Main     Puzzle 8N37 by Richard Pavlicek    

High Stakes Rubber

“Watson! Remember the break-in at Aerotec Research a few months back? We finally have a lead. Yesterday I was contacted by courier that Professor Moriarty wished to meet with me at the Portland Club. Once there he offered me a deal: He would reveal the location of the missing bomb-sight specs, if I would play one rubber of bridge against him for £100 a point.”

“Good heavens, Holmes! You couldn’t afford to lose 10 points.”

“Yes, but he allowed me to confer with Lestrade, who convinced Scotland Yard to back me. They wouldn’t risk putting you in the game, so Lestrade had to be my partner. Moriarty played with his chief goon, Colonel Moran. As West I won the cut for deal and bid and made a contract, and so did Moran, Lestrade and Moriarty on each successive deal — and that was it. The fourth deal ended the rubber.”

“Sorry, Holmes. Moriarty closing it out with a vulnerable game was a quirk of fate, but I guess he was bound to get lucky sometimes. I hope it didn’t cost the Yard too heavily.”

“You’re jumping to conclusions, Watson. The Yard didn’t lose a tuppence, and in fact acquired a substantial sum (less my fee of course) as well as a roll of microfilm with blueprints of the bomb sight. Whether to arrest Moriarty was an issue, but he had come forth voluntarily and was a man of his word, so Lestrade let him walk.”

“So how much did you win? What was the score?”

“Scotland Yard has sworn me to secrecy until month’s end, but I can tell you the results were remarkably different. Each level, strain, number of overtricks, and number of tricks won was unique. We also covered the entire gamut of risks, although one had to occur twice perforce.”

“Amazing, Holmes. What about honors? Let’s see… 100 honors would be…, um…, wait… £10,000 right there!

“Watson, your mathematical prowess makes me wonder why you chose the medical profession. But no; honors were never claimed. Indeed just about the opposite, as with all the police surveillance and commotion, we were claimed to be dishonorable members of the Portland Club.”

How much can you win for Scotland Yard?


Duncan Bell Wins Again!

In March 2017 this puzzle was presented as a challenge — with no help provided — inviting anyone who wished to submit a solution. Participation was solid with 66 persons giving it a try, of which 20 found the maximal margin of victory (3190) which dictated 31 tricks for Holmes. I cut the listing there, since the next highest possible margins (3140 3090 3020) were submitted by only five persons, and it fell rapidly after that. In fact, three entries had Holmes losing the rubber. Ouch! Or maybe this was an anti-Holmes rally in disguise.

Congratulations to Duncan Bell, England, who was first in the clubhouse with the winning solution. This is Duncan’s second win in a row, and his third in the last four puzzles — a new force to be reckoned with, as I alluded last month. Tim Broeken, Netherlands, my all-time leading winner, had to settle for second place; and Joseph DiMuro, California, completed the medal group in third.

Countrywise, I will unabashedly flaunt that the leaderboard reveals a rare win for my homeland: United States (7), Canada (6), England (3). So there! Our bidding and play may be suspect, but we damn sure know how to keep score.

1Duncan BellEngland319031
2Tim BroekenNetherlands319031
3Joseph DiMuroCalifornia319031
4Dan BakerTexas319031
5Foster TomBritish Columbia319031
6Gareth BirdsallEngland319031
7Tina DenleeQuebec319031
8Jamie PearsonOntario319031
9Dan GheorghiuBritish Columbia319031
10Ray LiuOntario319031
11Lin MurongOntario319031
12Julien ReichertFrance319031
13Stephen MerrimanNew Zealand319031
14Charles BlairIllinois319031
15Hendrik NigulEstonia319031
16Tom SlaterEngland319031
17Jonathan SiegelMaryland319031
18Jonathan WeinsteinMissouri319031
19Radu VasilescuPennsylvania319031
20Jon GreimanIllinois319031



Jon Greiman: Clearly, Moriarty took a pounding. You could also say he was ‘schooled’ — as long as it was Elementary.

Which brings me to the epilogue of the story:

“Watson! Despite the Yard’s substantial winnings, we were tricked. Moriarty reneged on his deal. Instead of handing over the microfilm as promised, he swallowed it. Lestrade arrested him immediately.”

“I knew you couldn’t trust that guy! So how will you recover the blueprints now, Holmes?”

“Alimentary, my dear Watson.”

A change of pace this month with a puzzle more about arithmetic than bridge, but all is fair that challenges the mind. The rubber-bridge element was also an alien encounter for some, as many of today’s players have never even played it. Most money-bridge clubs now play Chicago style (4-deal rounds) — including the one in my story:

Richard Morse: As a member of the Portland Club, I found this very amusing, and I alerted the Chairman to your problem. We now play Chicagos, re-cutting for partners after each round, but doubtless they played straight rubber bridge in Holmes’s day.

The reason for the demise of rubber bridge is obvious: A rubber only ends when one side completes two games, which could require anywhere from two to an infinite number of deals (in theory a rubber could never end). Therefore, the time required for each rubber varies greatly, which makes partnership changes/rotations awkward for clubs.

