Main     Puzzle 8K53 by Richard Pavlicek    

The Three Trick Gainer

Most players are familiar with the progressive squeeze that gains two tricks. But is it possible for a squeeze to gain three tricks without losing the lead? Yes, as illustrated by this deal from New Zealand Bridge, circa 1974. South can win only 10 tricks at the start (note the blockages) but West is squeezed for all 13.

7 NT S A 9 8 2
H A 9 8 2
D 7
C 5 4 3 2
S Q J 10
H Q J 10
D J 10 9 8
C 10 9 8
Table S 7 6 5 4 3
H 7 6 5 4 3
D 6
C 7 6
Club lead S K
H K
D A K Q 5 4 3 2
C A K Q J

Alas, the deal is flawed because 7 NT only makes with a club lead. Make that grossly flawed, because in real life a club would be West’s last choice of leads. A challenge for you:

Construct a deal where South can make 7 NT against any defense without a finesse, when able to win only 10 tricks at the start.

“At the start” means after West’s best lead; so if West were endplayed at trick one, any trick gained by the lead already belongs to declarer. “Without a finesse” means that declarer cannot vary his play based on a defender’s play in the same suit, except as South on the opening lead.

Multiple solutions exist. An additional goal is to achieve the fewest North-South HCP.

TopMain

Edouard Bonnet Wins

In July-August 2011 this puzzle was presented as a contest, with 46 participants from 17 locations. Thanks to the brave souls who entered, and congratulations to the 10 who constructed a valid deal to fulfill the conditions. Ties are broken by the fewest N-S HCP, the lowest N-S card sum, and lastly by date and time of entry.

The first of three to submit the optimal solution was Edouard Bonnet of France. That’s “Edward” to most of us, but the French continue to stockpile their consonants and distribute vowels of mass confusion. I mean, look at what they did to a simple “yes” with oui, and I doubt that anyone can spell hors d’oeuvre without looking it up first as I just did. All the makings of an anti-Conehead conspiracy, but I digress.

USA in Top 10 as England falls to Canada

Politically correct subtitles will hopefully distract people from the ranking list. Oh, well. Thanks to Mississippi and its newfound resident (formerly of California) for saving our country from omission.

RankNameLocationHCPSum
1Edouard BonnetFrance25188
2Pavel StrizCzech Republic25188
3Dan DangBritish Columbia25188
4Tim BroekenNetherlands25189
5Jonathan MestelEngland25189
6David BrooksAustralia31192
7Paul BardenEngland31194
8Gareth BirdsallEngland32193
9James LawrenceEngland33196
10Jim MundayMississippi34204

TopMain

Solution

A few solvers tried to skirt the essence of the puzzle by having West endplayed at trick one, thereby gaining at least one trick, so the squeeze only needed to gain one or two tricks. Sorry, no shortcuts. While I neglected to define “at the start” in the original puzzle (now clarified) the great majority of respondents correctly interpreted its intent.

A three-trick-gaining squeeze (without losing a trick) is extremely rare, because if declarer has three losers, the defender has three extra cards, which allows him to retain stoppers in three suits. The crux of this rare squeeze is a blocked position in two suits, as represented by the majors in the example. Either blocked suit, at declarer’s discretion, can produce a hidden winner, which is effectively like declarer having only two losers. Further, all three threat suits require an extended threat (two or more threat cards) so the initial squeeze gains two tricks. The third trick can then be gained in typical fashion by a simple squeeze.

To succeed against any defense, the main construction hurdle is to prevent the removal of either blocked threat-suit position. That is, instead of A-9-8-2 opposite a blank king, where one lead destroys it, declarer needs something like A-x-x-x-x opposite K-Q, where the lead is harmless. A valid construction:

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

S K Q
H K Q
D A K 10 5 4
C A Q 4 3

Many play sequences exist, but to verify the success after any lead, Tricks 1-4 consist of winning a top card in each suit. At Trick 5 West is triple-squeezed, and the path shown assumes he abandons spades; so cash the S Q then overtake the H Q. The good spades are then cashed, and at Trick 10 West is squeezed in the red suits. Obviously, if West abandons hearts early, the play is mirrored. If West instead abandons diamonds, running that suit will squeeze him in the majors.

