Main     Puzzle 8F17 by Richard Pavlicek    

Little Deuce Coupe

“You don’t know what I got…”

Anyone can win aces. Novices soon learn to win picture cards, and competent players are adept at winning spot cards. The little deuce, however, is a special coupe, I mean case. In notrump, a deuce can win a trick only if led and the other three hands are out. How many tricks can be won by deuces on a single deal? Three seems a logical guess and is indeed correct at suit contracts; but at notrump it is possible for every deuce to win a trick. Determining how this might happen is the underlying theme of this challenge:

Construct a deal where South can make 3 NT and all four deuces win a trick.

Note that I did not say against best defense or any defense. You only need a path to nine tricks via legal plays (no revokes or leads out of turn) no matter how unlikely or absurd. In other words, you can dictate the play of all four hands as desired to achieve the goal. Two further conditions must also be met:

West has exactly five spades.
No other hand has a suit over four cards.

Any deal meeting the requirements is correct (there are many) but fewer HCP for North-South is better. Correct solvers will be ranked by the total N-S HCP, fewest being best.

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Zhi Bang Lim Wins!

In September-October 2010 this puzzle was presented as a contest, ending appropriately on Halloween. Thanks to the 55 brave souls who submitted a solution, of which 14 were correct. Surprisingly, almost all the correct solvers produced a deal with the absolute minimum North-South HCP of 6. Well done! Whether this indicates extraordinary talent or total dementia is a matter for debate; but having presented this problem, I have my own marbles to worry about.

The effective tie for places 1-13 is broken by date and time; Zhi Bang Lim was the first to submit an entry. If he continues this habit in the future, I will rename him — from the words of Ralph Kramden if anyone remembers — to Bang Zoom Lim. Congratulations to the successful solvers:

RankNameLocationHCP
1Zhi Bang LimMalaysia6
2Chih-Cheng YehTaiwan6
3Charles BlairIllinois6
4Jean-Christophe ClementFrance6
5Jacco HopNetherlands6
6James WillsonTexas6
7Pavel StrizCzech Republic6
8Richard MorseEngland6
9David KenwardEngland6
10Andrei SecheleaRomania6
11Jim MundayCalifornia6
12Julian WightwickEngland6
13Ales VavpeticSlovenia6
14Bogdan VulcanRomania8

Solution

Before showing any deals, it might be wise to explain the little-known theory of winning deuces. In notrump, it is easy to see how three deuces might win a trick; but what about four? All gather ‘round for a bridge seminar:

Deucology 101

To understand how four tricks can be won by deuces, ask yourself what the other three hands will discard when the last deuce wins a trick. No two hands can discard the same suit, else the deuce of that suit could not have won an earlier trick; so each hand must pitch a different suit, which must be the suit in which that hand won its deuce. Therefore, each hand must win one deuce, and the first three deuce winners must have another card in that suit to pitch on the final deuce.

Given the conditions that West has five spades, and no other hand has more than four in any suit, it is apparent that West must have the S 2, which must be the first winning deuce. Further, spades must split 5-3-3-2 so West can win the deuce on the fourth round, leaving himself another spade to fulfill the mission stated above. It can also be proved that the second winning deuce must be in West’s doubleton, because it must be won on the third round to leave its leader with a surplus card in the suit, and West has no early opportunity to discard.

For declarer to win nine tricks in this fashion, both enemy deuces must be won with the absolute minimum trick loss; i.e., one trick to relinquish the lead and one trick to score their deuce, for both West and East. This also shows that North-South must win the first two tricks, since the earliest West could win the S 2 is at trick four, after gaining the lead at trick three.

Enough about theory. Let’s look at some solutions, starting with the winning entry:

3 NT S Q 9 6
H Q 9 6 2
D 5 4 3
C 5 4 3
Trick
1. W
2. N
3. N
4. W
5. W
6. N
7. N
8. N
9. S
10. S
11. S
12. S
13. E
Lead
S 8
S Q
S 6
S 2!
H 7
H Q
H 2!
D 5
D Q
D 2!
C 10
C 7
C 2!
2nd
9
10
H A
D 3
9
J
D K
7
J
C A
8
J
D 6
3rd
7
4
5
H K
8
5
C 6
9
4
C 3
4
5
S A
4th
3
J
K
H 3
4
10
D A
8
10
C K
9
Q
H 6
S A K J 8 2
H 10 7
D A J 8
C A J 8
Table S 10 7
H A K J 8
D K 10 7
C K Q 9 2
Zim Bang Lim
Malaysia

S 5 4 3
H 5 4 3
D Q 9 6 2
C 10 7 6

Zhi Bang’s deal not only has the minimal HCP but also the minimal spot cards for declarer to succeed. (Julian Wightwick also accomplished this feat in a similar deal.) One of many possible play sequences is shown, but I wouldn’t recommend studying it, lest you become a candidate for the looney bin.

On the next deal, Chih-Cheng (do I hear a cash register?) composed his 6 HCP of a king, queen and jack:

3 NT S J 10 7
H K 7 3 2
D 10 9 6
C 9 8 6
Trick
1. W
2. N
3. N
4. N
5. N
6. N
7. W
8. W
9. N
10. N
11. S
12. S
13. E
Lead
S 6
S J
H 7
D 10
C 9
S 7
S 2!
H J
H 2!
C 8
C 2!
D 8
D 2!
2nd
10
5
5
4
5
Q
C 6
K
C K
10
D J
Q
C 7
3rd
8
4
4
3
3
H 8
H A
Q
D 7
Q
D 6
9
S A
4th
3
9
6
5
4
K
H 9
10
C A
J
D K
A
H 3
S A K 9 6 2
H J 6
D Q J 5
C A J 4
Table S Q 8 5
H A Q 5
D A K 4 2
C K 10 5
Chih-Cheng Yeh
Taiwan

S 4 3
H 10 9 8 4
D 8 7 3
C Q 7 3 2

A different order of play is also shown to emphasize that many play sequences are possible on a single deal. Curiously, dummy wins seven tricks, the maximum possible, as each hand must win at least two tricks by definition.

The final deal utilizes an equal balance of two queens and two jacks for its 6 HCP. Call it a Dutch treat:

3 NT S J 10 6
H 7 6 5
D 6 5 4
C Q 10 3 2
Trick
1. W
2. N
3. N
4. W
5. W
6. S
7. E
8. E
9. S
10. S
11. S
12. N
13. N
Lead
S 3
S J
S 6
S 2!
D 7
D 10
D 2!
H 4
H Q
H 2!
C 5
C Q
C 2!
2nd
10
8
Q
D 4
5
Q
C 4
10
9
C A
8
J
D A
3rd
9
7
D 8
H A
3
6
H K
8
7
C 3
10
6
H 3
4th
5
4
K
D 9
J
K
H 5
6
J
C K
7
9
S A
S A K 4 3 2
H K 9 8
D Q 7
C A 9 8
Table S Q 9 8
H A J 4
D A K 3 2
C K J 7
Jacco Hop
Netherlands

S 7 5
H Q 10 3 2
D J 10 9 8
C 6 5 4

Well, there you have it, folks. If anyone can produce a deal with 5 HCP, I’d be most impressed; but all my attempts indicate it’s impossible. Warning! Leave now before this all sinks in, else nobody will be able to save you.

Richard Morse: Most enjoyable challenge, and slightly easier than the beer one.

See what happens when you ignore warnings?

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Acknowledgments to The Beach Boys and their 1963 song hit.
© 2010 Richard Pavlicek