Puzzle 8S25   Main

High Cards Amiss

by Richard Pavlicek

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, 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 bridge diagram written on a drink napkin.
“What board? Where are the high cards?”

“Ah, but that’s the puzzle. When you joined this club, you didn’t expect to be cured overnight, did you?”

“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.”

 4 South 10 9 8 7 6 5 4 3 2 5 4 3 2 10 9 8 7 6 8 7 6 5 4 3 2 10 9 West leads 4 3 2 10 9 8 7 6 5

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.

 4 South 10 9 8 7 6 5 4 3 2 5 4 3 2 10 9 8 7 6 8 7 6 5 4 3 2 10 9 East will getthe remainingfour high cards West leads 4 3 2 10 9 8 7 6 5

To see if your solution is successful click

Duncan Bell Wins

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.

Winner List
RankNameLocationTricks WonFreakness
1Duncan BellEngland1205
3Jean-Christophe ClementFrance1205
4Grant PeacockMaryland1208
5Dan GheorghiuBritish Columbia1208
6Bob OndoMichigan1208
7Tim BroekenNetherlands1211
8Alon AmselBelgium1211
9Ufuk CotukEngland1211
10Dustin MillerOhio1213
11Nicholas GreerEngland1213
12Jim MundayMississippi1213
13Venk NatarajanUtah1100
14Gordon HoHong Kong1100
15Richard MorseCalifornia1107
16Levi KatrielCalifornia1112
17Alessandro RapuanoItaly1124
18Gonzalo GodedSpain1003
19Samuel PahkMassachusetts1003
20Joe ClarkEngland1007
21Jarno SunMassachusetts1014

 Puzzle 8S25   Main Top   High Cards Amiss

Solution

The solution with the least deal freakness by far was submitted by Venk Natarajan and Gordon Ho:

 4 South A Q 10 9 8 7 6 K J 5 4 3 2 Trick1 W2 S3 N4 S5 N6 N7 S8 N9 S10 S11 W Lead 8 2 6 3 A 7 6 8 9 7 7 2nd255J8J9K 6A4 3rdQ1010Q4QJA 35A 4thK627K324 935 W-LW1W2W3W4W5W6W7W8W9L1L2 K J 5 4 3 2 A Q 10 9 8 7 6 8 7 6 K J 5 4 3 2 A Q 10 9 Win 11Frkn 0 4 3 2 A Q 10 9 8 7 6 K J 5 Win the rest

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 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:

 4 South A J 10 9 8 7 6 5 A J 4 3 2 Trick1 W2 E3 N4 S5 N6 N7 S8 N9 S10 N11 S Lead J 3 6 Q J 2 2 7 3 8 9 2nd575K10 65JKK 10 3rdA910A5 Q10QAA 3 4th6 929677384 2 W-LL1W1W2W3W4W5W6W7W8W9W10 K 5 4 3 2 K J 10 9 K 8 7 6 8 7 6 K J 5 A Q 4 3 2 10 9 Win 12Frkn 8 Q 4 3 2 A Q 10 9 8 7 6 Q 5 Crossruff the rest

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 (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:

 4 South A J 10 9 8 7 6 A J 5 4 3 2 Trick1 W2 N3 S4 E5 N6 S7 N8 S9 N10 N11 S Lead 5 6 5 7 3 6 4 7 J 7 A 2nd9563109A108J J 3rd692K 4J QA 8Q8 4th229A7283 34K W-LW1W2L1W3W4W5W6W7W8W9W10 K 5 4 3 2 K 10 9 K J 8 7 6 8 7 6 K J 5 4 3 2 A Q 10 9 Win 12Frkn 5 Q 4 3 2 A Q 10 9 Q 8 7 6 5 Win the rest

No matter what West leads, declarer routinely wins four trumps, two club ruffs, four hearts and two diamonds.

More Power

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.

 Puzzle 8S25   Main Top   High Cards Amiss