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Third Best Blues

“Low from odd, third from even” is a popular leading strategy against suit contracts, particularly among U.S. experts. The main purpose is to distinguish between three and four cards, e.g., Q-7-3 versus Q-9-7-3, since standard leads dictate the same card from both. Besides showing count, a third-best lead will often clarify high-card locations to partner. Consider the following deal:

4 S South S 9 7 4
H 8 5 4
D K 10 6 2
C A 5 4
N-S Vul

West

Pass
All Pass


North

2 S


East

Pass


South
1 S
4 S
S J 3
H Q 7 3
D Q 9 7 3
C Q 9 7 3
Table S 10 8 5
H J 10 9 6 2
D A J 8
C J 8
Lead: D 7 S A K Q 6 2
H A K
D 5 4
C K 10 6 2

Suppose West leads the D 7 and declarer plays low from dummy. East can deduce to finesse the eight, which will win if the lead is third-best, and won’t matter (versus the jack) while retaining communication if partner has led from a doubleton. If East wastefully won the jack, declarer could later finesse the 10 to establish the D K as his 10th trick.

Using standard fourth-best leads, West would lead the D 3, and East would have an insoluble problem. Finessing the eight would work as the cards lie, but it could just as easily cost a trick when South has 9-x.

While “third and low” is generally superior to fourth-best, it will occasionally come back to haunt. Suppose West chose to lead a club instead. The standard lead of the C 3 is fine, as the jack forces the king, and declarer is routinely set. But if West leads the C 7, declarer can succeed by capturing the jack and leading the C 10 to force West to cover, then the club spots establish a trick after trumps are drawn. Ouch!

Win some, lose some.

This brings me to the puzzle, which is a two-parter. West must lead from (1) D K-9-5-3, and (2) C Q-10-4-2. Assuming only declarer will lead the suit subsequently (from either hand) with plenty of entries, for each case:

Construct a suit layout where leading third-best loses a trick versus leading fourth-best.

A further goal is to make the sum of East’s card ranks as high as possible.

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Dan Gheorghiu Wins!

In August 2015 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Thirty-one persons gave it a try, but only the 13 listed created valid layouts for both parts. Thanks to all who participated.

Congratulations to Dan Gheorghiu, who was the first of only two solvers to find both optimal layouts (East having the highest card sum). Dan is on a roll, providing maximal solutions every month since winning The Case of the Four Aces back in May. Or to paraphrase Henry Higgins: By Gheorgh, he’s got it! Also finding both best layouts was Grant Peacock, winner of Ever More in February and [checking my archives] Keep the Ship Afloat more than 13 years ago.

RankNameLocationPart 1Part 2Total
1Dan GheorghiuBritish Columbia332962
2Grant PeacockMaryland332962
3Tim BroekenNetherlands322961
4Charles BlairIllinois322961
5Jonathan MestelEngland322961
6Jamie PearsonOntario292958
7Foster TomBritish Columbia292958
8Adam DickinsonScotland222850
9Leigh MathesonAustralia242448
10Richard SteinCalifornia202848
11Gareth BirdsallEngland192443
12Nicholas GreerEngland182442
13Jim MundayMississippi172441

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Solution

First let’s consider Part 1 where West leads from D K-9-5-3, and the third-best five must lose a trick versus the fourth-best three. Three solvers produced the following valid layout, which gives East a hefty tie-breaking sum of 32.

3 NT D 10 6 Leader
1. W
2. E
3. N
4. W
5. S
Lead
D 5
-
D 10
-
D 8
2nd
6
-
J
-
9
3rd
A
-
Q
-
.
4th
2
+
K
+
7
D K 9 5 3 Table D A J 7
D Q 8 4 2

In practice East would usually return the jack at trick two, which changes nothing, but conditions stated that only declarer will lead the suit subsequently, so assume East exits in another suit — the invisible Trick 2 won by the North (plus sign). The D 10 is then led, covered around to West, who also exits safely. Declarer next leads the D 8 to force the nine and squash the seven, establishing the four. This would not have been possible if West had led the three originally.

