Puzzle 8M07   Main

# The Case of the Four Aces

by Richard Pavlicek

“Come, Watson, come! The game is afoot. Inspector Lestrade seeks our aid to build a case against a conniving perpetrator from across the pond. To set the stage suppose you hold 5 K-Q-J-10-5-4 J-6-5-4 K-5.”

“I say, Holmes, a perfect hand! Twenty-nine points with the right jack, but way too many cards.”

“Not cribbage, you bumbling quack; bridge. North on your left opens a spade, which by foreign convention shows five, and South bids two notrump showing a balanced hand of 12 to 14 high-card points. Your side is vulnerable. What would you do?”

“Can I meld my heart sequence and four fives?”

“I think not. You know my methods, Watson. I would pass, because I have a promising lead and a vulnerable three-heart bid is dangerous. North raises to three notrump, which ends the auction. What would you lead?”

“Ah! Now I remember bridge! In my youth I penned a book on it, Play of the Hand.
Wait a second, I’ll fetch it… Here it is on Page 165, fourth best… ten of hearts!”

 3 NT SouthE-W Vul ?  ?  ?  ? WestPassPass NORTH1 3 NT EastPassPass South2 NTPass 5 K Q J 10 5 4 J 6 5 4 K 5 Q J 10 9 6 6 K 9 3 9 8 6 2 Lead: K ?  ?  ?  ?

[Sigh] “Close enough. According to Lestrade, the East-West players claimed that South played like he could see through the backs of the cards, spending four aces on the first four tricks, to win nine tricks against any defense. If we could fathom his course of play, it could be crucial evidence. This hotshot yankee declarer must be held accountable to the Crown’s justice.”

“I’m sorry, Holmes. I don’t see any North-South cards.”

“Would it matter if you did? Never mind. That’s all the information Lestrade has. After filing charges, the East-West players mysteriously vanished, and the Yard fears the worst. South has lawyered up and won’t speak to anyone, and North conveniently swears he remembers nothing. The court will need further evidence for a conviction. I cannot offer it my full attention because of the Moriarty kidnapping case, and you’re obviously no help. Oh, wait! I’ll get Pavlicek to put it on his web site.”

Always happy to assist Scotland Yard!

Construct a North-South layout consistent with the facts of the case.

The primary goal is to have the most aces (Tricks 1-4) won out of necessity (i.e., no other option will suffice). A secondary goal is to equalize the North and South rank sums (ace = 14, king = 13, queen = 12, jack = 11, ten = 10, etc.).

## Dan Gheorghiu Wins

In May 2015 this puzzle was presented as a contest, inviting anyone who wished to submit a solution. Only 19 people were brave enough to give it a try, of which the 12 listed below constructed a layout to fit the conditions, and where South can make 3 NT against any defense after the K lead. Thanks to all who participated, though “all” may be an overbid.

Congratulations to Dan Gheorghiu, British Columbia, who was the only solver to submit the maximal solution for necessity* and equalize the N-S totals. (Charles Blair got the hard part but was off a few points in the totals.) Dan is a new participant this year and he’s been on a tear: eighth in February, fourth in March, third in April, and now first in May. That’s the good news! The bad news is that he may be out for June, having to testify for Scotland Yard.

*Maximum is 7, scored as follows: One point if winning the first trick is necessary. At tricks 2-4, one point if necessary to lead the chosen suit, and one point if necessary to play the ace. Also, the defense cannot deprive you of winning an ace by making an alternate play that doesn’t gain. For example, if South leads a club and North has the ace, West will play the king unless playing low would gain a trick for the defense.

Winner List
RankNameLocationNecessityNorthSouth
1Dan GheorghiuBritish Columbia7104104
2Charles BlairIllinois7102106
3Julien ReichertFrance5104104
4Nicholas GreerEngland4104104
5Jamie PearsonOntario3104104
7Jon GreimanIllinois3104104
8Grant PeacockMaryland3104104
9Tim BroekenNetherlands2104104
10Jacco HopNetherlands2100108
11Jim MundayMississippi1104104
12Christina SyrakopoulouGreece193115

 Puzzle 8M07   Main Top   The Case of the Four Aces

## Solution

After winning the A, it is almost mandatory to keep West off lead, so the A should be in dummy behind West’s king. Further, there’s no reason why South can’t have five clubs in his required balanced hand. This leads to a straightforward solution:

 3 NT South K 8 7 4 3 A 8 7 10 7 2 A 7 Trick1 W2 N3 S4 N Lead K 3 Q 10 2ndA6KK 3rd6AAA 4th2524 W-LW1W2W3W4 5 K Q J 10 5 4 J 6 5 4 K 5 Q J 10 9 6 6 K 9 3 9 8 6 2 Lead: K A 2 9 3 2 A Q 8 Q J 10 4 3

Jon Grieman: [At Trick 2, lead a spade] to the ace; Q covered and won; 10 covered and won. Clubs are then established to provide a ninth trick with the Q entry. This might count as zero aces won out of necessity, though once each suit is led, ducking would certainly be hazardous to the contract.

