Puzzle 8M13   Main

Diamonds in Distress

by Richard Pavlicek

The diamond suit gets more than its fair share of accidental contracts, because artificial use in many contexts causes misunderstandings. Case in point: The multi 2  opening, showing a weak two-bid in either major, is firmly ingrained in Europe but new to the American scene. As the following deal illustrates, our mastery of this convention is a long way off, though results are promising — alas, not for bridge but for puzzle material.

6 South Both Vul	A K 6 3 K A Q 9 A Q 4 3 2			West Pass Pass Pass Pass	North 2 NT 4 NT 6	East Pass Pass Pass	SOUTH 2 3 5 Pass
Q J 10 10 6 4 K J 10 8 J 8 5			9 7 5 J 7 5 7 6 5 4 10 9 7
	8 4 2 A Q 9 8 3 2 3 2 K 6

South opened a multi 2 and North responded 2 NT, an artificial inquiry to clarify South’s hand; 3 then showed a maximum with hearts (3  would show spades, and three of a major would be natural and minimum). Oops! North forgot they were playing multi and interpreted 3  as a feature with a natural weak 2  bid; hence key-card Blackwood to the hopeless slam in diamonds. Well, almost hopeless.

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

Declarer took the first nine tricks with top cards, finessing diamonds en route (West perforce splitting) to reach the following ending:

win 3	6		Trick	Lead	2nd	3rd	4th
	—		10 N	6	9	3	J
	A 9		11 S	3	J	4!	5
	4		West is endplayed
J		9
—		—
K J 8		7 6 5
—		—
	—
	9 8 3
	3
North leads	—

A spade was ruffed in hand as both opponents helplessly followed, then a heart achieved a trump promotion: West chose to ruff high, so declarer pitched the club and won the last two tricks with a finesse. Believe it or not!

(Smart money is on not.)

The above deal shows the weakest trump suit to make a slam against any defense. As verification, note that if North-South held only four trumps, one opponent must have at least five, so at most eight side tricks could be won, which requires declarer to win four trump tricks. The only way to accomplish this is to have at least A-K-J-7* in one hand, which adds to 45. The above case (A-Q-9-3-2) adds to 40. Trying to reduce the strength of five trumps is also futile, though an alternate arrangement of A-Q-9-3 opposite 2 is possible.

*Thanks to Martin Sinot, Netherlands. I had original written A-K-J-9, but A-K-J-7 is sufficient if West has 6-5-4-3-2 and East has Q-10-9-8. Declarer wins eight side tricks, ruffs with the seven as West perforce underruffs, then exits to score A-K-J.

Perhaps more widely known is the weakest trump suit to make a grand slam (say 7 ), which is A-J-9 opposite K-2, adding to 49. Each opponent is 3=3=4=3 with Q-10 onside; so regardless of the lead declarer can cash nine side winners, finesse the 9, and crossruff the last three tricks. Alternately, trumps can be A-K-J-9 opposite 2, allowing a finesse and a coup.

Now, for the current puzzle:

What is the weakest diamond suit to make 5 against any defense?

Weakness (strength) is determined by the sum of card ranks (ace = 14, king = 13, queen = 12, jack = 11, ten = 10, etc.). Further goals (tiebreakers for the June contest) are for all North-South cards to be as low as possible, and secondarily for the North and South sums to be as close to equal as possible.

Tim Broeken Wins

In June 2015 this puzzle was presented as a contest, inviting anyone who wished to submit a solution. Interest took an upturn this month with 44 participants. Go figure; last month a normal 3 NT draws all of 19, and now a ridiculous 5  more than doubles the turnout. Could this reflect the status of bridge? Only 26 of the 44 entries were valid (5  makable against any defense) of which only the 16 listed below achieved a trump-suit total of 35 or less.

Congratulations to Tim Broeken, Netherlands, who was the first of four to submit the maximal, or more appropriately, minimal construction for both tie-breakers. He’s back in the saddle again… This is Tim’s seventh puzzle-contest win, far and away my leading solver.

Winner List

Rank	Name	Location	Diamonds	N-S Sum	Disparity
1	Tim Broeken	Netherlands	32	178	0
2	Tom Slater	England	32	178	0
3	Dan Gheorghiu	British Columbia	32	178	0
4	Grant Peacock	Maryland	32	178	0
5	Adam Dickinson	Scotland	32	182	0
6	Nicholas Greer	England	32	182	0
7	Jamie Pearson	Ontario	32	182	0
8	Dan Baker	Texas	32	184	0
9	Xavier Dantan	France	32	184	8
10	Charles Blair	Illinois	32	184	20
11	Dean Pokorny	Croatia	32	186	0
12	Leigh Matheson	Australia	33	175	7
13	David Kenward	England	33	179	1
14	Wayne Somerville	Northern Ireland	33	192	0
15	Julien Reichert	France	34	182	6
16	John Portwood	England	35	196	0

