Main     Puzzle 8N07 by Richard Pavlicek

# Fewest HCP Notrump

Bridge is on the rise in America! Statistics show that more people are declaring “No Trump” than ever before! Personally, I’m not so easily influenced and would always vote for Donald — Duck, that is, which might be the first write-in landslide in history. Meanwhile, back at the bridge ranch: How many high-card points do you need to make 1 NT?

Winning the majority of tricks should require the majority of HCP, or more precisely, by simple arithmetic: 7/13 × 40 = 21.5 approximately. Logically this should be rounded up to 22, since the defense usually has an advantage with the lead, let alone that you can’t hold half a HCP. The requirement is also affected by positional status, spot cards, suit lengths and communication, so it varies from deal to deal.

In the most extreme case, 1 NT can be made with only 3 HCP:

 1 NT South 7 6 5 4 3 2 — 8 7 6 5 4 3 2 — — A K J 5 4 3 2 — A K Q 4 3 2 A K Q J 10 9 8 — A K Q J 10 9 — Lead: any — Q 10 9 8 7 6 — J 10 9 8 7 6 5

No matter how West defends, he can win only his six high cards. Note that South’s Q is necessary, as a reduction to J-10-9-8-7-6 would allow West to establish his long heart before South could establish clubs.

The entire scenario, of course, is a reductio ad absurdum (remind you of American politics?) requiring a muzzle on East-West and a demented South, but only the latter is required if East bids the obvious 7  and South saves in 7 NT× (profitable except at unfavorable). Also kind of spooky is that West, who cannot beat 1 NT on lead, makes 7 NT as declarer.

On the above deal, 1 NT must be played by South, as North would be swept clean. For the puzzle it must make both ways. An obvious fix is to duplicate the S-W arrangement between North and East, which becomes 6 HCP, but it’s possible to go lower than 6. Can you?

 What is the fewest HCP required to make 1 NT by North or South against any defense?

Construct a deal to illustrate your answer. As a further goal (tiebreaker for the October 2016 contest) try to keep the sum of all North-South card ranks as low as possible, and between them, South as low as possible.

## Try it now

Enter a deal and click Verify to find out what, if anything, is wrong. Use the help provided to make corrections and repeat. See how many tries it takes you to discover the winning deal. North-South’s longest fit must be in diamonds, and their shortest fit in clubs.

 1 NT Northor South East will get what remains East orWest leads

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## Tina Denlee Wins!

In October 2016 this puzzle was presented as a challenge — with no help provided — inviting anyone who wished to submit a solution. Exactly 50 persons gave it a try, up slightly from last month, but only the 16 listed were able to make 1 NT North or South with 4 HCP, the fewest possible. Of the remaining entries, 20 required 5 HCP, and 14 (with 4-5 HCP) were flawed allowing the defense to prevail.

Congratulations to Tina Denlee, who was the first of 11 to submit the optimal solution, minimal card-rank sum for North-South and South. Tina is a dynamo, with six perfect solutions, and the rest pretty close, since joining this series a year ago. Also winning Notrump Ship Down, she’s evidently hip on the No Trump theme, and fortunately Canadian to witness our farce from the outside — or maybe she’s just lobbying against a wall around Canada.

Rounding out the top three are two stars who go a long way back: Leif-Erik Stabell won The House on Phantom Lane and Our Finest Gifts We Bring in 2003; and Charles Blair won Have Cards, Will Double (2003), Let Your Heart Be Light (2004) and Mission: Implausible (2005), as well as Two-Way Finesses in my 2010 puzzle series. Charles is also the only person who has been a regular participant in my monthly challenges from the start (September 2000).

RankNameLocationHCPN-S SumSouth Sum
1Tina DenleeQuebec415152
2Leif-Erik StabellZimbabwe415152
3Charles BlairIllinois415152
4Duncan BellEngland415152
5Hendrik NigulEstonia415152
6Tom SlaterEngland415152
7Ray LiuOntario415152
8Leigh MathesonAustralia415152
9Dan GheorghiuBritish Columbia415152
11Jamie PearsonOntario415152
12Grant PeacockMaryland415548
13Nicholas GreerEngland415949
14Tim BroekenNetherlands415949
15Martin VodickaSlovakia415949
16Jim MundayMississippi416563

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## Solution

A few months ago when looking for puzzle ideas, I perused my old “Fewest HCP To Make Contract” page (now split into separate pages for Suit and Notrump) and thought: Wouldn’t it be great if I could find an improvement on one of them? Chances were slim, since the page had been up for many years, and eagle-eyed viewers would usually have spotted it. In my best Gomer Pyle imitation: Shazam! The page stated 5 HCP as the minimum to make 1 NT North or South; and like all successful solvers this month, I realized I could name that tune in 4 notes. The faulty deal has been removed and will be replaced by the winning entry in this contest.

