Puzzle 8M01   Main

Fail Safe

by Richard Pavlicek

One of the goals of the opening leader is not to lose a trick in the suit led. Often this dictates leading from a safe holding such as an honor sequence or worthless suit. Alas, like the haunting conclusion of the Sydney Lumet film, a fail-safe approach can backfire. Almost anything short of a completely solid sequence can fail, as shown by the following deal, on which South opened 1 NT and North raised to game.

3 NT South	10 5 2 A Q 8 2 A 10 8 6 5 4			Trick 1 W 2 S 3 E 4 E 5 S 6 E 7 W 8 S 9 N 10 S 11 N 12 S	Lead J 3 K Q 8 3 K 4 8 6 10 4	2nd 2 2 7 A 2 6 10 3 6 10 7 7	3rd 3 4 10 4 2 J 9 A K Q Q 8!	4th K 9 5 6 J 5 A 5 J 5 8	W-L W1 L1 L2 W2 L3 L4 W3 W4 W5 W6 W7
K J 8 4 J 10 9 7 J 3 2 10 2			Q 9 3 5 3 9 7 6 5 K Q J 9
Lead: J	A 7 6 K 6 4 K Q 4 A 8 7 3

Rather than risk leading from spade honors or a dangerous minor suit, West chose the fail-safe lead of a heart, as would any expert. Safe? Hardly; in fact it’s the only lead to let 3 NT make. Declarer won in hand, but before committing in hearts did some radar reconnaissance in the other suits; plus there was a chance he could establish a club trick. When it came to crunch time, nothing unexpected had shown up on the radar, so declarer double-finessed in hearts for his ninth trick — or would have if West didn’t prolong the play by splitting.

Along similar lines in an old puzzle I posed: Against 3 NT you lead the nine from 9-8-7-5, which lets declarer win an extra spade trick. Assuming no voids or entry problems, what is dummy’s spade holding? Curiously, only Q-J-10-6 (declarer having A-x-x) prevents the lead from being safe. Note that K-J-10-6 facing A-x-x is not a valid solution, because declarer can win four tricks on his own by felling the queen (double-dummy play is presumed for puzzle purposes); nor is Q-J-10-6-x opp. A-x, because declarer cannot profit with just one card to lead from hand.

This brings me to the current puzzle, which is a two-parter. Against 3 NT West leads (1) high from 10-9-8-4, and (2) low from 8-5-4-3. Assuming declarer has plenty of entries, and there are no singletons or voids, for each case:

Construct a suit layout where the lead allows declarer to win an extra trick in that suit.

A further goal is to make the North holding in each case as strong as possible (counting ace = 14, king = 13, … two = 2). Give it a try before reading the best solutions below.

Tim Broeken Wins Again

In April 2015 this puzzle was presented as a contest, inviting anyone who wished to submit a solution. Thirty-nine people gave it a try, of which only the 18 listed constructed a valid suit layout for each part. Thanks to all who participated, but I’ll direct my “well done” to Europe and Canada, since I can’t remember a past contest without a U.S. player in the top 10.

Congratulations to Tim Broeken, Netherlands, for achieving back-to-back wins and a record sixth time atop the leaderboard. Tim and fellow countryman Jacco Hop were the only solvers to submit the optimal solution (strongest North hand) for both parts — and their entries came near the end of the month, so there was plenty of opportunity for anyone to win.

Winner List

Rank	Name	Location	Part 1	Part 2	Total
1	Tim Broeken	Netherlands	41	34	75
2	Jacco Hop	Netherlands	41	34	75
3	Dan Gheorghiu	British Columbia	41	33	74
4	Gareth Birdsall	England	41	28	69
5	Nicholas Greer	England	41	28	69
6	Jamie Pearson	Ontario	41	28	69
7	Leigh Matheson	Australia	41	28	69
8	Jonathan Mestel	England	36	33	69
9	Dean Pokorny	Croatia	41	27	68
10	Peter Nixon	British Columbia	41	26	67
11	Nate Munger	California	36	28	64
12	Grant Peacock	Maryland	36	28	64
13	Xavier Dantan	France	36	28	64
14	Christina Syrakopoulou	Greece	36	27	63
15	Tom Slater	England	30	28	58
16	Adam Dickinson	Scotland	30	28	58
17	Radu Vasilescu	Pennsylvania	30	27	57
18	Jim Munday	Mississippi	30	26	56

Puzzle 8M01   Main

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Solution

No safety in spades

In Part 1, West holds 10-9-8-4 and leads the 10. For this to cost a trick, West’s holding must become finessable after the lead, but not if declarer must breach the suit on his own. Hence a North holding of K-Q-J-7 facing x-x-x is not a solution, because declarer can triple-finesse the seven by himself — albeit double-dummy, but that’s presumed for these puzzles.

The most common solutions involve Q-J-x-(x) in North, with the king East and the ace South. Maximizing North’s length and spot cards for tie-breaking advantage yields the following layout, submitted by four solvers:

3 NT	Q J 7 6		Trick 1 W 2 S 3 N	Lead 10 5 7	2nd J 8	3rd K Q	4th A 2
10 9 8 4		K 3 2
West leads	A 5

If declarer has to play the suit himself, only two tricks can be made, since West does not have to burn a spot if the queen is led (covered). Note that giving South A-x-x and East K-x is not a valid solution, because declarer could win three tricks on his own by cashing the ace and ducking the second round. Five respondents missed out by overlooking this.

