Puzzle 8M31 Main

The Jackson Four

by Richard Pavlicek

Professor Freebid strolled into the Hilton coffee shop for breakfast, Daily Bulletin in hand, perusing the carryovers to the day’s final sessions of the Life Master Pairs. Across the room he spots Timothy and heads for his table.

“Good morning, Timothy. May I join you?”

“Sure. Nice game last night! I see you and Marlon are in fourth place. Grover and I were borderline to qualify until I made seven notrump on our last board to cinch it. What happened at your table? It was Board 4.”

“Ah, yes, the four-by-one jacks. We defended the same contract.”

“Darn,” said Timothy. “I lost my hand record and can’t remember the exact cards. Grover and I each had 18 high-card points, and our auction was tortuous. West led the jack of… um, I forgot… and I made it on a squeeze.”

“Well done, but you would have gone down at my table.
[The Professor scribbles two hands on his clipboard]
I led the only jack to beat it!”

7 NT South	?  ?  ?  ?
J 10 9 8 J J 10 9 8 J 10 9 8		7 10 9 8 7 6 7 6 5 7 6 5 4
Which Jack do you lead?	?  ?  ?  ?

“Wow!” exclaimed Timothy. “That’s a real tossup… well, except for Grover who would lead his singleton.”

Complete the diagram with North-South hands to fit the story.

Multiple solutions exist. A further goal (tie-breaker for the September 2015 contest) is for the North-South combined point count to be as low as possible, based on the Kaplan-Rubens method. Use my Bridge Hand Evaluator to find the K-R points of any hand.

Tim Broeken Wins

Congratulations to Tim Broeken, who was the first of only three solvers to find the optimal layout (lowest total Kaplan-Rubens point count). No surprises at the top! This is Tim’s eighth win in 25 puzzle contests, his first being my Yarborough Fair fantasy puzzle back in January 2011. Also in the groove, Dan Gheorghiu has submitted optimal solutions for five months in a row.

Winner List

Rank	Name	Location	North K-R	South K-R	Total K-R
1	Tim Broeken	Netherlands	16.65	17.9	34.55
2	Dan Gheorghiu	British Columbia	17.9	16.65	34.55
3	Jean-Christophe Clement	France	17.9	16.65	34.55
4	Dean Pokorny	Croatia	18.25	16.65	34.9
5	Tom Slater	England	18.25	16.65	34.9
6	David Brooks	Australia	18.75	17.35	36.1
7	Leigh Matheson	Australia	18.75	17.35	36.1
8	Jamie Pearson	Ontario	17.35	19.45	36.8
9	Jonathan Mestel	England	17.55	19.25	36.8
10	Nicholas Greer	England	19.25	17.55	36.8
11	Benjamin Kristensen	Minnesota	19.45	17.35	36.8
12	Grant Peacock	Maryland	17.8	19.3	37.1
13	Leif-Erik Stabell	Zimbabwe	18.75	18.55	37.3
14	Zla Khadgar	Ohio	17.8	19.6	37.4
15	Dan Baker	Texas	17.55	20	37.55
16	Wayne Somerville	Northern Ireland	19.9	17.8	37.7
17	Reint Ostendorf	Netherlands	18.6	19.9	38.5
18	Jim Munday	Mississippi	18.3	21.1	39.4
19	Charles Blair	Illinois	23.7	23	46.7
20	Rob Dixon	England	23	26.8	49.8
21	Adam Dickinson	Scotland	25.7	24.9	50.6

Puzzle 8M31 Main

Top The Jackson Four

Solution

Before showing the optimal solution, let’s examine a few curiosities. Only one solver submitted a layout in which only a spade defeats 7 NT, which required extreme shape, which in turn dictated a high Kaplan-Rubens count. Nonetheless, this solver is never bothered by the number of ‘which’ occurrences. Call it the Blair Which Project:

	A K Q 6 5 4 3		West	North	East	South
Both vul	4 3 2		Pass	2	Pass	3
	—		Pass	3	Pass	4
	A K Q		Pass	5	Pass	7 NT
J 10 9 8		7	Pass	Pass	Pass
J		10 9 8 7 6
J 10 9 8		7 6 5
J 10 9 8		7 6 5 4
	2
	A K Q 5
Lead: J	A K Q 4 3 2
7 NT South	3 2

Imagine holding the South hand and seeing partner open 2 ! Six bids to 7 NT might in practice be just two. (Auctions shown are for interest sake and have no bearing on the puzzle solutions.)

Without a spade lead, declarer simply cashes his club-heart tops ending in hand, and West is squeezed in the pointed suits. The  J, however, severs communication. No entry in a threat suit means no squeeze.

Charles Blair: It would be fun if only the J worked, then the Professor would have led the only card to set 7 NT.

Now that you mention it, constructing a deal that only Grover would set is a tall order. In fact I thought about saving it for a future puzzle, but the layout I discovered has rigid constraints, leaving little scope for ranking. To wit:

Grover wrecks the pentagon

On the following deal only the J lead defeats 7 NT, and it does so by scuttling a pentagon squeeze. First let’s see how a normal lead, say the J, allows 7 NT to be made.