Before showing the optimal solution, let’s look at a few others. My good friend Jim Munday submitted this deal on a Friday, which probably accounts for his 16th-place finish. Never lose sync with The Mamas and Papas! Nonetheless, he did manage to reduce North-South to 4 HCP, which most respondents did not. I especially liked his duplicated theme of jack catching.

 1 NT Northor South 2 2 Q 10 9 8 7 6 Q 10 9 8 7 A A A K 5 4 3 2 A K 6 5 4 K Q J 10 9 8 K Q J 10 9 J J 7 6 5 4 3 8 7 6 5 4 3 — 3 2

Jim Munday: Good luck in reaching 1 NT! Fortunately, creating a realistic auction was not part of the conditions.

On a major-suit lead and low-diamond shift, declarer must hop queen, or he won’t take a single trick. After that, the club play is just deja vu, and declarer is assured seven tricks with North-West deadlocked.

Also capitalizing on the “hop or bust” theme is the following layout, submitted by three solvers. Suppose West wins his major tops and shifts to a low club. Zero or seven! (In my ward that’s a Roman key-card response, but not here.)

 1 NT Northor South 7 8 7 10 9 8 7 6 5 K J 10 9 A K A A K A 8 7 6 5 4 3 2 Q J 10 9 8 K Q J 10 9 Q J Q Lead: any 6 5 4 3 2 6 5 4 3 2 4 3 2 —

Nicholas Greer: West makes his six top tricks; North makes four diamonds and three clubs. There’s no grand slam, but 6 NT will be a good sacrifice against the making 6  if E-W are vulnerable. With N-S also vulnerable, an exciting way to gain 1 IMP!

### Southward bound

The weakest possible South hand (rank sum 48) was submitted by Grant Peacock (migrating south for the winter?) though the combined North-South sum (155) overran the mark in the priority tiebreaker. Even so, I liked the necessity of South holding a key spade spot to avert a squeeze against North.

 1 NT Northor South 10 9 7 10 9 8 7 6 5 — K J 9 3 A K A K A A Q 10 8 7 6 5 4 Q J Q J K Q J 10 9 8 7 6 5 — Lead: any 8 6 5 4 3 2 4 3 2 4 3 2 2

If West cashes his major tops, 1 NT is easily made; but a diamond lead squeezes North: A club pitch allows West to establish clubs before either major can be developed; a spade allows West to cash his major tops and eventually score two club tricks; so North must pitch a heart. Then if West persists with clubs, N-S win three clubs and four spades (North unblocking 10-9); or if West cashes out to avoid the repeated endplays, North enjoys heart winners instead. Note that switching the 8 and 7 allows the defense to prevail.

Of the 11 perfect solvers, all produced essentially the same layout, which the tiebreaking conditions virtually forced in order to be optimal. Curiously, 10 of the 11 solvers gave West the best hand (four aces). So who was the odd man out? Oops, make that woman, as Tina Denlee shows once again that she marches to a different drummer. Well then, let’s check out her drumsticks! Apologies if that sounds crude, but for Campaign 2016 it fits the bill; and now that I think about it, 20 years ago it fit the Bill.

 1 NT Northor South 8 7 10 9 8 7 6 5 4 K J 9 2 K Q J 10 9 K Q J 10 9 8 Q J — A A A K A Q 10 8 7 6 5 4 3 Lead: any 7 6 5 4 3 2 6 5 4 3 2 3 2 —

Tina Denlee: I usually open 2 with the South hand (weak, both majors) asking North to bid his better major, but here North would pass. Unfortunately, 2  fails, so from now on I’ll be more sensible and open 1 NT.

Jamie Pearson: With East-West completely blocked, forced club leads will give North at least two clubs, and North will establish five diamond tricks before [East] can establish clubs.

Tom Slater: I’m not sure if this puzzle is a trick or a treat — the sense of deja vu is very strong, and I fear my attention is being diverted from something better. It would be nice to give North a fifth club to weaken the triple stop; but if [East] has only eight clubs, what can his other card be?

Aside from suit identity and swapping the East-West hands, the optimal solution allows two trivial variations: North’s 8 can be swapped with West’s 8, and/or North’s 7 can be swapped with South’s 7. In either or both cases, the rank sums are unchanged.

Charles Blair: You didn’t use freakness as a tiebreaker!

Interesting thought. Had I stipulated “lowest freakness” as a third tiebreaker, it would have eliminated any variation in the optimal solution (7-4-1-1 = freakness 8, while 7-4-2-0 = 9). But look at the bright side: I could have put you in last place by stipulating “not from Illinois.”

### Election central

Jamie Pearson: A win for No Trump with only 10 percent of the votes! I mean, points. No wonder certain people think this whole thing is rigged!

Tina Denlee: No Trump wins again! Looks like I’ll have to move to Canada… No, wait… I already did.

Dan Gheorghiu: The “muzzle” requirement you mention in the example seems most appropriate this year.

Yes, if enforced it could produce the first truthful debate:

Donald T:

Hillary C:

Write in: Donald Duck. I’m Richard Pavlicek and I approved this message.

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