Puzzle 8K59 Main

Top of Nothing

by Richard Pavlicek

Last night at the club I was West as my opponents routinely bid their balanced hands to 3 NT. A club lead was unattractive, especially at matchpoints, and my majors bore omens of “never lead from a jack.” Therefore, I led top of nothing in diamonds. Argh! Turns out my lead let declarer win a diamond trick that he could not have won on his own. Oh, well.

I would have lost at poker, too! My full house (jacks over eights) would easily beat North but would have lost to South.

South deals	?		West	North	East	South
None vul	?					1 NT1
	?		Pass	3 NT	Pass	Pass
	?		Pass
J 7 4		?
J 6 4		?	1. 15-17
8 7 4		?
A J 8 6		?
	?
	?
Lead: 8	?
3 NT South	?

Now it’s time for you to do some detective work:

Construct a layout consistent with the auction and story.

Multiple solutions exist. A further goal is for North to have the worst poker hand, and South the best poker hand, in that order of priority. Whether the contract makes or not is irrelevant.

Tim Broeken Wins

Well, what can I say. Another record broeken, as Tim comes through with a fourth win in this series. Not only that, but he topped the construction I had in mind when creating this puzzle. Evidently, my plan to flood the dikes using kelp beds isn’t working, so it may be time to send Fritz on a Dutch bridge holiday — the sight of him alone could reduce bridge interest by 70 percent.

Winner List

Rank	Name	Location	North	South
1	Tim Broeken	Netherlands	3322A	TTTTA
2	David Brooks	Australia	3322A	5555A
3	Aurelien Boutin	France	3322A	5555A
4	Leigh Matheson	Australia	3322A	5555A
5	Dan Dang	British Columbia	3322A	KKKAA
6	Jim Munday	Mississippi	3322A	KKKQQ
7	Tony Norris	Massachusetts	6655A	TTTTA
8	Richard Stein	California	QQ33A	KKKAA

Puzzle 8K59 Main

Top Top of Nothing

Solution

When preparing the entry forms for this contest series, I try to keep the presentation brief so the page will redisplay in place, rather than require scrolling. Occasionally this leaves an ambiguity I didn’t foresee: “Let declarer win a diamond trick that he could not have won on his own” was intended to imply after any other plausible lead, but I didn’t say that. Some respondents assumed that if one or two of the other leads would stop the extra diamond trick, the solution was valid.

Rather than enforce my intention, I decided to compromise. If at least two of the three alternate leads ( 4, 4 and 6) would prevent the extra diamond trick, I accepted the solution as valid. But if only one of them would do it, the solution was… well, accepted for deposit in my new trash can.

Bridge Poker 101

Choosing the best poker hand from a bridge hand has some curious quirks, as many — excuse me, few — probably discovered with this puzzle, most notably in regard to the worst poker hand. I’ll yield the helm to “Diamond Jim” on the mighty Mississip’:

Jim Munday: Given that straights and flushes are legal, North must be 4-4-3-2 or 4-3-3-3 and have no fives or tens. Thus, two pair must be present at a minimum; twos and threes the worst. I could not construct a 10-HCP hand without an ace [kicker] or a higher two pair.

Jim correctly asserts that a bridge hand of “twos and threes” (two pair) cannot hold a five or a ten. For example, A-K-Q-J-9-8-7-5-4-3-3-2-2 might seem like two pair, but a closer look reveals a low-ball straight; replacing the five with a six makes it valid.

As to the best poker hand, the West cards allow only two straight flushes ( A-K-Q-J-10, K-Q-J-10-9) which are ruled out by the puzzle’s basic condition. Next best is four-of-a-kind, of which kings, queens, nines, threes and twos would ruin North’s worst hand, the primary goal. Therefore, South should have four tens, four fives, or a full house higher than West’s.

Bridge Poker Trivia

What is the highest poker hand that cannot be the best in a bridge hand? [Answer at end]

While lacking in the poker department, I liked the following construction because of its delicate diamond layout, and even more because Richard Stein explains it well. Now if he could just explain why he missed my Toughest Beer in Bridge fest, clearly in his honor.

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

Richard Stein: North and East cover the 8 as South wins. Spade up, then a club; Q-K-A. West exits in a major, dummy wins, then a club to the 10. West can win the J or duck; it doesn’t matter, but suppose he wins and returns the J to South’s ace to reach:

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

Richard Stein: Declarer runs his majors then endplays West in clubs. West is forced to break diamonds, which are frozen thanks to the 8 lead. Making four. This is declarer’s best line of play, I believe. On an original major lead, declarer can still make 10 tricks by stripping the majors early, but this is unlikely to happen in practice.

Richard’s concern about declarer making 10 tricks on another lead does not dilute the solution. All that matters in the puzzle is how many diamond tricks declarer can win, and only a diamond lead allows three. For example, on a major lead West can accept his loss in clubs, yielding no extra diamond trick.

Dan Dang also incorporated the frozen-suit endplay but with J-9-6 opposite A-K-3, and he improved on the poker hands. Ah, memories of Roger Miller, “Dang me… they oughta take a rope and hang me.”

Four fives down under

The next deal, submitted by Leigh Matheson (Australia), is essentially the solution I intended. Besides giving North the worst possible poker hand, it gives South four fives. The diamond lead costs a trick outright, allowing declarer to establish a spot by force.

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

The first two diamond leads are covered all around, and East exits safely with a heart. The  6 is then led to force out East’s remaining honor and smother the four. What does this leave? That’s right, the 3 is high. If West doesn’t cover the 5 at Trick 2, declarer simply rides it and smothers the seven later.

While only a diamond lead lets declarer score an extra diamond trick, 3 NT can be made on any lead. After a spade lead (best) South wins and leads the 9, ducking when West covers. Eventually, the Q will establish (losing three clubs). The defense has the option to prevent this by shifting to diamonds, which reverts to the original theme (or East can be endplayed if he leads the K upon winning the K).

Two others produced the same poker hands (deuces and treys opposite four fives) but with different diamond layouts. Aurelien Boutin had A-J-9-6 opposite Q-5, and David Brooks had J-9-6 opposite A-Q-5, each with a valid deal construction.

Forty miles of railroad track

I did not think four tens for South could be achieved (opposite the weakest two pair in North) because my attempts focused on card combinations, rather than tempo and entries. Therefore, the following optimal solution was not in the windmills of my mind — until it arrived from a broeken windmill in nether-nether land. Bravo!

3 NT South	A K Q		Trick	Lead	2nd	3rd	4th
	9 8 7		1. W	4	7	2	10
	J 6 3 2		2. S	5	4	A	2
	4 3 2		3. N	2	9	10	4
J 7 4		9 8 3 2	4. S	6	7	K	3
J 6 4		5 3 2	5. N	3	A	5	7
8 7 4		A Q 9	6. E	8	10	J	Q
A J 8 6		Q 7 5		Declarer fails
	10 6 5
	A K Q 10
	K 10 5
	K 10 9

Tim Broeken: Going for style points. [Besides a diamond], only the lead of the A costs a trick.

With a diamond lead, declarer has an easy time to develop three diamond tricks. But with any other lead the task is impossible, because the defense is a tempo ahead. Note that East will duck the first diamond (or cover the jack with the queen) then hop with the ace on the second round. This way, all the spade entries will be wiped out before declarer can unblock the K.

Trivia answer: The highest poker hand that cannot be the best in a bridge hand is three tens.
All triplets require 11 different card ranks, and if any is a five or ten, there must be a straight.

Puzzle 8K59 Main

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© 2011 Richard Pavlicek