While the above solution is excellent, East’s rank sum can be improved a notch to 33 with an entirely different layout, giving North the length instead of South. Only two solvers discovered this optimal construction:

3 NT D J 10 4 2 Leader
1. W
2. E
3. S
4. E
5. N
Lead
D 5
-
D 8
-
D J
2nd
2
+
9
-
7
3rd
A
-
10
-
.
4th
6
-
Q
+
K
D K 9 5 3 Table D A Q 7
D 8 6

As before, the play involves establishing the four after West has wasted the five. East must win an honor at Trick 1 lest South win the eight, then exits safely. The D 8 is covered around, and East exits safely again. Finally, the jack squashes the seven to establish the four.

Dan Gheorghiu: East-West win four tricks when the lead is fourth-best, but only three tricks when the lead is third-best.

Dan (and Nicholas Greer) also pointed out that if North held A-10-4-2 and South Q-8-6, the five would be the only card to blow the entire suit; i.e., West could lead the king or nine, as well as the three, and eventually score a trick.

Part 1 had 53 valid solutions, too numerous to list, though many were just slight variations (e.g., East having A-Q-6 and South 8-7). The most common case was East having the queen (similar to the winning layout of Part 2). East can even have a singleton (e.g., North A-Q-J-4-2, East 10, South 8-7-6) albeit the worst for tie-breaking purpose.

Bring on the puppy feet

Part 2 was more difficult. Rarely does a four-spot contain rank value, but wasting that card from C Q-10-4-2 must cost West a trick. Declarer or dummy clearly must have the three, which eventually is promoted, but even so the precise construction is hardly obvious. Essentially there are two feasible layouts (each with slight variations), the first of which gives East the jack. Four solvers found it.

3 NT C A 9 8 3 Leader
1. W
2. S
3. N
Lead
C 4
C 5
C 9
2nd
8
10
7
3rd
J
A
.
4th
K
6
Q
C Q 10 4 2 Table C J 7 6
C K 5

Nicholas Greer: Only the C 4 lead blows a trick. It is covered all around, then the C 5 must be covered by West to force the ace. Next leading the C 9 establishes the three. Do I get extra credit for third-best being the only lead to cost a trick?

Nicholas alludes to the fact that West could safely lead the queen or 10, as well as the two, after which declarer cannot negotiate a third trick with any play. Somehow, leading “third best” is beginning to look like the maiden voyage of the Titanic.

Also worth noting is that with dummy in view (even A-9-8-7) the proper lead is the queen if you expect South to have a doubleton. Besides catering to K-x, this also impedes development of a trick when South has J-x by requiring an additional entry to dummy.

The above layout was not a boon in the tie-breaking department, as the sum of East’s ranks is only 24. This can be improved to 29 by giving East the king in the following optimal solution:

3 NT C 6 5 Leader
1. W
2. S
3. N
4. W
5. S
Lead
C 4
-
C 6
-
C J
2nd
5
-
7
-
Q
3rd
K
+
8
-
.
4th
A
-
10
+
9
C Q 10 4 2 Table C K 9 7
C A J 8 3

Jamie Pearson: East plays high to force the ace. After leading from dummy and covering East’s card, declarer next leads high to establish the three. Leading fourth-best would preserve the four to beat South’s three.

Foster Tom: The third-best lead gives declarer a second trick. East plays high to force the ace, then declarer finesses a middle card on the second round. Leading the second middle card next smothers East’s spot, and finally the C 3 beats the two.

Part 2 had 13 valid solutions: four with East having J-7-6 or J-7-5 (North-South A-K switchable) and nine with East having K-9-7, K-9-6, K-9-5, K-8-7, K-8-6 or K-8-5 (N-S spots sometimes switchable).

Dog day blues  

In case nobody noticed, August 31 was National Dog Day, recently changed from August 26 to avoid conflicting with Women’s Equality Day. Wise move! If I understand this correctly, in future ACBL women’s events all hands must be biddable. No more dogs.

Charles Blair: I wonder whether the blue dog is a clue.

No, completely clueless… but thanks for the segue to U.S. politicians:

Donald Trump: Sorry, I abstain from notrump problems. Now if you were to make clubs or diamonds trumps, I have the perfect solution: Build a wall around dummy and make declarer pay for it. No way he could finesse me then.

And if elected, we anxiously await the renaming of Denali to Trump Tower II.

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© 2015 Richard Pavlicek