Puzzle 8M19   Main

# Queens Around

by Richard Pavlicek

“Timothy! Come over here. I’ve got a hand to show you.”

The Professor had just completed his lecture at the Puzzlers Anonymous meeting, expounding his theory that queen leads, unlike other honors, yield a high degree of variance in trick production. Why? Because queens magnetically deflect bosons, causing extreme eccentricity in the quantum model.

[The Professor jots down a bridge hand on his trusty clipboard]
“Timothy, suppose Grover is declarer in six notrump, and it’s your lead.”

Timothy looks over briefly and asks, “What was the bidding?”

“With Grover at the table, it could be anywhere from lame to insane, but I’ll give you the auction — and I’ll even show you the dummy. [The Professor scribbles on his clipboard.] There you have it; all the kings on your left with neatly ordered spot cards.”

 6 NT South K 5 4 A K 6 5 4 K K 7 6 5 WestPassPassPass NORTH1 4 6 Pass EastPassPassPassPass South2 4 6 NT Q J 10 9 8 Q J 10 9 Q J 10 Q ?  ?  ?  ? ?  ?  ?  ?

“Wow, spooky. Am I supposed to guess which queen to lead?”

“No. You’d probably pick your longest suit, but it’s not a lead problem. I will even tell you the result, which supports my boson deflection theory. Generally, four queens induce a positive spike for one, while the other three are synchronous, analogous to a 3-1 suit break being more likely than 2-2. But here we have equilibrium: Two queens will set the contract and two will not, assuming best play all around.”

“Hmm…” Timothy ponders, “Grover is going nowhere in the majors, so he’ll need a bunch of club tricks; or he could have a long diamond suit instead; or maybe both. But what if the slam is cold with any lead? Wouldn’t that disprove your… whatchacallit, bozo theory?”

“No. [sigh] You weren’t paying attention to my lecture. The theory is probabilistic, like the Law of Total Quarks. Collisions occur when bosons become positively charged by the absorption of pions, causing trick production to stabilize asymptotically. Extreme charges (greater than 0.7425 electron volts) can even surrender an overtrick.”

“I see, um, whatever… so if I choose a queen at random, I have a fifty-fifty chance to set the contract.
‘Best play all around’ must mean a fill-in for Grover, as otherwise even a wrong queen should suffice.”

“Timothy, you’re beginning to see the light. Now see if you can solve the puzzle:”

Construct the missing hands so that exactly two queen leads will defeat 6 NT.

Multiple solutions exist. A further goal (tiebreaker for the July contest) is to make the East hand as strong as possible, judged by the sum of all card ranks (ace = 14, jack = 11, ten = 10, etc.).

## Dean Pokorny Wins

In July 2015 this puzzle was presented as a contest, inviting anyone who wished to submit a solution. Only 29 brave souls gave it a try, of which the 13 listed produced a layout where exactly two queen leads will defeat 6 NT.

Congratulations to Dean Pokorny, Croatia, who was the first of two solvers to submit the optimal solution of East having a rank sum of 96. Dean is a longtime successful participant, winning my Dead Man's Deal puzzle contest in 2011 and my Distribution Most Foul play contest way back in 2004. The other perfect solver was Dan Gheorghiu, who has maxed my last three puzzles, though only the first to submit the winning solution to The Case of the Four Aces.

Winner List
1Dean PokornyCroatia Q or Q96
2Dan GheorghiuBritish Columbia Q or Q96
3Nicholas GreerEngland Q or Q93
4Tim BroekenNetherlands Q or Q93
5Tom SlaterEngland Q or Q89
6Jon GreimanIllinois Q or Q84
7Jamie PearsonOntario Q or Q82
8Charles BlairIllinois Q or Q82
10Peter YehTaiwan Q or Q80
11Gareth BirdsallEngland Q or Q80
12Grant PeacockMaryland Q or Q80
13Leigh MathesonAustralia Q or Q76

 Puzzle 8M19   Main Top   Queens Around

## Solution

A natural presumption is that South holds the three missing aces, as would any sane player contracting for 12 tricks in notrump. Did I mention Grover? His multifaceted bidding is world-renowned; alas, none of his ‘facets’ seem to be sanity. Most solvers caught on to the advantage in giving East an ace, not only because it allows a quick set on the right lead, but to beef up the East hand for tie-breaking purposes. Here’s one such solution from across the pond.

 6 NT South K 5 4 A K 6 5 4 K K 7 6 5 Trick1 W2 N3 N4 S5 S6 S7 S Lead Q K 5 10 9 8 4 2ndKJ 4 9 10 J 9 3rd32A67 4 5 4th2Q 8 5 3 6 7 W-LW1W2W3W4W5W6W7 Q J 10 9 8 Q J 10 9 Q J 10 Q A 7 6 3 8 7 8 7 6 5 4 3 J Lead: Q 2 3 2 A 9 2 A 10 9 8 4 3 2

Tom Slater: A minor-suit lead allows declarer to unblock the K and run the clubs. West has to discard all his spades, then is given a heart trick to set up the hearts. A spade lead is instantly fatal, while a heart lead prevents the duck at the end.