7 NT South	A K Q 6 5 4 3 2		Trick	Lead	2nd	3rd	4th
	A 5 4		1. W	J	A	7	2
	K Q		2. N	K	5	2	8
	—		3. N	Q	6	3	9
J 10 9 8		7	4. N	4	6	K	J
J		10 9 8 7 6	5. S	A	10	2	7
J 10 9 8		7 6 5	6. S	A	8	3	4
J 10 9 8		7 6 5 4	7. S	K	9	4	5
	—		8. S	Q	10	5	6
	K Q 3 2			continued below…
	A 4 3 2
	A K Q 3 2

Declarer must pitch a club at Trick 1, unblock diamonds, cross in hearts and cash the A. Cashing the top clubs is optional but desirable to reach this ending:

NT win 5	K Q 6		Trick	Lead	2nd	3rd	4th
	A 5		9. S	2	J	A	7
	—		10. N	K	8	4	8
	—		11. N	Q	?
10 9 8		—		East is squeezed
—		10 9 8 7
J		—
J		7
	—
	Q 3 2
	4
South leads	3

Next the 2 squeezes West in three suits, and his only hope is to let go the club, which East still guards. Finally, North’s top spades squeeze East in hearts and clubs.

The squeeze works the same with either minor-suit lead, but the J ruins communication.

Back to the real world

The popular solution, found by 12 of the 21 successful solvers, keyed on a club lead to break up a double squeeze. Each included a common suit of A-K-Q-x opposite x, so one lead removed the crucial entry. Various layouts produced Kaplan-Rubens counts from 36.1 to 39.4, of which the best (lowest) is shown below.

	Q 6 5 4 3 2		West	North	East	South
Both vul	A K Q		Pass	1	Pass	2 NT
	A K 2		Pass	3	Pass	4
	2		Pass	4 NT	Pass	5
J 10 9 8		7	Pass	7	Pass	7 NT
J		10 9 8 7 6	Pass	Pass	Pass
J 10 9 8		7 6 5
J 10 9 8		7 6 5 4
	A K
	5 4 3 2
Lead: J	Q 4 3
7 NT South	A K Q 3

David Brooks: The J breaks the double squeeze; other jacks allow 13 tricks.

Leigh Matheson: Club lead kills the entry required for the double squeeze.

After a non-club lead, declarer wins A-K-Q, A-K and A-K. Next the Q forces West to unguard clubs to keep a spade stopper, then the Q squeezes East in hearts and clubs. Alternatively, declarer could win two diamonds, three spades (pitch a diamond) and three hearts, then the K effects a simultaneous double squeeze.

Jim Munday: Kostal’s Law, “They never lead clubs,” comes into play here. Leading a jack at random would be my best hope of guessing the right one.

Why only a jack? My experience has been any card.

Crisscrossing the globe

Three solvers managed to construct the best layout, or should that be the worst, capitalizing on a blank king, a doubleton K-Q, A-K-Q tight and a queen-high suit, all of which portend gloom in Kaplan-Rubens evaluation. The layout is unique, except the North-South hands can be switched. Only our winner gave the long spades to North, which makes South more likely to declare notrump.

7 NT South	Q 6 5 4 3 2		Trick	Lead	2nd	3rd	4th
	A K Q		1. W	J	2	7	A
	K		2. S	K	8	2	4
	A 3 2		3. S	Q	9	3	5
J 10 9 8		7	4. S	2	J	A	6
J		10 9 8 7 6	5. N	K	7	3	10
J 10 9 8		7 6 5	6. N	Q	8	4	J
J 10 9 8		7 6 5 4		continued below…
	A K
	5 4 3 2
	A Q 4 3 2
	K Q

With communication knotted by three blocked suits, appearances suggest no squeeze will work with any lead; but squeezes sometimes defy tradition. Hidden in the midst is an extended threat crisscross, unless West finds the killing lead.

Suppose West errs by leading a spade. Declarer wins the A, K-Q and A-K-Q to reach this position:

NT win 7	Q 6 5 4 3		Trick	Lead	2nd	3rd	4th
	—		7. N	A	6	5	?
	K			West is squeezed
	A
10 9 8		—
—		10 9
J 10 9 8		7 6 5
—		7 6
	K
	5
	A Q 4 3 2
North leads	—

Next comes the A (South pitching a heart) and West has no answer; whichever suit he lets go, declarer will unblock its top card and cross to the opposite hand which is high.

Only an original diamond lead stops it.

Dan Gheorghiu: The squeeze against West is a morphing of a crisscross, in that the pointed suits are overloaded with [threat] cards, which are necessary because the A-Q or Q (depending on West’s discard) may not win tricks. A diamond lead kills the squeeze and sets 7 NT. Interestingly, North-South can be swapped with the same result. Finding the optimal K-R count was, to say the least, painful.

I can attest that Dan is an expert on morphing. I noticed that Dan Dang, winner of The Law of Total Trash in 2011, is the same person. Dang, I suspected that was phony; but I can understand it. Imagine how many times he’s had to spell “Gheorghiu” for people.

K-R chronicles

Tom Slater: I never have any idea what the K-R outcome will be when I swap cards.
Singleton kings and doubleton queens seem to be a good idea though.

Jonathan Mestel: I don’t understand this K-R stuff, but you found a way to rectify the count in a grand slam.

Jamie Pearson: I’m glad you’re easing us into the K-R count… can’t wait until jacks, 10s and nines are in play.

Yes, you never know. Some day you might win a trick with a jack and have only Edgar and Jeff to thank.

Puzzle 8M31 Main

Top The Jackson Four

© 2015 Richard Pavlicek