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One Foggy Christmas Eve

Santa came to say, “Rudolph with your nose so bright,
did you set this slam tonight?”

“Six spades they tried,” the reindeer replied,
“I led a heart, and declarer was fried.”

“I see,” said the master of Yuletide arts,
“But what if the slam is played in hearts?”

“Ouch!” cried Rudolph, about to stampede,
“That’s even set if a trump is the lead.”

Rudolph
6 S S Q 9 8 7 6 5
H K 10 9 8 7
D
C 9 2
Both Vul

West

Pass
Pass
Pass


North

2 D
3 H
6 S


East

Pass
Pass
All Pass


South
2 C
2 NT
4 S



 
Table


 
Heart lead S A K J
H Q J 6 5
D A K J 10 5
C A

Construct a West hand to match the dialogue. Multiple solutions exist; optimal solution is for West to have the same HCP total (6) and spot-card sum (55) as East.

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Eddy Choi Wins!

In November-December 2010 this puzzle was presented as a contest, with 76 participants from 42 locations. Thanks to those who entered, and congratulations to the 25 successful solvers. Ranking is by the most even division of E-W HCP, secondarily by the most even spot-card sums, and tiebroken by date and time of entry. The top 11 solvers all produced perfect solutions.

RankNameLocationHCPSpots
1Eddy ChoiHong Kong6-655-55
2Charles BlairIllinois6-655-55
3Jonathan MestelEngland6-655-55
4Gareth BirdsallEngland6-655-55
5Manuel PauloPortugal6-655-55
6John AuldEngland6-655-55
7Jim MundayCalifornia6-655-55
8Pavel StrizCzech Republic6-655-55
9Leif-Erik StabellZimbabwe6-655-55
10Julian WightwickEngland6-655-55
11Howard LiuIllinois6-655-55
12Mike ChanterEngland6-653-57
13James LawrenceEngland6-648-62
14Ufuk CotukEngland5-758-52
15Sid IsmailSouth Africa2-1055-55
16Barry RigalNew York2-1055-55
17Paul GilbertEngland10-259-51
18Jan KuipersNetherlands10-251-59
19Richard SteinCalifornia2-1062-48
20Barry GoodmanEngland2-1062-48
21Wayne SomervilleNorthern Ireland2-1063-47
22Ryan LeungSingapore10-247-63
23Sebastien LouveauxBelgium2-1063-47
24Harold GoodmanGeorgia (US)2-1063-47
25Rik ter VeenNetherlands2-1063-47

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Solution

Per the dialogue, the object is to construct a West hand, and thereby a complete deal, in which a heart lead will defeat bothS and 6 H. It is easy to see how 6 S can be defeated by a heart ruff; but if such a layout exists, how could a heart lead defeat 6 H? Indeed, it could not, so a more devious approach is needed.

Ten solvers (places 16-25) went for the brazen solution of a 10-0 club division, allowing the defense to get a club ruff against either 6 H or 6 S. While undeniably valid, this construction requires the H A to be in the same hand as the long clubs for a heart lead to beat both slams, which means a lopsided HCP division and therefore no contention for the top spot. (A 10-card club suit also makes the auction implausible, but this has no bearing on the puzzle solution.)

A more subtle construction allows an equal arrangement of 6 HCP for West and East, as well as an equal spot-card sum of 55. The first to submit such a solution was the winner:

6 S S Q 9 8 7 6 5
H K 10 9 8 7
D
C 9 2
Trick
1. W
2. E
Lead
H 4
H 3
2nd
7
3rd
A
4th
5
S
H 4 2
D Q 9 8 7 6 4 3 2
C K J 10
Table S 10 4 3 2
H A 3
D
C Q 8 7 6 5 4 3
Eddy Choi
Hong Kong

S A K J
H Q J 6 5
D A K J 10 5
C A

Six spades is defeated by the defense shown, since declarer has no way to reach dummy to draw the last trump; so East must score the S 10. Six hearts is defeated in more straightforward fashion by the imminent spade ruff after any lead (including a trump).

Eddy’s solution was duplicated by Jonathan Mestel, John Auld (can I add Lang Syne? as I write this just an hour into 2011), Julian Wightwick and Howard Liu. Alas, this was not my intended solution — translation: I missed it — but of course equally correct.

Howard Liu: Neat puzzle! Nice usage of the C 9 in dummy to enforce the tiebreaker, in the spirit of dividing the twelve points of Christmas.

Not sure how to interpret that, but next time I’m makin’ a list and checkin’ it twice. The solution I expected was submitted by the remaining six of the top 11, including the winner of my last contest:

Rudolph
6 S S Q 9 8 7 6 5
H K 10 9 8 7
D
C 9 2
Trick
1. W
2. E
Lead
H 2
D 3
2nd
7
3rd
A
4th
5
S
H 4 3 2
D Q 9 8 7 6 4 2
C K J 10
Table S 10 4 3 2
H A
D 3
C Q 8 7 6 5 4 3
Charles Blair
Illinois

S A K J
H Q J 6 5
D A K J 10 5
C A

Six spades is defeated by the defense shown, similarly garnering a trump trick for East. It is also curious, though irrelevant to the puzzle, that a heart lead is required to beat 6 S, whereas any lead suffices on the previous deal. Six hearts is defeated with any lead.

Charles Blair: Of course, using the Four Aces Point Count (ace = 3, king = 2, queen = 1, jack = 0.5) this fails the “equal HCP” goal.

Hmm. Instead of the traditional three wise men, I get one wise guy.

And finally, to avoid any possible ambiguity in the instructions, this devious solver scoured the African plain to leave no antlers unturned:

6 S S Q 9 8 7 6 5
H K 10 9 8 7
D
C 9 2
Trick
1. W
2. E
Lead
H 2
D 9
2nd
7
3rd
A
4th
5
S
H 4 3 2
D Q 8 7 6 4 3 2
C K J 5
Table S 10 4 3 2
H A
D 9
C Q 10 8 7 6 4 3
Leif-Erik Stabell
Zimbabwe

S A K J
H Q J 6 5
D A K J 10 5
C A

Leif-Erik Stabell: Assuming high cards are included in counting spots: ace = 1, king = 13, queen = 12, jack = 11. If high cards are included but ace = 14, swap the D 9 and D 3, in which case E-W cannot have the same spot count. If high cards are not included in the spot count, swap the C 10 and C 5 in addition.

I think I see it. Three cases covered, two subtle swaps, and a partridge in a pear tree.

Pavel Striz: Merry Christmas, reindeer!

Then all the reindeer loved him, as they shouted out with glee,
“Rudolph the red-nosed reindeer, you’ll go down in his–tor–y!”

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Acknowledgments to Rudolph the Red-Nosed Reindeer lyricist Johnny Marks.
© 2010 Richard Pavlicek