An eerie ambiance pervades the Friday night meeting of Puzzlers Anonymous, being the 13th of April. Professor Freebid bores everyone as usual with his opening lecture, then he sits down to enjoy a glass of burgundy. Glancing toward the door he sees a familiar disheveled face. Timothy, youre late! Come here, I want to show you a board from last night.
Timothy walks over to the Professors table and stares at a bridge diagram written on a drink napkin.What board? Where are the high cards?
Ah, but thats the puzzle. When you joined this club, you didnt 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 youd 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 in the diagram so that N-S can win at least 10 tricks in spades.
As apparent to reach 13 cards, each hand must receive exactly four high cards. Further, each hand must total 10 HCP, which can be of any composition (possibilities are AKQJ, AAJJ, AQQQ, KKQQ and KKKJ) and in any number of suits. Successful solvers will be ranked by the most tricks winnable in spades, with ties broken by the lowest freakness of the complete deal.
This puzzle contest, designated September 2018 for reference, was open for over a year. Participants were limited to one attempt, unlike my usual contests allowing entries to be revised with only the last one counting. There were 21 correct solutions, of which three were optimal.
Congratulations to Duncan Bell, who was the first to submit the optimal score. Duncan is a brilliant solver with many previous wins including The Twelve of Spades, Just Another Zero, High Stakes Rubber, Bridge with the Abbott, Cat O' Nine Tails and Pay No Taxes! and in the last he found a solution that even bettered my expected solution.
Ranking is by the most tricks won, least 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 finessers field day. With every card onside, declarer easily wins 11 tricks, losing just two aces.
Unfortunately, freakness was only 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: Im 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 (your feathered cousin?). You did better than most of the 12-trick winners but came up short or should I say long for the ideal solution.
Duncan Bell, Konrad Majewski and Jean-Christophe Clement produced identical layouts to win 12 tricks with the least possible freakness:
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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