Main     Puzzle 8M31 by Richard Pavlicek    

The Jackson Four

Professor Freebid strolled into the Hilton coffee shop for breakfast, Daily Bulletin in hand, perusing the carryovers for 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 Jackie are fourth. Marlon and I were borderline to qualify until I made seven notrump on the 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, I lost my hand record and can’t remember the exact cards. Marlon and I had the same number of high-card points, and our auction was tortuous, almost ending in six. West led the jack of… um, something, I forget… and I made it on a squeeze.”

“Well done, but you would have gone down at my table.”
The Professor jots down the E-W cards on his clipboard.
“I led the only jack to beat it.”

7 NT South  ?
 ?
 ?
 ?
S J 10 9 8
H J
D J 10 9 8
C J 10 9 8
Table S 7
H 10 9 8 7 6
D 7 6 5
C 7 6 5 4
West leads  ?
 ?
 ?
 ?

“Wow! A real tossup. Well, except for Grover who would lead his singleton.”

Complete the diagram with North-South hands that fit the dialogue.

Many 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 of evaluation. Use my Bridge Hand Evaluator to find the K-R points of any hand.

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Tim Broeken Wins!

In September 2015 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Forty-seven persons gave it a try, but only the 21 listed constructed a layout in which only one jack lead defeats 7 NT. Thanks to all who participated.

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.

RankNameLocationNorth K-RSouth K-RTotal K-R
1Tim BroekenNetherlands16.6517.934.55
2Dan GheorghiuBritish Columbia17.916.6534.55
3Jean-Christophe ClementFrance17.916.6534.55
4Dean PokornyCroatia18.2516.6534.9
5Tom SlaterEngland18.2516.6534.9
6David BrooksAustralia18.7517.3536.1
7Leigh MathesonAustralia18.7517.3536.1
8Jamie PearsonOntario17.3519.4536.8
9Jonathan MestelEngland17.5519.2536.8
10Nicholas GreerEngland19.2517.5536.8
11Benjamin KristensenMinnesota19.4517.3536.8
12Grant PeacockMaryland17.819.337.1
13Leif-Erik StabellZimbabwe18.7518.5537.3
14Zla KhadgarOhio17.819.637.4
15Dan BakerTexas17.552037.55
16Wayne SomervilleNorthern Ireland19.917.837.7
17Reint OstendorfNetherlands18.619.938.5
18Jim MundayMississippi18.321.139.4
19Charles BlairIllinois23.72346.7
20Rob DixonEngland2326.849.8
21Adam DickinsonScotland25.724.950.6

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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:

7 NT S A K Q 6 5 4 3
H 4 3 2
D
C A K Q
Both Vul

West
Pass
Pass
Pass
All Pass


North
2 C
3 S
5 C


East
Pass
Pass
Pass


South
3 D
4 H
7 NT
S J 10 9 8
H J
D J 10 9 8
C J 10 9 8
Table S 7
H 10 9 8 7 6
D 7 6 5
C 7 6 5 4
Lead: S J! S 2
H A K Q 5
D A K Q 4 3 2
C 3 2

Imagine holding the South hand and seeing partner open 2 C! 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 S J, however, severs communication. No entry in a threat suit means no squeeze.

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

Grover destroys the pentagon

Now that you mention it, constructing a deal that only Grover would set is a tall order. I thought about saving this for a future puzzle, but the layout I discovered has rigid constraints (only the red-suit tops are switchable) leaving little scope for ranking. Only the H J defeats 7 NT, and it does so by scuttling a pentagon squeeze. First let’s see how to make 7 NT, say, with the S J lead.

7 NT S A K Q 6 5 4 3 2
H A 5 4
D K Q
C
Leader
1. W
2. N
3. N
4. N
5. S
6. S
7. S
8. S
Lead
S J
D K
D Q
H 4
D A
C A
C K
C Q
2nd
A
5
6
6
10
8
9
10
3rd
7
2
3
K
S 2
S 3
S 4
S 5
4th
C 2
8
9
J
7
4
5
6
S J 10 9 8
H J
D J 10 9 8
C J 10 9 8
Table S 7
H 10 9 8 7 6
D 7 6 5
C 7 6 5 4
Lead: S J? S
H K Q 3 2
D A 4 3 2
C A K Q 3 2

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

Next the H 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 a minor lead, but the H J ruins communication.
South
leads
S K Q 6
H A 5
D
C
S 10 9 8
H
D J
C J
Table S
H 10 9 8 7
D
C 7
S
H Q 3 2
D 4
C 3

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 C A-K-Q-x opposite C 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.

7 NT S Q 6 5 4 3 2
H A K Q
D A K 2
C 2
Both Vul

West
Pass
Pass
Pass
Pass
All Pass


North
1 S
3 S
4 NT
7 S


East
Pass
Pass
Pass
Pass


South
2 NT
4 C
5 C
7 NT
S J 10 9 8
H J
D J 10 9 8
C J 10 9 8
Table S 7
H 10 9 8 7 6
D 7 6 5
C 7 6 5 4
Lead: C J! S A K
H 5 4 3 2
D Q 4 3
C A K Q 3

David Brooks: The C 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 D A-K-Q, S A-K and H A-K. Next the H Q forces West to unguard clubs to keep a spade stopper, then the S Q squeezes East in hearts and clubs. Alternatively, declarer could win two diamonds, three spades (pitch a diamond) and three hearts, then the D 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 I say 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. In fact the layout is unique, except that North-South can be flipped. Only our winner gave the long spades to North, which makes South more likely to declare notrump.

7 NT S Q 6 5 4 3 2
H A K Q
D K
C A 3 2
Both Vul

West
Pass
Pass
Pass
Pass
Pass
Pass


North
1 S
2 S
4 H
5 S
6 D
7 NT


East
Pass
Pass
Pass
Pass
Pass
All Pass


South
2 D
3 S
4 NT
5 NT
6 H
S J 10 9 8
H J
D J 10 9 8
C J 10 9 8
Table S 7
H 10 9 8 7 6
D 7 6 5
C 7 6 5 4
Lead: D J! S A K
H 5 4 3 2
D A Q 4 3 2
C 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 S A, C K-Q and H A-K-Q to reach the position at right.

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

Only an original diamond lead stops it.
North
leads
S Q 6 5 4 3
H
D K
C A
S 10 9 8
H
D J 10 9 8
C
Table S
H 10 9
D 7 6 5
C 7 6
S K
H 5
D A Q 4 3 2
C

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 D A-Q or S Q (depending on West’s discard) may not win tricks. A diamond lead kills the squeeze and sets 7 NT. Interestingly, the N-S hands can be swapped with the same result. Finding the optimal K-R count was, the least to say, painful.

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

The 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 system, but you’ve found a way of making us 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.

Yep, some day you might win a trick with a jack — and have only Edgar and Jeff to thank.

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© 2015 Richard Pavlicek