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 A-2 A-2 A-3-2 A-K-Q-J-10-9, an ironclad nine tricks. Note, however, that replacing the 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, A-K-Q-J A-K-Q-J A-K-Q-J K might win only one trick in notrump, and even that could require guessing the last card if LHO ran 12 clubs. Therefore, a stopper in every suit is mandatory; e.g., A-K-Q-J A-K-Q A-K 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 K-Q-J-10 A-J-10 A-K-Q-2 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 K-Q-J-9 A-K-Q-10 A-K-J A-Q, since LHO has to lead something.
So much for making 3 NT. For the puzzle lets look at the dark side. 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.
Enter a deal and click Verify. Use the help provided to make corrections and repeat. See how many tries it takes you to discover the winning deal. North-Souths longest fit must be in diamonds, and their shortest fit in spades.
In May 2016 this puzzle was presented as a contest, 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 arbitrarily chose 158 (South sum) as the listing cutoff, because the next best entry 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. (Ill speak to my wardmaster about going coed for future admissions.)
Ranking is by highest South sum, highest North sum, and date-time of entry, in that order of priority.
As usual, before showing the optimal solution Ill 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.
Nicholas Greer: Three notrump goes down on the lead of a low spade. The J must be kept to prevent declarer making three spade tricks, and the 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, and West cashes a fourth spade to leave this position:
West must not cash his last spade (it would squeeze East) but exit with a heart; jack, king. If declarer next wins the A and ducks a club, West grabs it and cashes his last spade to squeeze South before East, who pitches a club. If declarer instead cashed A-K before exiting, West would have three winners to cash.
If declarer tried to prevent this by stripping Wests hearts early (before leading the 10), West could reach East with the 7 instead of cashing the 8.
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 count theory.
Hendrik Nigul: South can hold at most 167 pips (4-3-3-3, 13 tops). Pip count reducers: non-top card by 1+; doubleton by 1; singleton by 3. To prove Souths pip count 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.
throw in a few bosons and you could go on tour with the Professor.
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 months winner, Tina Denlee, was not only the first to submit it, but she supplied an auction and story.
Tina Denlee: [Battle of Tribliana War Diary, Entry# Minus 1] East: Spiky torpedoes 1 and 2
Fire 1! Fire 2 ! South: Enemy shoot from starboard! Double speed! West: Fire 3 ! [South retreats to 3 NT harbor, but West fires again with the A and 9.] South: We got hit! Abandon ship! [South cashes out to minimize damage but the ship is sunk.] South: I had a 30-power warship! What the hell happened?
I worry about Tina. She 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 rather dull, this solver added some artistry:
Tom Slater: I cant 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 9-8-7 and a club void, so a jettison play was necessary to unblock and run the spade suit].
Our March winner found brewery advantage in 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!
Jamie Pearson: It feels like 161 must be possible, but theres only so many ways to jam an extra point into the South hand, and it never quite works. Frustrating!
John Doe: I showed your puzzle to my mentor
took him all of about five minutes to come up with a hand.
Never mind that. Are you a real person, or an unidentified body from the shipwreck? (Normally I reject entries without real names but let this one slide since I cant be sure.)
© 2016 Richard Pavlicek