Main     Puzzle 8M19 by Richard Pavlicek    

Queens Around

“Timothy! Come over. I’ve got a hand to show you.” The Professor had just finished his lecture at the P.A. meeting, expounding his theory that queen leads, unlike any other honor card, achieve a high variance in trick production because they magnetically deflect bosons, causing extreme levels of eccentricity in the quantum model. Timothy arrives at the table, and the Professor jots down a hand on his trusty clipboard. “Suppose Grover is declarer in six notrump, and it’s your lead.”

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

“With Grover at the table, it could be anywhere from lame to insane; but the bidding is irrelevant. This is a construction problem, and I will also show you dummy.” The Professor reaches for the clipboard and scribbles down another hand. “There you have it; all the kings on your left with neatly ordered spot cards.”

6 NT South S K 5 4
H A K 6 5 4
D K
C K 7 6 5
S Q J 10 9 8
H Q J 10 9
D Q J 10
C Q
Table  ?
 ?
 ?
 ?
West leads  ?
 ?
 ?
 ?

“Am I supposed to guess which queen to lead?” Timothy asks.

“No. You’d probably go with your longest suit, but it’s not a lead problem. I will even tell you the result, assuming best play all around, which epitomizes my boson deflection theory. Generally, four queens exhibit a variance for one with the other three synchronous (analogous to four missing cards being most likely to split 3-1) but in this case two queens will set the contract, and two will not.”

“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 were 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 from the absorption of pions, causing trick production to stabilize asymptotically. Extreme charges (greater than 0.7425 electron volts) produce an overtrick.”

“I see… um, I think. So if I choose a queen at random, I have a fifty-fifty chance to set the contract. I gather ‘best play all around’ means a substitute for Grover, as otherwise even the wrong queen would make me a favorite.”

“Finally, brilliance emerges.”

Construct the South and East hands so that exactly two queen leads will defeat 6 NT.

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

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Dean Pokorny Wins!

In July 2015 this puzzle was presented as a challenge, 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 card 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.

RankNameLocationLeads SetEast Sum
1Dean PokornyCroatiaH Q D Q96
2Dan GheorghiuBritish ColumbiaH Q D Q96
3Nicholas GreerEnglandS Q C Q93
4Tim BroekenNetherlandsS Q C Q93
5Tom SlaterEnglandS Q H Q89
6Jon GreimanIllinoisS Q C Q84
7Jamie PearsonOntarioS Q C Q82
8Charles BlairIllinoisS Q H Q82
9Adam DickinsonScotlandS Q C Q80
10Peter YehTaiwanS Q H Q80
11Gareth BirdsallEnglandS Q H Q80
12Grant PeacockMarylandS Q H Q80
13Leigh MathesonAustraliaS Q H Q76

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Solution

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

Tom Slater: A minor-suit lead allows declarer to unblock the D 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. 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.

Note that an original heart lead foils this, as declarer would then lose communication with dummy if he ducked a heart in the ending.
South
leads
S K
H A K 6 5 4
D
C
S Q
H Q J 10
D J 10
C
Table S A 7 6
H 8 7
D 8
C
S 2
H 3 2
D A 9
C 3

Restoring some sanity

The previous East hand totaled 89, including an ace (14 rank points). The next solution managed to top that without an ace, restoring some sanity to the final contract. The South hand might even make Grover proud, upgrading his bidding prowess from insane to egregious.

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

Nicholas Greer: On a red-suit lead declarer can set up diamonds, cash the C 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 C 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 C Q start makes it impossible to develop diamonds, and only nine tricks are available. The S Q start makes the squeeze impossible.

After a heart or diamond lead, the ending at the right is reached. 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, except North’s entry would be the S K instead of the H A.
South
leads
S 5
H A 6
D
C
S J
H J 10
D
C
Table S
H 8
D
C J 10
S 6
H 3
D 4
C

Beware the curse of Scotland

The ultimate solution, found by two solvers, gives East the highest possible card-rank total (96) while leaving South an abominable collection. No problem! “Abominable” would be most 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 D 9 grows if West fails to lead that suit.

6 NT S K 5 4
H A K 6 5 4
D K
C K 7 6 5
Leader
1. W
2. N
3. N
4. S
5. S
6. S
Lead
S Q
C K
C 5
C 9
C 8
C 4
2nd
K
10
J
S 9
D 10
H 9
3rd
6
2
A
6
7
S 4
4th
2
Q
S 8
D 3
D 4
D 5
S Q J 10 9 8
H Q J 10 9
D Q J 10
C Q
Table S 7 6
H 8 7
D A 8 7 6 5 4 3
C J 10
S A 3 2
H 3 2
D 9 2
C 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 S K-5-4 H A-6 D K] East should not take the D A, and the count is not rectified to squeeze West again. [If North instead keeps S K-5 H A-6-5 D K, East must win the D A and return a heart.]

On the given lead, the S K must be won, then clubs are led to reach the ending at right. The last club triple-squeezes West, who does best to pitch a diamond; North pitches a heart. The D 2 is then led to establish the nine, which will squeeze West in the majors to make 6 NT.

Variation: If East ducks the D K to avoid rectifying the count, diamonds are out of the picture, so declarer simply ducks a heart to establish that suit.
South
leads
S 5
H A K 6 5 4
D K
C
S J 10
H Q J 10
D Q J
C
Table S 7
H 8 7
D A 8 7 6
C
S A 3
H 3 2
D 9 2
C 3

Meanwhile, 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?

OK, OK. If you must know, it’s the probability of you being set on any given hand.

Jon Greiman: North dealt and opened 1 H. South was new to the concept of bidding boxes and used his cards instead, showing the D 2. North rebid 3 C, then South chose to show his S 3. North bid the obvious 3 NT, and South, lacking options, showed his D 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.

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© 2015 Richard Pavlicek