The Friday night meeting of Puzzlers Anonymous had an eerie ambiance. It was July 13th!
Professor Freebid bores everyone as usual with his lecture then sits down to enjoy a glass of burgundy.Glancing toward the door, he notices a familiar disheveled face enter the room.“Timothy, you’re late! Come here, I want to show you a board from last night.”
Timothy walks over to the Professor’s table and stares at a cocktail napkin with a diagram.“What board? I don’t see any high cards.”
“Ah, but that’s the puzzle. When you joined this club, you didn’t expect to be cured overnight, did you?”
“Come on, Professor, a puzzle needs more information than that.”
“Yes it does and there is: Each hand has 10 high-card points.South is declarer in four spades, and no defense can beat it.The deal irrefutably supports my boson deflection theory.”
“I knew you’d work your bozo theories into this some way. Does anybody really care?”
“Yes, Timothy, my friend Richard pays me big bucks for this stuff. Now shut up and solve it.”
Place the 16 high cards so declarer can make 4 .
Each hand must have exactly 10 HCP. A further goal (contest tiebreaker) is for North-South to win the most tricks. Before reading further, challenge yourself, or make your best guess:
1. Which North-South distribution was the winner?
Quit
*There was no multiple choice, so solvers had to construct their own holdings.
Congratulations to Duncan Bell, who was the first to submit the optimal solution. Duncan is a brilliant solver with many past wins including The Twelve of Spades, Just Another Zero, High Stakes Rubber, Bridge with the Abbott and Pay No Taxes! — and in the last his solution was even better my expected solution.
Ranking is by most tricks won, lowest deal freakness, and date-time of entry, in that order of priority.
The solution with the least deal freakness by far was submitted by Venk Natarajan and Gordon Ho:
This symmetric deal with zero freakness (all hands 4-3-3-3) is a finesser’s field day. With every card onside, declarer easily wins 11 tricks, losing just two aces.
Unfortunately, freakness was the secondary tiebreaker, so the “perfect zero” above loses out to the winning of another trick. It is possible to win 12 tricks as shown by this construction:
Grant Peacock: I’m hoping your “freakness” tiebreaker was a red herring and there is only one layout that makes 12 tricks.
Well, actually it was a red pheasant (maybe your feathered cousin?). You did better than most of the 12-trick winners but came up short, or should I say long, in the freakness department with 8.
Duncan Bell, Konrad Majewski and Jean-Christophe Clement produced identical layouts to win 12 tricks with a freakness of only 5:
No matter what West leads, declarer routinely wins four trumps, two club ruffs, four hearts and two diamonds.
Professor Freebid: Next month I plan to rent the large hadron collider, which will generate more power to accelerate the bosons to nearly light speed. With the proper settings we should be able to get a grand slam out these cards. Stay tuned! It could be your next puzzle.
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by Richard Pavlicek by Richard Pavlicek