Puzzle 8M25 Main

Third Best Blues

by Richard Pavlicek

“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 either. Besides showing count, a third-best lead will often clarify high-card locations to partner. Consider the following deal:

	9 7 4		West	North	East	South
None vul	8 5 4					1
	K 10 6 2		Pass	2	Pass	4
	A 5 4		Pass	Pass	Pass
J 3		10 8 5
Q 7 3		J 10 9 6 2
Q 9 7 3		A J 8
Q 9 7 3		J 8
	A K Q 6 2
	A K
	5 4
4 South	K 10 6 2

Suppose West leads the 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 K as his 10th trick.

Using standard fourth-best leads, West would lead the 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 occasionally comes back to haunt. Suppose West leads a club instead. The standard 3 lead is fine, as the jack forces the king, and declarer is routinely set. But if West leads the 7, declarer captures the jack and succeeds by later leading the 10 to force a cover, establishing his club spot. Ouch!

Win some, lose some — which brings me to the puzzle:

West must lead from (1) K-9-5-3, and (2) Q-10-4-2. For each case:

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

Assume only declarer will lead the suit thereafter, with adequate entries to lead from either hand to best advantage. Further, since tricks can only be won by the suit itself (not by ruffing) assume the contract is notrump.

A secondary goal (contest tiebreaker) is to make the sum of East’s card ranks as high as possible.

Dan Gheorghiu Wins

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, I think 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.

Winner List

Rank	Name	Location	Part 1	Part 2	Total
1	Dan Gheorghiu	British Columbia	33	29	62
2	Grant Peacock	Maryland	33	29	62
3	Tim Broeken	Netherlands	32	29	61
4	Charles Blair	Illinois	32	29	61
5	Jonathan Mestel	England	32	29	61
6	Jamie Pearson	Ontario	29	29	58
7	Foster Tom	British Columbia	29	29	58
8	Adam Dickinson	Scotland	22	28	50
9	Leigh Matheson	Australia	24	24	48
10	Richard Stein	California	20	28	48
11	Gareth Birdsall	England	19	24	43
12	Nicholas Greer	England	18	24	42
13	Jim Munday	Mississippi	17	24	41

Puzzle 8M25 Main

Top Third Best Blues

Solution

Part 1 (diamond suit)

If West leads from K-9-5-3, conditions require that leading the five (third-best) must lose a trick versus leading the three (fourth-best). Three solvers produced the following valid layout, which gives East a hefty tie-breaking sum of 32.

NT South	10 6
K 9 5 3		A J 7
West leads	Q 8 4 2

In practice East would usually return the jack at trick two, which changes nothing, but conditions state that only declarer will lead the suit thereafter, so assume East exits in another suit. When declarer leads the 10, it is covered all around to West, who also exits in another suit. Declarer then can lead the 8 to force the nine and crush 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 upped a notch to 33 with an entirely different layout, giving North the length instead of South. Only two solvers discovered this optimal construction:

NT South	J 10 4 2
K 9 5 3		A Q 7
West leads	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  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 many to list, though most were just slight variations (e.g., East A-Q-6 and South 8-7). The most common case was East holding the queen (similar to the winning layout of Part 2). Curiously, East can even have a singleton (e.g., North A-Q-J-4-2, East 10, South 8-7-6) albeit hopeless for tie-breaking purposes.

Part 2 (club suit)

Bring on the puppy feet! The club suit ( Q-10-4-2) is more difficult, as rarely does a four-spot contain rank value, yet wasting that card must cost West a trick. Declarer or dummy clearly must have the three, which is eventually promoted, but even with this knowledge a valid 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.

NT South	A 9 8 3
Q 10 4 2		J 7 6
West leads	K 5

Nicholas Greer: Only the 4 lead blows a trick. It is covered all around, then the 5 must be covered by West to force the ace. Next leading the 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:

NT South	6 5
Q 10 4 2		K 9 7
West leads	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 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 swappable) 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 swappable).

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. A wise move. If I understand this correctly, in future ACBL women’s events all hands must be biddable. No more dogs! At least that’s more tasteful than the female dog explanation.

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

No, it was clueless… which brings us to American politics:

The Donald: Sorry, but I object to the conditions, just like I object to the renaming of Mount McKinley to Denali.* Who leads third best? I am the best, and I lead the same way — from the top — and if elected I’ll build a wall around everyone who thinks otherwise.

*but of course… No doubt he prefers Trump Tower II.

Puzzle 8M25 Main

Top Third Best Blues

© 2015 Richard Pavlicek