Redoubled redress

Finding the winning solution is mostly a matter of trial and error. A good start is to give Holmes the highest possible nonvulnerable making score on Deal 1, which is 7 NT redoubled (1980). Then on Deal 3, the highest possible vulnerable making score that’s not a game, which is 1 C (or 1 D) redoubled with six overtricks (2580). Alas, both win 13 tricks, which is disallowed, so the least expensive change is to keep the grand and score just five overtricks (2180) on Deal 3.

Next we must give the bad guys two games (else the rubber won’t end), one of which must be doubled to cover the gamut of all risks, and each must score an overtrick or two since “zero overtricks” is already taken with Deal 1. To minimize their score, the nonvulnerable game (Deal 2) should be doubled with one overtrick, and the vulnerable game undoubled with two overtricks. Suppose we try 2 H doubled making three (270) on Deal 2, and 4 S making six (180) on Deal 4. Adding the rubber bonus gives Moriarty 950, which subtracted from 4160 nets a whopping 3210 points, or £321,000.

Oops, that doesn’t compute, because Deals 3 and 4 both win 12 tricks. The only undoubled game with two overtricks to win fewer than 12 tricks is 3 NT, but notrump was already used on Deal 1. Aha! Let’s change the redoubled grand slam to spades (1940) which costs just 40 points, then Deal 4 can be 3 NT making five (160). The totals are then 930 versus 4120 for a net 3190 points, or £319,000. This provides the maximal gain, and the parlay is unique, except the majors on Deals 1 and 2 can be swapped, and Deal 3 can be either minor.

Jonathan Siegel: No rigorous proof, but I think 3190 is the best you can do… It’s better to make a grand slam (Deal 1) and five redoubled overtricks (Deal 3) than a small slam and six redoubled overtricks. Since Holmes’s grand has zero overtricks, Moriarty’s side has to be at least +1 and +2 on its contracts, and it’s better for the doubled contract to be +1 on Deal 2, when they’re nonvulnerable.

Dan Baker: One diamond redoubled +6 vulnerable seems obvious for Lestrade, but a grand slam for Holmes scores more than the sixth overtrick in 1 D redoubled (and a small slam) would. That means Moran and Moriarty must get one and two overtricks, both making game. The only two-overtrick undoubled game that gets fewer than 12 tricks is 3 NT, and since this limits the one-overtrick game to 10 tricks (and 3 NT can’t be used twice) it must be doubled (2 H× +1). This also allows Holmes’s grand slam (7 S) to be redoubled.

Could it really happen?

Over the years my veracity has often been questioned, but you can trust me explicitly. I solemnly swear that everything I write is the truth, the whole truth, or anything but the truth.

Jamie Pearson: Next contest: Construct a deal where 1 C redoubled making six has any chance of happening.

You won’t have to wait…
Just follow the bouncing ball:

Tina Denlee: On Deal 1 Holmes bid 7 H, which Moriarty doubled with the S A; but it was elementary from Moran’s behavior that he had no spade to lead, so Holmes redoubled. On Deal 2, trying to recoup their losses, Moran and Moriarty switched to the Tribliana bidding system: 2 D (majors) × ×× (you pick) P; 2 S × AP, stealing a cheap game. On Deal 3 Lestrade opened 1 C, Moriarty doubled, and Holmes redoubled; Moran had nothing to say so he passed. Oops! This was a penalty pass per Tribliana standards, so Moriarty also passed, regrettably to watch Lestrade score five overtricks. After that, Moriarty designated Moran as permanent dummy to achieve a normal result of 3 NT making five.

Foster Tom: Holmes reached 7 S with S Q-J-10-9-8-7 H — D — C K-Q-J-10-9-8-7 opposite S A-K H 8-7-6-5-4-3-2 D 5-4-3-2 C —, doubled by Moriarty on his right with S 3-2 H A-K-Q-J D A-K-Q-J-10-9 C A, and redoubled by Lestrade with the A-K of trumps.

Interesting layout. I suppose Moran (forever in doubt) led a trump. When Holmes led the C K, he had a prevision of Zia’s dubious tip, “If they don’t cover, they don’t have it,” so he ruffed as his only chance, dropping the bare ace from Moriarty. Bravo! No doubt Holmes also observed that ‘Moran’ is just one letter off from ‘moron’ which surely supports his deduction.

Worthless bits

Grant Peacock: How do I collect my fee?

Charles Blair: I would have thought Holmes would partner the Honorable Ronald Adair.

The Donald: Your reference to a “tuppence” being nearly worthless has inspired my latest bill before Congress. I propose that the U.S. Mint replace the half-dollar with a new coin called a trump-pence. The Vice President and I both believe this will greatly boost the economy.

That makes sense… from nearly worthless to totally. I’ll take the tuppence.


Acknowledgments to Sir Arthur Conan Doyle (1859-1930)
© 2017 Richard Pavlicek