Grant Peacock: I wanted to compete for the rule-interpretation award with K-Q-7-6 opposite J-5, with which declarer can gain a trick after the 10 lead if West can be thrown in in another suit; but your “old puzzle” explanation seems to invalidate this.

Wise decision, as I jump at the chance to disqualify members of the feathered persuasion (Gareth Birdsall, be warned too).

Grant was on the right track, however, because a valid layout does exist with North having K-Q, and the jack as well for a tie-breaking bonanza. Nine solvers, all in the top 10, discovered it:

3 NT	K Q J 5		Trick 1 W 2 E 3 S 4 N 5 S	Lead 10 - 7! - 3	2nd J + 9 - 4	3rd A - Q + 5	4th 2 - 6 -
10 9 8 4		A 6
West leads	7 3 2

Tricks 2 and 4 in the play diagram indicate leads in a different suit, first to force declarer to play the spade suit himself, and then for declarer to lead a spade from the necessary hand.

Dan Gheorghiu: Declarer wins three tricks with the 10 lead, but only two if he has to play spades himself.

Leigh Matheson: The 10 is covered and won by East. Leading the 7 then pins the six, double finessing West for the rest.

Jamie Pearson: After the 10 lead, declarer covers then can finesse twice against West for three tricks… Without the 10 lead,…any lead from South will be covered as cheaply as possible by West, then East’s six [spoils any] finessing attempt.

Part 1 had 19 valid solutions: 11 with Q-J-x opp. A-x-x-x, seven with Q-J-x-x opp. A-x, and the above layout which is unique (not a single spot card can be changed).

Eight, skate and donate

Losing a trick when leading from an honor sequence is painful enough, but in Part 2 we now have to worry about leading from an eight spot. Holding 8-5-4-3, leading the three must give declarer a trick that he could not obtain by himself. While this may seem unlikely, a variety of layouts exist, most of which require 4-3-3-3 distribution. The popular choice was the following, found by nine solvers:

3 NT	Q 9 7		Trick 1 W 2 S 3 N	Lead 3 - Q	2nd 7 -	3rd 10 +	4th J -
8 5 4 3		K 10 2
West leads	A J 6

Jamie Pearson: After the 3 lead, North plays the seven, South covering if necessary; East is then finessed for all three tricks. Without the 3 lead, East will cover the queen or nine on any finessing attempt, with West’s eight stopping declarer from taking three tricks.

Leigh Matheson: The free finesse of the 7 allows the suit to be picked up without loss.

Tom Slater: This doesn’t look like it will be the best solution, but it’s hard enough to get North and South the right way around this early in the morning.

Yes, I catch the drift from last month, and your harbinger of a better solution is on the mark.

Only three solvers were able to fit four cards into the North hand. Here is one of them:

3 NT	Q 10 9 2		Trick 1 W	Lead 3	2nd 2	3rd	4th
8 5 4 3		K J 6
West leads	A 7

Dan Gheorghiu: Declarer wins three tricks with the 3 lead, but only two if he has to play hearts himself.

It is easy to see how three tricks are won after West leads — just duck in dummy — but it takes a closer look to see why declarer cannot do this on his own. If declarer starts by leading the 10, East covers (jack, ace); then West covers the seven (eight, nine, king) leaving West the fourth-round boss. Note that North’s spot must be the two, else declarer could always win three tricks.

The above can be improved a notch by juxtaposing the top three honors. Our winner and runner-up were right on target with this optimal layout (North sum 34).

3 NT	K 10 9 2		Trick 1 W	Lead 3	2nd 2	3rd	4th
8 5 4 3		A J 6
West leads	Q 7

The play is essentially the same; declarer can win only two tricks on his own, but the lead avails three. It is also worth noting that ducking in dummy is the proper play in practice (not double-dummy) because it gains against A-8-x-x or 8-x-x-x, while declarer cannot win three tricks (barring an endplay) against any other four-card holding.

Jacco Hop: The next time partner complains that you finessed him out of an honor, show him these examples of how you could have finessed him out of a six-spot.

Part 2 had 24 valid solutions: four each with Q-9-x opp. A-J-x, J-9-x opp. A-K-x, J-9-x opp. A-Q-x, 10-9-x opp. A-K-x, and 10-9-x opp. A-Q-x; plus the unique cases of J-10-9-2 opp. A-7, Q-10-9-2 opp. K-7, and the two preceding layouts.

Off the Radar

Jonathan Mestel: I usually find it much easier to blow a trick when I’m on lead.

Peter Nixon: Bobby Fischer said, “If you find a good move, look for something better.” You are asking me to find a bad lead and then look for something worse. My leads are bad enough as it is, so I’m going to settle for mediocrity.

Well, with 10th place you achieved your goal. Alas, your tricky namesake wasn’t so lucky:

Richard Nixon: Why am I not on the leaderboard? I had both winning solutions recorded, until Pat left her friggin’ cigarette on my desk in the Oval Office, and a fire destroyed the tapes. Trust me on this. I am not a crook! And I’d make a damn good Pope, too!

Puzzle 8M01   Main

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© 2015 Richard Pavlicek