Puzzle 8K10 Main

One Foggy Christmas Eve

by Richard Pavlicek

Santa came to say…

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

“Six spades they tried,” Rudolph replied. “I led a heart and Blitzen was fried.”

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

“No!” said Rudolph about to stampede. “That’s even set if a trump is my lead.”

When the fog finally lifted on Christmas Day, only the North-South cards remained:

	Q 9 8 7 6 5			West	North	East	South
Both vul	K 10 9 8 7			Rudolph	Donner	Ronald	Blitzen
	—						2
	9 2			Pass	2	Pass	2 NT
				Pass	3	Pass	4
				Pass	6	Pass	Pass
				Pass
	A K J
	Q J 6 5
	A K J 10 5
6 South	A

Help! Nobody can remember the other hands, so it’s up to you:

Construct a West hand for Rudolph that fits the dialogue.

Multiple solutions exist. Further goals (tiebreakers for the December contest) are for the HCP total and spot-card sum (in that order of priority) for West and East to be as near to equal as possible.

Eddy Choi Wins

Kudos to Eddy Choi, Hong Kong, who was the first of 11 solvers to give East-West identical HCP totals and spot card sums.

Winner List

Rank	Name	Location	HCP	Spots
1	Eddy Choi	Hong Kong	6-6	55-55
2	Charles Blair	Illinois	6-6	55-55
3	Jonathan Mestel	England	6-6	55-55
4	Gareth Birdsall	England	6-6	55-55
5	Manuel Paulo	Portugal	6-6	55-55
6	John Auld	England	6-6	55-55
7	Jim Munday	California	6-6	55-55
8	Pavel Striz	Czech Republic	6-6	55-55
9	Leif-Erik Stabell	Zimbabwe	6-6	55-55
10	Julian Wightwick	England	6-6	55-55
11	Howard Liu	Illinois	6-6	55-55
12	Mike Chanter	England	6-6	53-57
13	James Lawrence	England	6-6	48-62
14	Ufuk Cotuk	England	5-7	58-52
15	Sid Ismail	South Africa	2-10	55-55
16	Barry Rigal	New York	2-10	55-55
17	Paul Gilbert	England	10-2	59-51
18	Jan Kuipers	Netherlands	10-2	51-59
19	Richard Stein	California	2-10	62-48
20	Barry Goodman	England	2-10	62-48
21	Wayne Somerville	Northern Ireland	2-10	63-47
22	Ryan Leung	Singapore	10-2	47-63
23	Sebastien Louveaux	Belgium	2-10	63-47
24	Harold Goodman	Georgia (US)	2-10	63-47
25	Rik ter Veen	Netherlands	2-10	63-47

Puzzle 8K10 Main

Top One Foggy Christmas Eve

Solution

Per the dialogue, the object is to construct a West hand, and thereby a complete deal, in which a heart lead will defeat both 6 and 6 . It is easy to see how 6 can be defeated by a heart ruff; but if such a layout exists, how could a heart lead defeat 6 ? 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 or 6 . While undeniably valid, this construction requires the 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 solver to submit an optimal solution was Eddy Choi (Hong Kong):

6 South	Q 9 8 7 6 5		Trick	Lead	2nd	3rd	4th
	K 10 9 8 7		1. W	4	7	A	5
	—		2. E	3
	9 2			Lose 1 trump trick
—		10 4 3 2
4 2		A 3
Q 9 8 7 6 4 3 2		—
K J 10		Q 8 7 6 5 4 3
	A K J
	Q J 6 5
	A K J 10 5
	A

Six spades is defeated by the defense shown, since declarer cannot reach dummy to draw the last trump; East must score the 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 overlooked it — but of course equally correct.

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

Not sure how to interpret this, but next time I’m makin’ a list and checkin’ it twice!

Intended solution

The solution I intended was submitted by the six of the top 11, including the winner of my last contest, Charles Blair (Illinois):

6 South	Q 9 8 7 6 5		Trick	Lead	2nd	3rd	4th
	K 10 9 8 7		1. W	2	7	A	5
	—		2. E	3
	9 2			Lose 1 trump trick
—		10 4 3 2
4 3 2		A
Q 9 8 7 6 4 2		3
K J 10		Q 8 7 6 5 4 3
	A K J
	Q J 6 5
	A K J 10 5
	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 , whereas any lead suffices on the previous deal. Six hearts is defeated with any lead.

African safari

And finally, to avoid possible ambiguity in the instructions, Leif-Erik Stabell (Zimbabwe) left no antlers unturned as he rafted the mighty Zambezi River, producing this layout:

6 South	Q 9 8 7 6 5		Trick	Lead	2nd	3rd	4th
	K 10 9 8 7		1. W	2	7	A	5
	—		2. E	9
	9 2			Lose 1 trump trick
—		10 4 3 2
4 3 2		A
Q 8 7 6 4 3 2		9
K J 5		Q 10 8 7 6 4 3
	A K J
	Q J 6 5
	A K J 10 5
	A

Leif-Erik Stabell: This assumes high cards are included in counting spots as ace = 1, king = 13, queen = 12, jack = 11, etc. If high cards are included but ace = 14, swap the 9 and 3, in which case East-West cannot have the same spot count. If high cards are not included in the spot count, swap the 10 and 5 in addition.

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

Gleefully yours

Charles Blair: Using the Four Aces Point Count (ace = 3, king = 2, queen = 1, jack = 0.5) your solution fails the “equal HCP” goal.

Wonderful. Instead of a Yuletide with three wise men, I get one wise guy.

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!”

Puzzle 8K10 Main

Top One Foggy Christmas Eve

Acknowledgments to song lyricist, Johnny Marks
© 2010 Richard Pavlicek