Main     Problem 8S01 by Richard Pavlicek    

Two-Way Finesses

The sure-trick problem is a classic bridge poser, the object being to find a line of play to guarantee a contract against any layout of the unseen cards. Early bridge writing is rife with these problems, mostly illustrating safety plays that are taken for granted today. Unlike bridge puzzles, which usually have no practical value — anyone who tried my “Little Deuce Coupe” will attest to that — sure-trick problems are almost always instructive. Here’s a new one to challenge you:

3 NT S A 8 3 2
H Q J 2
D A 3 2
C K J 10
Trick
1. W
2. E
Lead
H 10
D 10
2nd
J
J
3rd
K
K
4th
3
?


Lead: H 10
 
Table


 
S K 10 9
H A 5 4 3
D Q J 5 4
C A 2

As South, how do you guarantee making 3 NT after the start shown?

Decide whether you will win or duck at Trick 2, and devise a plan. Assume all enemy plays will be optimal for the defense. That is, any finesse you take will lose, no suit you test will split 3-3, and no helpful lead or discard will be made by an opponent.

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Charles Blair Wins!

In October-November 2010 this problem was presented as a contest, with 58 participants from 37 locations. Thanks to those who entered, and congratulations to the seven successful solvers. Ranking is based entirely on date and time of entry.

Headlines of yesteryear! Charles Blair was a three-time winner in my 2000-06 contest series, and one of only three persons out of 7000+ to participate in all my old polls and contests. Can’t lose this guy no matter how much I downscale! Seriously, Charles has become a good cyber friend with our common interest in puzzles (bridge, chess, math).

Other solvers are also familiar. Jim Munday and Julian Wightwick are regulars, sharp as a tack, but that seventh-place Missouri gal has the most impressive record. Ding-Hwa dominated my 2007 series, winning three times and topping the final leaderboard in all 17 stat categories. I once referred to her as “the Tiger Woods of RPbridge.” Oops! I just removed that reference from an old page, lest I get my face slapped.

Congrats also and thanks to N. Scott Cardell (Washington) with whom I tested this problem shortly after creating it. (Scott is a longtime correspondent and a composer of many interesting bridge problems.) Actually, I was a little irked that he came back with the solution too fast. Wise guy. Scott didn’t enter the contest of course because of this prior exposure.

RankNameLocation
1Charles BlairIllinois
2Pavel StrizCzech Republic
3Jim MundayCalifornia
4Chris ChambersEngland
5Julian WightwickEngland
6Charles BrennerCalifornia
7Ding-Hwa HsiehMissouri

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Solution

After the first seven plays (forced) you have eight top tricks. Knowing how the cards lie would yield an easy 10 tricks, and it might be possible to win as many as 12; but never mind that. Nine is the number you need, and a series of wrong views and misbehaving suits could leave you with the same eight you started with. The keys to success are to develop a red-suit threat behind East and to rectify the count (lose four tricks).

3 NT S A 8 3 2
H Q J 2
D A 3 2
C K J 10
Trick
1. W
2. E
Lead
H 10
D 10
2nd
J
J
3rd
K
K
4th
3
?


Lead: H 10
 
Table


 
S K 10 9
H A 5 4 3
D Q J 5 4
C A 2

First, you must duck at Trick 2. West must return a red suit (a club or a spade would gift a trick), and you must block both red suits, winning the H A and D Q in whichever order West determines. Next run the S 10 to East. If East has no more red cards, he is endplayed in the black suits. Otherwise, you will have a red-suit threat behind East (or a suit breaking 3-3) after winning his exit (say, a diamond) in dummy.

Now you must make the unusual play to finesse spades the opposite way, leading a spade to the nine. A butcher job in the spade suit if there ever was one, but you must focus on the forest not the trees. Assume West wins and exits safely with a spade, won in dummy as East follows.

This leaves the ending at right, where you know East guards diamonds (the suit he returned when he won a spade trick). Next cash the H Q (important before the last spade) to discover who guards hearts. If West, the S 8 inflicts a double squeeze (neither opponent can protect clubs). If East guards hearts (and diamonds), the S 8 will force him down to one club; then C A and finesse is a lock. North
leads
S 8
H Q
D
C K J 10



 
Table


 
S
H 5 4
D 5
C A 2

Here’s how some of the correct solvers played it:

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


Lead: H 10
 
Table


 
Charles Blair
Illinois

S K 10 9
H A 5 4 3
D Q J 5 4
C A 2

Charles Blair: Next cash the H Q and S 8. If East began with heart length, take the club finesse. If West had hearts, there is a double squeeze. I think this might be called a “double Gallagher finesse.”

Yes, making 3 NT is like smashing watermelons. Oh, wait; wrong Gallagher. Charles refers to Ann Gallagher, a movie actress circa 1940, who purportedly would take a finesse one way; then if it worked, take it the other way saying, “Now let’s see if I’m really lucky.” I’m guessing she was more lucky at picking up suitors than suits this way.

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


Lead: H 10
 
Table


 
Jim Munday
California

S K 10 9
H A 5 4 3
D Q J 5 4
C A 2

Jim Munday: The idea is to endplay East for Trick 6; he cannot return a black suit or I have nine tricks. Whichever red suit East returns (he must have 4+ or I’m home already) will give me a positional threat over him. I need both top red honors in dummy to win the return and lose the spade to West. At Trick 8, West cannot return a club so must lead a heart or spade. From the diagram, I will win the H Q and lead the fourth spade… [ending described].

Jim is the only solver to earn top marks in artistic merit, ducking Trick 2 with the three to parallel Trick 1.

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


Lead: H 10
 
Table


 
Chris Chambers
England

S K 10 9
H A 5 4 3
D Q J 5 4
C A 2

Chris Chambers: Next win the D A and lead the S A. If East has four diamonds (as well as four hearts), he is forced to a singleton club, so C A and a finesse will win (same for 2=4=4=3). If West has four diamonds,…a double squeeze… This is symmetrical about the red suits, but if East didn’t have four of either, he was endplayed in the blacks at Trick 6. In the stated line, if West exits with a diamond, North can lead the S A and S 8 to effect the same double squeeze.

Chris neatly illustrates the mirror layout of East having heart length. This is why I had both opponents lead the 10, i.e., to be consistent with either a long or short suit.

Julian Wightwick: Partner didn’t like my spade plays, but cheered up at the end. A spectacular hand… I like the deliberate red-suit blockages and the spade compression. It is instructive to play for a position with a red-suit menace over East.

Thanks. Maybe I’ll be inspired to work on more realistic problems and less nonsense like winning deuces. Nah.

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© 2010 Richard Pavlicek