Main   Puzzle 8M79 by Richard Pavlicek  

Notrump Ship Down

From time to time, almost everyone picks up a hand that should make 3 NT on its own, and sometimes this is true against any distribution. An obvious case would be S A-2 H A-2 D A-3-2 C A-K-Q-J-10-9, an ironclad nine tricks. Note, however, that replacing the C 9 with the eight would lose the guarantee, though still remain a big favorite.

Many hands do not qualify because of an unstopped suit. For example, S A-K-Q-J H A-K-Q-J D A-K-Q-J C K might win only one trick in notrump, and even that could require guessing the last card if LHO runs 12 clubs. For practical purposes, however, let’s assume a stopper in every suit; so S A-K-Q-J H A-K-Q D A-K C J-10-3-2 will ensure nine tricks, as with a club lead and return, you can split to leave the defense helpless.

Sometimes declarer is assured the ability to establish one or more tricks, which requires a double stopper in each of the other suits. For example, making 3 NT is guaranteed with S K-Q-J-10 H A-J-10 D A-K-Q-2 C A-K, but replacing either 10 with the nine would allow distributions where you would fail.

In rare cases the lead must be helpful, e.g., 3 NT is a lock with S K-Q-J-9 H A-K-Q-10 D A-K-J C A-Q, since LHO has to lead something.

So much for making 3 NT. For the puzzle let’s look at the darker side, as this ship is going down:

What is the strongest hand with every suit stopped that can be defeated in 3 NT?

Assume South is declarer with the strong hand. Valid solutions must show a deal where 3 NT can be defeated no matter how declarer plays. The main goal is to create the strongest South hand. A secondary goal (tiebreaker for the May 2016 contest) is for North to be as strong as possible. Hand strength is judged by the sum of all card ranks: Ace = 14, King = 13, Queen = 12, Jack = 11, etc.

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 spades.

3 NT South S

Table  East will get
what remains
West leads S


Tina Denlee Wins!

In May 2016 this puzzle was presented as a challenge — with no help provided — inviting anyone who wished to submit a solution. Participation set another high mark for the series, with 68 persons offering a try, of which the 32 listed were deemed to be correct. I chose 158 as the arbitrary cutoff for South’s rank sum, because the next highest dropped to 155.

Congratulations to Tina Denlee, Quebec, who was the first of 12 to find the optimal solution. Tina has been a regular, keen participant since October, and this is her fourth perfect solution, though her previous best finish was second by date/time. She also exudes a bit of craziness, which may be a prerequisite for solving these puzzles. (I’ll speak to my wardmaster about going coed for future admissions.)

RankNameLocationSouth SumNorth Sum
1Tina DenleeQuebec160117
2Martin VodickaSlovakia160117
3Tom SlaterEngland160117
4Duncan BellEngland160117
5Audrey KuehEngland160117
6Jonathan MestelEngland160117
7Jon GreimanIllinois160117
8Mike WenbleEngland160117
9Ray LiuOntario160117
10Jamie PearsonOntario160117
11Dan GheorghiuBritish Columbia160117
12Tim BroekenNetherlands160117
13Dean PokornyCroatia160116
14Julien ReichertFrance160115
15Grant PeacockMaryland160115
16Hendrik NigulEstonia160114
17Leigh MathesonAustralia160113
18Foster TomBritish Columbia160112
19John MayneCalifornia160112
20Jean-Christophe ClementFrance160111
21Marcin KrawczykPoland160109
22Dustin MillerOhio160107
23Jacco HopNetherlands159118
24Jurijs BalasovsLatvia159108
25Richard SteinCalifornia159107
26David BrooksAustralia159105
27Ivan LoyHong Kong159105
28Wayne SomervilleNorthern Ireland159102
29Jeff YutzlerVirginia158117
30John DoeOhio158111
31Nicholas GreerEngland15899
32Dan BakerTexas15898


As usual, before showing the optimal solution I’ll look at a few others that caught my fancy. This one from Nicholas Greer was well out of contention for the top spot but contained some interesting play elements.

3 NT South S 9
H 9 8 7 6 5
D 10 9 8 7 6
C 8 7
1. W
2. S
3. S
4. W
S 3
S 10
S 8
D 8
D 6
D 7
C 9
S J 8 4 3 2
H 3 2
D 3 2
C Q 4 3 2
Table S Q 7 6 5
H Q J 4
D Q 5 4
C J 6 5
S A K 10
H A K 10
C A K 10 9

Nicholas Greer: Three notrump goes down on the lead of a low spade. The S J must be kept to prevent declarer making three spade tricks, and the S 8 is needed to maintain flexibility in who wins the fourth spade (without it declarer can succeed on an endplay).

