Main   Puzzle 8S59 by Richard Pavlicek

# Jolly Old Saint Nicholas

Jolly old Saint Nicholas
Held South’s cards today.
All he had was just a queen,
Not much else to say.

Any-one with half a brain
Knows to pass is proper.
Saint Nick took another view,
“Gotta show my stopper!

“Notrump is the place to be;
Heart tricks I should gain.
But I must admit to you,
I’ve been found insane!”

 3 NT×× South — A J 10 9 8 7 6 5 4 7 6 7 6 N-S Vul West1 3 DblAll Pass North1 3 Pass East2 3 Pass South2 NT!3 NT!Rdbl! — K K Q J 10 9 8 K Q J 10 9 8 A K J 10 9 8 7 6 5 4 Q A A Lead: K Q 3 2 3 2 5 4 3 2 5 4 3 2

So there rests “Crazy Claus” in 3 NT, redoubled and vulnerable, with only 7 combined HCP. On most occasions this would be a debacle, maybe never winning a trick (minus 5200), but a bright “Star in the East” shone through. After losing the first four tricks, Nick had the rest. Plus 1000!

Note that 3 NT is cold against any defense, and from either side (North or South declarer). I am confident that 7 HCP is the fewest possible HCP to achieve this feat. Now for my holiday puzzle, let’s lower the contract by a notch:

What is the fewest possible HCP to make two notrump against any defense from either side?

Construct a deal to illustrate. Ties for fewest HCP will be broken by the sum of (1) North-South’s card ranks* and (2) South’s card ranks, in that order of priority. Lower is better.

*Ace = 14, King = 13, Queen = 12, Jack = 11, etc.

## Solution

Use the form below if you wish to submit a solution. Entries will be accepted until Christmas Eve (December 24) at midnight GMT — just in time for Jolly Insane Nick to drop down your chimney. You may submit multiple times, but only the latest one counts. The best solutions will be published here on Christmas Day, with acknowledgment to the top solvers.

 2 NT South East will getwhat remains Click to view deal

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