Jon was pessimistic, as he scores 3 on my necessity scale. It was indeed necessary to win the A, A and A after starting the suit. Points were lost because Trick 1 could have been ducked, and a different suit could have been led at Tricks 2-4. The play, however, would certainly arouse suspicion with declarer grabbing the first trick as if he knew hearts were 6-1.

“How con-veen-ient,” said the Church Lady.

### Julienne cut french fries

Our French connection took no chance of losing the first necessity point with this solution, as only a revoke could prevent winning Trick 1. The layout embraces most of the intended puzzle theme.

 3 NT South K 7 4 3 2 A 10 8 7 A J 7 4 Trick1 W2 N3 S4 S5 N Lead K 2 A Q 8 2ndA94KK 3rd6A7A2 4th25325 W-LW1W2W3W4L1 5 K Q J 10 5 4 J 6 5 4 K 5 Q J 10 9 6 6 K 9 3 9 8 6 2 Lead: K A 8 9 8 7 3 2 A Q 2 Q 10 3

Julien Reicher: At Trick 2 a low spade must be led, and even if playing the six does not cost a trick, I will assume East splits to force the ace. After that, a high heart can still enable nine tricks; but the only other option is the A, then a high club (or the last spade first but it’s not what I want), covered by West (normal play) and necessarily won. Then a diamond from dummy kills East, giving me nine tricks (three clubs, two diamonds, one heart and three spades after a throw-in).

Julien’s analysis is spot on, losing a point as noted for the optional high heart at Trick 3 and another for the optional spade at Trick 4. Still, 5 out of 7 necessity points topped all but two solvers. North and South totals are perfect at 104 each.

### The seven [percent] solution

Holmes was too busy with the Moriarty case to solve this one, but the Commonwealth did. The Queen will be pleased that Canada came through on her behalf. On the following layout, winning the first trick and playing each suit including its ace through Trick 4 are absolutely essential to succeed — 7 out of 7 necessity score — and the North-South hands each total 104.

 3 NT South K 8 7 3 2 A 7 10 8 7 A 7 4 Trick1 W2 N3 S4 S Lead K 8 A Q 2ndA94K 3rd6A7A 4th2532 W-LW1W2W3W4 5 K Q J 10 5 4 J 6 5 4 K 5 Q J 10 9 6 6 K 9 3 9 8 6 2 Lead: K A 4 9 8 3 2 A Q 2 Q J 10 3

Dan Gheorghiu: There are eight top tricks, and spades, diamonds or clubs could provide a ninth. Trick 3 is critical for the defense: If East follows with a low diamond, a diamond is led at Trick 5 toward the queen, then East is eventually endplayed with the fourth club to take two spade tricks (ducked) then North scores the 7. If East instead unblocks the K (preserving an entry to partner’s jack), declarer cashes his clubs and K, then exits with a heart to endplay West [assuming he keeps J-x] to win the last two tricks in diamonds. If the court would call on me, I will do my civic duty to testify against South.

Dan’s evidence looks pretty damning, so that rogue South player should pay for his evildoings. Pack your bags, Dan! Lestrade has booked you first class — but don’t head for the airport… it’ll be rail to New York, then RMS Mauretania to London. Bon voyage!

This puzzle was inspired by an old double-dummy problem, Ace Grabber.

## Club Closure

Charles Blair: “…wherever else that red herring [ 7] led your pack.” -The Adventure of the Priory School

Wouldn’t that be a black herring? Charles alludes to East’s curious 9-8-6-2, which suggests a solution of Q-J-10-7 opp. A-4-3, allowing four club winners (shades of last month’s Fail Safe). I wish I could claim such deviousness, but I went for symmetry not subterfuge. In case nobody noticed, the E-W spot cards were chosen to equalize their rank sums (104 each).

Meanwhile, back at the Yard…

Inspector Lestrade: You know, Holmes? In another life you’d have made an excellent criminal.

Sherlock Holmes: Yes. And you, sir, an excellent policeman.

 Puzzle 8M07   Main Top   The Case of the Four Aces

Acknowledgments to Sir Arthur Conan Doyle (1859-1930)