s 5 contract. Partner plays the jack and declarer the trey. Plan the defense.

As is our custom, the free play will go to the correct answer from the RBC member with the fewest masterpoints. “Correct” is a relative term; just saying “I lead X,” even if X is the winning play, will probably not qualify unless nobody else finds the play. You need to furnish some justification for your play. (On the other hand, a briefly or poorly stated but correct justification will fully qualify, if I can figure it out. This is not an essay contest.) I am the sole judge of what constitutes “correct.” Send answers to JohnCTorrey@aol.com.

---

April Problem

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

West Pass All Pass

North 2

East Pass

South 1 6

Trick 1. W 2. N 3. W

Lead 8 K ?

2nd K 7

3rd 5 2

4th 6 A

You led the eight of trump to South’s six spade contract. South won the king in dummy and led the king of clubs, discarding a diamond as you won the ace. Plan the defense.

This teaser was held over when nobody answered correctly in April. Solvers had trouble picturing a hand where South would discard a diamond rather than a heart.

South is unlikely to make the hand with fewer than seven trumps. If South has four hearts to the AQ and two diamonds, a diamond looks right, but he can’t make the hand if we return a heart, because we still own the third round of the suit. If South has AQx in both red suits we will beat the hand with a diamond return, but South with this holding would have discarded a heart rather than a diamond. This leaves the above distribution, where only a heart return is a winner. South’s play was well conceived; if he had thrown a heart on the club, a heart return would have been automatic.

Audery Ventura and Ashok Damle both answered correctly. Ashok Damle wins the free play.

---

May Problem

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

West Pass Pass

North 3 3 NT

East Pass All Pass

South 2 NT 3

The defense takes the first four tricks in hearts. Plan the play.

I might call this “The case of the poisoned spade.” You have three tricks in spades and two in diamonds, so it seems you need four in clubs, or three plus another in spades or diamonds. You need the club finesse even if the spades break or the queen of diamonds drops in two rounds. You can take three club tricks when the suit breaks badly, and four if it divides 3-3. BUT one extra trick in spades or diamonds is useless, because cashing that trick squeezes the south hand in clubs: if you throw the seven you can’t repeat the finesse, and if you throw the jack the defense can cover the ten. Better to just hope for 3-3 clubs!

So saving the fourth spade is not a good idea. But saving four diamonds is a good idea, because a defender with the queen of diamonds and four clubs gives up two diamond tricks if he discards the queen. Win the diamond shift, cash South’s winners in spades and diamonds, and lead a spade to the ace. The dummy is down to two diamonds and two clubs; South has four clubs, and East has…the queen of diamonds and three clubs. Now finesse the clubs and make the rest.

Strangely, it does not matter who has the ten of diamonds, provided the queen is not doubleton. And if West has two extra small clubs and the queen of diamonds instead of the eight (with East having the spade length), the squeeze still works.

This teaser drew several thoughtful answers; everybody saw the problem in the club suit. Erik Secan was the only one to project the squeeze: he wins the free play.