Puzzle 8M13   Main

Top   Diamonds in Distress

Solution

There are various ways to define the “weakest” trump suit, though for this contest it was stated to be the lowest rank sum, regardless of the number of cards. For curiosity sake, the fewest number of trumps required to make 5  is only three, which can be as weak as A-Q-9, as John Portwood showed in the following deal:

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

After any plain-suit lead, declarer can reach this ending:

win 3	5		Trick	Lead	2nd	3rd	4th
	—		9 N	5	J	8	2
	—		10 E	7	9	3	4
	7 6 5 4		11 S	5	4	5	8
—		—	Declarer succeeds
—		—
6 5 4 3 2		K J 10 8 7
—		—
	—
	5
	A Q 9
North leads	8

North leads a spade, and East must ruff with the 10 or jack (else South scores the 9 and exits to win two more trumps); South discards. The forced diamond return is won cheaply, then South exits with his last plain-suit card to win the last two. The trump promotion is even easier after an original diamond lead, needing to endplay East only once.

Another possible definition of “weakest” is fewest trumps in one hand, for which two opposite two (four trumps total) can be achieved with surprisingly low ranks. I’ll leave the minimal construction up to the (crazed) reader, but this solver is right on target:

Tom Slater: The weakest four-card trump suit I can manage is A-J opposite 8-2.

Lowest rank sum

All that mattered for this contest was the sum of card ranks, no matter how many trumps were held or how they were divided. The above solutions (A-Q-9 and A-J-8-2) each add to 35, which is well off the mark. Various improvements are possible (as shown in the Winner List) of which the lowest is 32, or specifically K-10-4-3-2. Nicholas Greer demonstrates this with a N-S total of 182, perfectly split at 91 each.

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

Nicholas Greer: This trump suit can be held to two losers by leading a plain suit after cashing all the side tricks, having used three spade entries and the K to take all four rounded-suit finesses.

Diamonds must be led once, forcing West to split his honors to force the king, though the exact order of play can be varied to reach this ending:

win 1	5		Trick	Lead	2nd	3rd	4th
	—		11 S	2	Q	5	6
	10 4		Declarer succeeds
	—
—		—
—		—
A Q 8		9 7 6
—		—
	—
	2
	3
South leads	3

A heart or club lead then promotes the 10 into an 11th trick. If West chose to win the A early and return the queen, the 10 is still promoted but in a two-card ending.

The ultimate solution

Not surprisingly, the top solvers found a way to reduce the N-S rank sum. Rather than require the 10 for finessing in each side suit, a ruff enables a third trick in one. This requires the N-S trumps to be divided 4-1, else two rounds of trumps would be fatal. The following deal submitted by Tim Broeken has the lowest possible N-S sum of 178, split evenly 89-89.

5 South	5 4 3 2		Trick	Lead	2nd	3rd	4th
	A Q 10		1 W	J	4	7	Q
	K 10 4 3		2 S	2	Q	K	5
	5 4		3 N	5	8	A	9
8 7 6		K J 9	4 S	2	10	3	K
K J 9		8 7 6	5 N	2	9	10	6
A Q J 9		8 7 6 5	6 S	2	9	10	6
J 10 9		K 8 7	7 N	3	J	Q	7
	A Q 10		8 S	3	J	Q	7
	5 4 3 2		9 N	A	8	4	K
	2		10 N	4	K	A	8
Lead: J	A Q 6 3 2		11 S	6
			Declarer succeeds

Tom Slater: …Ruffing in the hand with longer trumps removes the need for the 10.

Dan Gheorghiu: There are eight side tricks by finessing everything, the K by leading a diamond from South, one club ruff in North, and the 10 is promoted at Trick 11 or 12.

Grant Peacock: South recently talked his partner into playing Roman 2 and is eager to try it out.

Mathematical Proof

Charles Blair: I wonder whether the answer would be different if you used a sum from ace = 13 to two = 1, which, I think, is a more natural summation by rank.

Only to Charles would “two = 1” be natural. The only difference I see is a reduction from 32 to 27, but with acknowledgments to Professor Irwin Corey, the answer three cubed would be a blunder of exponential proportion. Clearly, the weakest ruffing* suit to lose only two tricks in five diamonds must be two to the fifth power. Case closed.

*Due to derogatory remarks against the Mexican people, the use of ‘trump’ suit is henceforth abolished.

Leigh Matheson: It’s a diamond heist and they can’t stop it!

Puzzle 8M13   Main

Top   Diamonds in Distress

© 2015 Richard Pavlicek