The following ending is reached before South leads the last club:

 NT win 5Success K A K 6 5 4 — — Trick8 S9 S Lead 3 2 2nd Q10 3rd K4! 4th 67 W-LW1L1 Q Q J 10 J 10 — A 7 6 8 7 8 — South leads 2 3 2 A 9 3 Win the rest

West obviously cannot let go a red card, so his only hope is to pitch his last spade, as does North. A heart is ducked, then declarer has the rest with any return. An original heart lead of course foils this, as declarer would lose communication in the ending.

With six clubs an absolute laydown, Grover’s final bid was crazy, but then so is he.

### From insane to egregious

The previous East hand totals 89, including an ace. The next solution manages to top that without an ace, albeit with a far-fetched layout, but anything is possible with Grover at the table. Indeed, psyching 2  with the South hand might even signify a rehabilitation, upgrading Grover’s bidding skill from insane to egregious.

 6 NT South K 5 4 A K 6 5 4 K K 7 6 5 Trick1 W2 N3 N4 S5 S6 W7 N8 N9 S10 S Lead Q K 5 A 3 Q K 4 6 5 2ndK72JQK3 4 9 10 3rd72A 4 57 3A 6 7 4th210Q892 89 8 9 W-LW1W2W3W4L1W5W6W7W8W9 Q J 10 9 8 Q J 10 9 Q J 10 Q 7 8 7 9 8 7 J 10 9 8 4 3 2 Lead: Q A 6 3 2 3 2 A 6 5 4 3 2 A

Nicholas Greer: On a red-suit lead declarer can set up diamonds, cash the K, then squeeze West in hearts and spades. On a club lead 6 NT goes several off, because there aren’t enough entries to set up diamonds. A spade lead also defeats 6 NT, because another spade when declarer gives up a diamond prevents declarer from cashing the K while there is still an entry to the South hand.

Tim Broeken: With a red queen start, South can develop diamonds and squeeze West in spades and hearts for a 12th trick. The  Q start makes it impossible to develop diamonds, and only nine tricks are available. The  Q start makes the squeeze impossible.

After a heart or diamond lead, the following ending is reached:

 NT win allSuccess 5 A 6 — — Trick11 S Lead 4! 2nd? 3rd 4th W-L J J 10 — — — 8 — J 10 South leads 6 3 4 — West is squeezed

The last diamond forces West to part with his stopper in one of the majors. If West had led a second heart at Trick 6, a similar ending is reached, with North’s entry the K instead of the A.

### Beware the Curse of Scotland

The ultimate solution, found by two solvers, gives East the highest possible rank total (96) while leaving South an abominable collection. No problem! ‘Abominable’ would be fitting for Grover’s tombstone, which, by the way, the city has pre-purchased in hopes of immediate utilization. Only two queen leads will defeat 6 NT despite only 10 apparent tricks. Watch how the 9 grows if West leads a black suit.

 6 NT South K 5 4 A K 6 5 4 K K 7 6 5 Trick1 W2 N3 N4 S5 S6 S Lead Q K 5 9 8 4 2ndK10J 9 10 9 3rd62A67 4 4th2Q 8 3 4 5 W-LW1W2W3W4W5W6 Q J 10 9 8 Q J 10 9 Q J 10 Q 7 6 8 7 A 8 7 6 5 4 3 J 10 Lead: Q A 3 2 3 2 9 2 A 9 8 4 3 2

Dan Gheorghiu: Either red queen sets 6 NT. A diamond is obvious. On a heart lead, the triple squeeze is there, but [if North retains K-5-4 A-6 K] East should not take the A, and the count is not rectified to squeeze West again. [If North instead keeps K-5 A-6-5 K, East must win the A and return a heart.]

On the given lead, the K must be won, then clubs are led to reach this ending:

 NT win 6Success 5 A K 6 5 4 K — Trick7 S8 S9 E10 N11 S Lead 3 2! 8 5 9! 2nd JQ27? 3rd 4K10A 4th 6AK10 W-LW1L1W2W3 J 10 Q J 10 Q J — 7 8 7 A 8 7 6 — South leads A 3 3 2 9 2 3 West is squeezed

The last club triple-squeezes West, who does best to pitch a diamond; North pitches a heart. The 2 is then led to establish the nine, which will squeeze West in the majors to make 6 NT. Variation: If East ducks the K to avoid rectifying the count, diamonds are out of the picture, so declarer simply ducks a heart to establish the suit.

## Back at the Asylum

Dan Gheorghiu: I wonder what Richard hints us with 0.7425? Is it somehow East’s or South’s suit lengths?

No, just a red herring… or the probability of you being set on any given deal.

Jon Greiman: North dealt and opened 1 . South was new to the concept of bidding boxes and used his cards instead, showing the 2. North rebid 3 , then South chose to show his 3. North bid the obvious 3 NT, and South, lacking options, showed his 6, which North corrected to 6 NT. East deliberated so long that West led out of turn.

Yep, that explains it. Case closed — or more appropriately, closed by a nut case.

 Puzzle 8M19   Main Top   Queens Around