September Problem, again

		5 A Q 7 5 A 7 3 K J 10 8 4
A J 10 8 7 4 K 4 2 9 5 6 3
N-S Vul
You West 2 Pass Pass	North Dbl Pass 5		East 3 4 Dbl	South 4 Pass All Pass
Trick 1. W 2. W	Lead A ?		2nd 5	3rd 3	4th 2

The September contest is over, but before I announce the winner I’d like to ask the RBC community to consider a slightly different version of the problem. (Maybe I’m wondering if I should have posed the problem differently…) In this version, instead of playing the king on your ace-of-spades lead, partner plays the three. Now, what do you continue? Let me know at johnctorrey@aol.com. I do look forward to your responses.

You, West, opened two spades and ended up on lead against five hearts doubled. You led the ace of spades and partner played the king. What now? (In problems like this the assumption always is that partner is a good player with whom you do not have nonstandard agreements. In this case, you have agreed to play “standard leads and signals” but have had no further discussion on the subject.)

As is our custom, the free play will go to the correct answer from the player 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.

For no extra credit (but to satisfy my own curiosity) you can answer the following

Bidding problem

	None Vul
		5 3 A 8 6 5 3 A K J A 8 4

West 1 3 3 NT

North Pass Pass Pass

You East 1 3 ?

South 2 Pass

In your partnership 4NT would be natural and invitational; 4 would be Gerber.

August Answer

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

You West Dbl

North All Pass

East 1

South 2

Trick 1. W 2. W

Lead A ?

2nd 5

3rd 2

4th 7

This problem drew a good number of thoughtful answers. As several responders noticed, it is possible to count the unseen hands. Partner has three clubs and thus (having opened one club) should not have four diamonds. Likewise partner should not have four spades, given that our double invited a spade response. (If partner is 4-4-2-3 then declarer is 1-6-5-1, an unlikely and hard to defeat shape.) And partner should not have five hearts, having opened a cheesy Jxx of clubs. The only distribution that works is 3-4-3-3 for partner and 2-6-4-1 for declarer. If we do not take our spades NOW, they will go away on dummy’s club winners. So most responders select the double-dummy jack of spades, protecting against declarer having the ten.

That is just what I did when I held the hand, about 15 years back. When I posted the problem I had forgotten the unhappy conclusion: my thoughtful, careful partner was not playing double-dummy and played the ace. I could have had J109xx in spades, after all – and leading the jack from KJ is an unexpected, unusual play. And if I did have the king, we would still have our two spade tricks. All true, except that now we can’t beat the contract. So this month’s winner is Audrey Ventura, the only entrant who leads the king of spades and then the jack, planning to shift to diamonds if the jack holds. Partner may have to do some endplay avoidance if declarer covers the jack, but I haven’t come up with any situations where this is impossible or even really difficult.