Puzzle 8F17   Main

# Little Deuce Coupe

by Richard Pavlicek

### “You don’t know what I got…”

Anyone can win aces. Novices soon learn to win picture cards. Competent players become adept at winning spot cards. The little deuce, however, is in a special coupe, I mean group. Except for the case of ruffing with the deuce of trumps, a deuce can win a trick only if led, and the other three hands are out of that suit.

How many tricks can be won by deuces on a single deal? Three seems the logical answer, and is true in suit contracts; but in notrump the answer is four. Determining how this might happen is the underlying theme of my puzzle:

Construct a deal where South makes 3 NT, and all four deuces win a trick.
West must have five spades, and no other hand has a suit over four cards.

Note that I did not say against best defense. All you need is 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 to achieve the goal.

Multiple solutions exist. A further goal (tiebreaker for the October contest) is to give North-South the fewest HCP.

## Zhi Bang Lim Wins

In October 2010 this puzzle was presented as a contest, ending appropriately on Halloween. Thanks to the 55 brave souls who submitted solutions, of which 14 were correct. Surprisingly, all but one of the correct solutions produced 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.

Congratulations to Zhi Bang Lim (Malaysia) who was first to submit the optimal solution. If he continues this habit in the future, I will have to rename him — from the words of Ralph Kramden, if anyone remembers — to Bang Zoom Lim.

The top 13 places are ranked by date-time of entry.

Winner List
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

 Puzzle 8F17   Main Top   Little Deuce Coupe

## Solution

Before showing any deals, it might be a good idea 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 requirements that West has five spades, and no other hand has more than four in any suit, it is apparent that West must have the 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, the defensive deuces must be won with the absolute minimal trick loss: One trick to gain 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 2 is at trick four, after gaining the lead at trick three.

Enough theory. Let’s get the cards on the table!

### Deuces never loses

Below is the clever solution from our winner, Zhi Bang Lim (Malaysia):

 3 NT South Q 9 6 Q 9 6 2 5 4 3 5 4 3 Trick1 W2 N3 N4 W5 W6 N7 N8 N9 S10 S11 S12 S13 E Lead 8 Q 6 2! 7 Q 2! 5 Q 2! 10 7 2! 2nd910 A 39J K7J A8J 6 3rd745 K85 694 345 A 4th3JK 3410 A810 K9Q 6 W-LW1W2L1L2W3W4W5W6W7W8W9L3L4 A K J 8 2 10 7 A J 8 A J 8 10 7 A K J 8 K 10 7 K Q 9 2 Lead: 8 5 4 3 5 4 3 Q 9 6 2 10 7 6

Zhi Bang’s deal not only has the minimal HCP but also the minimal spot cards for 3 NT to be made. (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 soon become a candidate for the looney bin.

### Do I hear a cash register?

Chih-Cheng Yeh (Taiwan) composed the 6 HCP of a K-Q-J instead of three queens:

 3 NT South J 10 7 K 7 3 2 10 9 6 9 8 6 Trick1 W2 N3 N4 N5 N6 N7 W8 W9 N10 N11 S12 S13 E Lead 6 J 7 10 9 7 2! J 2! 8 2! 8 2! 2nd105545Q 6K K10 JQ 7 3rd84433 8 AQ 7Q 69 A 4th39654K 910 AJ KA 3 W-LW1W2W3W4W5L1L2W6W7W8W9L3L4 A K 9 6 2 J 6 Q J 5 A J 4 Q 8 5 A Q 5 A K 4 2 K 10 5 Lead: 6 4 3 10 9 8 4 8 7 3 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.

### Dutch treat

Jacco Hop (Netherlands) shows another way to compose 6 HCP with two queens and two jacks:

 3 NT South J 10 6 7 6 5 6 5 4 Q 10 3 2 Trick1 W2 N3 N4 W5 W6 S7 E8 E9 S10 S11 S12 N13 N Lead 3 J 6 2! 7 10 2! 4 Q 2! 5 Q 2! 2nd108Q 45Q 4109 A8J A 3rd97 8 A36 K87 3106 3 4th54K 9JK 56J K79 A W-LW1W2L1L2W3L3L4W4W5W6W7W8W9 A K 4 3 2 K 9 8 Q 7 A 9 8 Q 9 8 A J 4 A K 3 2 K J 7 Lead: 3 7 5 Q 10 3 2 J 10 9 8 6 5 4

So there you have it, folks. The Little Deuce Coupe with “four on the floor” — deuces, that is. If anyone can produce a deal with fewer than 6 HCP for N-S, I’d be most impressed; but all my attempts indicate it’s impossible.