Puzzle 8K53 Main

The Three Trick Gainer

by Richard Pavlicek

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

7 NT South	A 9 8 2		Trick	Lead	2nd	3rd	4th
	A 9 8 2		1. W	10	2	6	J
	7		2. S	A	8	3	7
	5 4 3 2		3. S	K	9	4	3
Q J 10		7 6 5 4 3	4. S	Q!	10	5	4
Q J 10		7 6 5 4 3	5. S	K	J	2	5
J 10 9 8		6	6. S	K	10	A!	3
10 9 8		7 6	7. N	A	6	2	Q
	K		8. N	9	7	3	J
	K		9. N	8!	4	4	?
	A K Q 5 4 3 2			West is squeezed (again)
	A K Q J

Alas, the deal is flawed because 7 NT only makes with a club lead. Indeed 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, with only 10 cashable tricks at the start.

“At the start” means after West’s best lead; so if West were endplayed at Trick 1, anything gained by that lead already belongs to declarer. “Without a finesse” means declarer cannot vary his play based on a defender’s play in the same suit, except as South at Trick 1.

Multiple solutions exist. Further goals (tiebreakers for the August contest) are to have the fewest North-South HCP, and lowest North-South rank sum, in that order of priority.

Edouard Bonnet Wins

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 insert vowels of mass confusion. I mean, look at what they did to a simple “yes” with oui, and I don’t think anyone can spell hors d’oeuvres without looking it up first (as I did). All the makings of a Conehead conspiracy, but I digress.

USA in Top 10! Czech Republic loses!

After my carefully crafted subtitle, readers might have no need to peruse the leaderboard. Oh well. Thanks to Mississippi and its newfound resident (formerly of California) for saving us from oblivion.

Winner List

Rank	Name	Location	HCP	Sum
1	Edouard Bonnet	France	25	188
2	Pavel Striz	Czech Republic	25	188
3	Dan Dang	British Columbia	25	188
4	Tim Broeken	Netherlands	25	189
5	Jonathan Mestel	England	25	189
6	David Brooks	Australia	31	192
7	Paul Barden	England	31	194
8	Gareth Birdsall	England	32	193
9	James Lawrence	England	33	196
10	Jim Munday	Mississippi	34	204

Puzzle 8K53 Main

Top The Three Trick Gainer

Solution

A few solvers tried to skirt the essence of the puzzle by having West endplayed at Trick 1, thereby gaining a trick, so the squeeze only needed to gain 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 my 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, such that declarer would have only two losers if he could reach all his winners. 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 hurdle is to prevent the removal of either blocked threat-suit position. Rather than A-9-8-2 opposite a blank king, where one lead will destroy it, declarer needs something like A-x-x-x-x opposite K-Q, where one lead is harmless. Below is a valid construction by David Brooks (Australia):

7 NT South	A 5 4 3 2		Trick	Lead	2nd	3rd	4th
	A 5 4 3 2		1. W	J	2	6	K
	3 2		2. S	K	8	2	6
	2		3. S	A	6	2	J
J 10 9 8		7 6	4. S	A	K	2	5
J 10 9 8		7 6	5. S	Q	8	3	6
9 8 7 6		Q J	6. S	Q	9	3	7
K		J 10 9 8 7 6 5	7. S	Q	9	A	7
	K Q		8. N	A	7	3	10
	K Q		9. N	5	8	4	10
	A K 10 5 4		10. N	4	9	4	?
	A Q 4 3			West is squeezed

Many play sequences exist. 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 Q then overtake the 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 decide to set your priorities straight, PavCo Attorneys Down Under will represent you pro bono.

Leaping tall buildings in a single bound

I especially liked the next entry from Tim Broeken (Netherlands), this series’ most formidable solver with 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 South	A J		Trick	Lead	2nd	3rd	4th
	A J		1. W	K	A	5	3
	A Q J 10 4 3 2		2. N	A	7	2	K
	A 2		3. N	A	7	2	K
K		10 9 8 7	4. N	A	8	3	4
K		10 9 8 7	5. N	Q	6	4	5
K		9 8 7 6 5	6. N	J	7	5	6
K Q J 10 9 8 7 6 5 4		—	7. N	10	8	6	7
	Q 6 5 4 3 2		8. N	J	9	3	8
	Q 6 5 4 3 2		9. N	J	8	Q	9
	—			Declarer succeeds
	3

If West leads a non-club, North wins the A, A and A in any order. Note that North’s 4-3-2 comprises a triple extended threat, as do South’s low cards in each major, provided the club is discarded on the A. Therefore, when the 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!

A similar layout, but majors A-Q-x-x-x-x opposite J-x with blank kings offside, was submitted by this brilliant solver:

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, their rank 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 this winning entry from Edouard Bonnet (France) shows:

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

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. The  A then triple-squeezes West, who does best to give up a major (spades as shown) and North pitches the opposite major. Cash the last diamond winner to pitch a heart, unblock spades, then cross to dummy with the A.

Next run spades to reach this ending before the last spade is led:

NT win 3	5		Trick	Lead	2nd	3rd	4th
	Q 6		11. N	5	J	2	?
	—			Declarer succeeds
	—
—		—
10 9		—
8		—
—		K Q J
	—
	J
	3 2
North leads	—

On the 5 South sheds a diamond, and West is squeezed in an unusual matrix. If he pitches a heart, the Q drops the jack and the 6 wins the last. If he pitches his diamond, declarer crosses to the J to win the good diamond. This specific ending is not required to make 7 NT but is just one of various successful finishes.

If West leads a major, the play is easier. North wins the A and A, then a club to the ace squeezes West, who must pitch a diamond to avoid losing three tricks at once; then the good diamonds finish him off. North also makes 7 NT, as the only difference could be a club lead, which squeezes West immediately.

The same optimal layout was submitted by Pavel Striz and Dan Dang.

Odd man out

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

Entering these contests might suggest otherwise, but not to worry… I used to be an odd person, but thanks to diligent study and research, I have graduated to a nut case.

Puzzle 8K53 Main

Top The Three Trick Gainer

© 2011 Richard Pavlicek