To illustrate, suppose declarer follows the above line, leading out spades. West wins Trick 3 and cashes a fourth spade but must not cash his last, which would squeeze East.

From the diagram at right, West must exit with a heart; jack, king. If declarer next plays C A-K and a club, West has three winners to cash. If he ducks a club, West can grab it and cash his last spade (East pitching a club) to squeeze South before East.

If declarer tries to prevent this maneuver by stripping West’s hearts early, West can reach East with the S 7.
H 9 8 7 6 5
D 10 9
C 8 7
S 4
H 3 2
D 3 2
C Q 4 3 2
Table S
H Q J 4
D Q 5 4
C J 6 5
H A K 10
C A K 10

Pipology 101

The next deal, from Hendrik Nigul, achieves the optimal rank sum for South (160) but the arrangement leaves only enough wood for North to have 114. While the deal itself is unspectacular (3 NT is routinely set with a club lead) I was amused by his pip theory… throw in a few bosons and he could go on tour with The Professor.

3 NT South S 9 8 7
H 9 8 7
D J 10 9 8 7
C J 10
S 4 3 2
H 4 3 2
D 2
C A 8 7 6 5 4
Table S Q 6 5
H Q 6 5
D 6 5 4 3
C 9 3 2
S A K J 10
H A K J 10

Hendrik Nigul: South can hold at most 167 pips (4-3-3-3, 13 tops). Pipcount reducers: non-top card by 1+; doubleton by 1; singleton by 3. To prove South’s pipcount is at most 160: South has no singleton (4-4-4-1 max = 159) and no 5-card suit (5-3-3-2 max = 159) so he must be 4-3-3-3 or 4-4-3-2. Case 1 (4-3-3-3): South must have a double stopper in each suit, and the lead must be into a solid suit (else ninth trick), so South must have some suit with only one top. If the suit is A-Q-J, ninth trick is there; if A-Q-10 or worse, then an extra trick in some other suit with two non-tops can be established. Case 2 (4-4-3-2): If doubleton is K-Q, ninth trick is there; if A-J or worse, there are nine tops; if A-Q or better, declarer can establish a ninth trick in the suit with two non-tops.

Can’t say I understand it… nonetheless, required reading at any nut house.

Abandon ship!

The optimal solution (rank sum of 160 South and 117 North) requires a specific South hand, except for suit designation, as indicated below. The North hand can be as shown, or with either seven moved to the J-10-9-8 suit (three cases). This month’s winner, Tina Denlee, was not only the first to submit it, but she also added an auction and story.

3 NT South S J 10
H J 10 9 8
D 10 9 8 7
C 9 8 7
N-S Vul


3 S
All Pass



2 S

3 NT
S A 9 2
H 7 6 5 4 3
C A 6 5 4 3
Table S 8 7 6 5 4 3
H 2
D 6 5 4 3 2
C 2
C K Q J 10

Tina Denlee: [Battle of Tribliana War Diary, Entry# Negative 1] East: “Spiky torpedoes 1 and 2… Fire 1! Fire 2 S!” South: “Enemy shoot from starboard! Double speed!” West: “Fire 3 S!” [South retreats to harbor in 3 NT, but West fires again with the S A and S 9.] South: “We got hit! Abandon ship!” [South cashes winners to minimize damage, but remaining torpedoes sink the ship.] South: “I had a 30-power warship! What the hell happened?”

I worry about you, Tina. You and Hendrik should collaborate on a book, and rest assured, nobody would know what happened.

Dan Gheorghiu: Placing two aces in East-West was counterintuitive, leaving South with no tenace in any suit; then a simple opening lead blows off the K-Q doubleton stopper, after which declarer can only cash eight tricks and surrender. Ugly, but effective.

With the North-South hands so dull, some solvers turned to East-West for artistry:

Tom Slater: I can’t find a way to make the North-South hands interesting, so at least this layout gives East-West a chance to have some fun. [Tom gave East S 9-8-7 and a club void, so a jettison play was necessary to unblock and run the spade suit].

Our March winner sought brewery advantage, swapping spades and diamonds:

Martin Vodicka: After a diamond lead (ace then nine), declarer has no chance. After [cashing his winners] he must lose the lead to West, then East will run his diamonds. Moreover, East wins a beer!

Bon voyage!

Jamie Pearson: It feels like 161 must be possible, but there’s only so many ways to jam an extra point into the South hand, and it never quite works. Frustrating!

John Doe: Showed your puzzle to my mentor… took him all of about five minutes to come up with a hand.

Never mind that! Just prove you’re a real person, not an unidentified body from the shipwreck. TopMain

© 2016 Richard Pavlicek