David Brooks: I am sure you can do better, but if I spend any more time on this I will be doing the next problem waiting in the divorce courts.

Probably a wise decision, but if you ever decide to set your priorities straight, PavCo Attorneys-Down-Under will represent you pro bono.

I especially liked the next entry from this series’ most formidable solver (three wins to date). Not only does it reduce North-South to a mere 25 HCP, but the squeeze occurs in a single bound — three tricks gained at once. Further, if West chooses, the squeeze can occur on the opening lead. East is the victim:

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

S Q 6 5 4 3 2
H Q 6 5 4 3 2
D
C 3

If West leads a non-club, North wins the S A, H A and D A in any order. Note that North’s D 4-3-2 comprises a triple extended threat; and South holds the same in each major, provided the club is discarded on the D A. Therefore, when the C A is led at Trick 4, whichever suit East abandons gives declarer three immediate tricks, so all that remains is to cash winners. If West makes his natural lead of a club, he achieves the ignominy of forcing partner to lose three tricks on the opening lead. Ouch! So much for partnership harmony.

A similar solution was submitted by another top solver but with each major A-Q-x-x-x-x opposite J-x, so the blank kings were offside:

Jonathan Mestel: If you won’t let us finesse, we’ll drop ‘em all.

Lowering the bar a notch

While the previous deal contains the absolute minimum N-S HCP, the card sum of 189 can be lowered to 188 — a trivial difference, but worth gold in these contests. This is accomplished by changing the enemy singleton king in declarer’s non-blocked threat suit (diamonds) to Q-J doubleton, as the winning entry shows:

7 NT S A Q 6 5 4 3
H A Q 6 5 4 3
D
C 4
Trick
1. W
2. S
3. S
4. S
5. S
6. S
7. N
8. S
9. N
10. N
Lead
D 5
D A
D 10
C A
D 9
S 2
S 4
H 2
S Q
S 6
2nd
S 3
4
6
S 7
7
8
C 8
7
C 9
C 10
3rd
J
H 3
C 4
H 4
H 5
A
J
A
C 2
C 3
4th
K
Q
C 5
6
C 7
K
9
K
10
H 8
S 10 9 8 7
H 10 9 8 7
D 8 7 6 5 4
C
Table S K
H K
D Q J
C K Q J 10 9 8 7 6 5
Edouard Bonnet
France

S J 2
H J 2
D A K 10 9 3 2
C A 3 2

The play is flexible after any lead. A diamond is most interesting, in which case only three diamonds can be won immediately, pitching one card from each suit in dummy. At Trick 4 the C A triple-squeezes West, who does best to give up a major (spades as shown) then North pitches the opposite major. The last diamond winner is cashed to pitch a heart, spades are unblocked, then dummy is reached with the H A.

Next run the spades. When the last spade is led (see ending) South sheds a diamond, and West is squeezed in an unusual matrix. If he pitches a heart, the H Q drops the jack and the H 6 is good. If he pitches his diamond, declarer crosses to the H J to score the good diamond. This specific ending is not required to make 7 NT but is just one of various successful finishes. North
leads
S 5
H Q 6
D
C
S
H 10 9
D 8
C
Table S
H
D
C K Q J
S
H J
D 3 2
C

If West leads a major, the play is simpler. North wins the S A and H A, then a club to the ace squeezes West, who must pitch a diamond to avoid losing three tricks at once; then the diamond suit finishes him off. North also makes 7 NT, as the only difference would be a club lead, which squeezes West immediately. Besides Edouard Bonnet, the same essential layout was submitted by Pavel Striz and Dan Dang.

Pavel Striz: I would prefer a solution with less than 25 HCP — I am not an odd person.

Entering these contests may suggest otherwise, but don’t worry. I used to be an odd person but not anymore! Thanks to diligent research I have graduated to nut case.

TopMain

© 2011 Richard Pavlicek