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The Law of Total Trash

A popular catchphrase of the times is “The Law of Total Tricks” which states: If each side were to play in its longest trump fit, the combined number of trumps is approximately equal to the combined number of winnable tricks. Unfortunately, the LOTT has dubious value because of its high variance. Consider this deal:

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

North-South have 8 spades and East-West have 9 hearts = 17 total trumps. With best defense N-S win 3 tricks and E-W win 7 = 10 total tricks. This deviation of seven puts quite a strain on “approximately equal,” which brings me to the puzzle:

What is the greatest possible deviation in the Law of Total Tricks?

Construct a deal that deviates from the LOTT by at least eight (the more the better). Longest trump fit must be distinct (no ties) as spades for N-S, and hearts for E-W.

Winnable tricks are determined at double-dummy using the more productive declarer (if it matters) for each side; e.g., if North wins more tricks than South in spades, North’s result applies.

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Dan Dang Wins!

In September-October 2011 this puzzle was presented as a contest, with 35 participants from 16 locations. As participation continues to decline, it may be time to end this series; else, who knows, I might hit zero and become undefined. Congratulations to the 17 who submitted correct solutions, ranked below by greatest absolute deviation. A plus sign indicates more tricks than the LOTT predicts; a minus sign indicates fewer. All achieved a 20-20 HCP split (designated tiebreaker) so ties are broken only by date and time of entry.

After almost a year of European domination, the winner has returned to American soil — at least continentally. Dang, we have a winner, eh? I learned to appreciate our northern neighbors in the heyday of Rick Moranis and Dave Thomas with their “Great White North” skits. My first contest in this series may have been a subconscious side effect, eh? And I might have reversed the declining turnout if I had followed Rick’s footsteps to title this one “Honey, I Shrunk the Law.”

RankNameLocationDeviation
1Dan DangBritish Columbia-12
2Hendrik NigulEstonia-12
3Ed BarnesAustralia-12
4Tim BroekenNetherlands+10
5James LawrenceEngland-10
6Tim BentleyMichigan-10
7Edouard BonnetFrance-10
8Reint OstendorfNetherlands-10
9Leigh MathesonAustralia+10
10Licai YeoSingapore-10
11Tony NorrisMassachusetts-10
12Richard SteinCalifornia+10
13Manuel PauloPortugal-9
14Jim MundayMississippi-9
15Wes EldredMinnesota+9
16Wayne SomervilleNorthern Ireland-8
17Jean-Christophe ClementFrance-8

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Solution

In this contest the longest trump fit for each side was required to be distinct; i.e., a deal was not allowed if two suits tied for a side’s longest fit. Besides simplifying the task, this eliminated the trivial case where each side has two 7-card fits and a pat hand like S A-K-Q-J-10-9-8 H — D A-K-Q-J-10-9 C — (14 trumps, 26 tricks). This restriction does not affect the solution, since the trivial case (+12) must be the greatest positive deviation, and negative deviations can produce the same (-12). Also note that if a side has two equal-length fits, it must be allowed to play in the more productive one, so an extreme negative deviation would be impossible.

A few respondents overlooked “by more productive declarers if it matters” on the entry form and submitted a deal that required a specific declarer to achieve a high deviation. Sorry, but just as in real life, it would be unfair to force a contract to be wrong-sided. For more insight on this subject check out my latest puzzle, Right-Sided Spades.

Three respondents produced deals with a negative deviation of 12, curiously based on entirely different layouts — or maybe that should be expected, since they live on three different continents. Let’s begin with our winner, who was not only first in the clubhouse but brought a touch of deja vu:

S by South S 6 5 4 3
H A K Q J 2
D 10 9 8
C 7
Trick
1. W
2. E
3. W
4. W
5. W
6. W
7. W
8. S
9. E
Lead
D 7
D 2
S A
S K
S Q
S J
C 8
C 2
D 3
2nd
8
Q
3
4
5
6
7
9
K
3rd
A
S 2
H 7
H 8
H 9
H 10
J
H 2
H 3
4th
J
9
7
8
9
10
A
Q
10
S A K Q J 2
H 6 5 4 3
D 7
C 10 9 8
Table S
H 10 9 8 7
D A 6 5 4 3 2
C K Q J
Dan Dang
British Columbia

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

Dan Dang: North or South plays in spades (8 spades); East or West plays in hearts (8 hearts). Nobody wins more than 2 tricks, so the LOTT is off by 12. The same result could be achieved if South held S x-x-x-x H — D A-K-Q C J-x-x-x-x-x, and East S — H x-x-x-x D J-x-x-x-x-x C A-K-Q.

Yes, and other holdings as well. My own construction, which inspired this contest, has the same deal pattern but with South holding S x-x-x-x H — D Q-J-10 C A-K-x-x-x-x, and East analogous.

Dan’s construction is also aesthetically pleasing with each hand having 10 HCP and one card of each rank — a far cry from the next deal, where a hail storm just wiped out the northeast.

Stoned in Estonia

Two nine-card fits are exploited in this version of the minus-12 genre, with each side being able to win only three tricks. Pretty weird, but the bridge world can’t hide from a composer gone amok:

S by North S 9 8 7 6 4 3
H
D 5
C 8 7 6 5 3 2
Trick
1. E
2. W
3. W
4. W
5. W
Lead
D 2
S A
S K
S Q
C A…
2nd
A
3
4
6
3rd
S J
H 2
H 4
H 5
4th
5
2
5
10
S A K Q J
H 10 9 3
D
C A K Q J 10 9
Table S
H 8 7 6 5 4 2
D 8 7 6 4 3 2
C 4
Hendrik Nigul
Estonia

S 10 5 2
H A K Q J
D A K Q J 10 9
C

Hendrik Nigul: Total trumps = 18; total tricks = 6; so the LOTT is off by 12. [No matter who declares], the defense can take the first four tricks with trumps (trick one is a ruff if the trump void is on lead) and then six side-suit winners.

Twelve the hard way

The creativity award (style points) for this month must go to the following exhibit, which bolsters the concept of its source down under. While North-South have the predictable 4-4 spade fit, the East-West heart fit is 8-0! Remarkably, the 8-bagger is blessed by a 3-2 trump break and can win only three tricks. On second thought, “blessed” should be replaced by “screwed.”

H by West S 9 8 7 6
H A Q 10
D A K Q J 10 9
C
Trick
1. N
2. N
3. N
4. N
5. N
6. N
7. S
8. N
Lead
D A
D K
D Q
D J
D 10
S 6
C 3
S 7
2nd
7
C 2
C 9
C 10
C J
10
H 3
J
3rd
8
S 2
S 3
S 4
S 5
H J
H 10
H K
4th
2
3
4
5
6
H 2
Q
H 4
S
H 9 8 7 6 5 4 3 2
D 6 5 4 3 2
C
Table S A K Q J 10
H
D 7
C A K Q J 10 9 2
Ed Barnes
Australia

S 5 4 3 2
H K J
D 8
C 8 7 6 5 4 3

Ed Barnes: This looks like minus 12, but it can’t be the [optimal] solution as I’ve failed to squeeze anyone out of a trick or two, or three.

A stylish comment, too. Hmm. Do I see a new puzzle? Trash the Law with a three-trick-gaining squeeze? Yes! And I see a perfect prize for the winner: a week with Ed in the outback. (Losers get two weeks.)

Richard Stein: Following the Law at the table has gotten me into more trouble than I care to admit, so it might be time, after 15 years or so, to put the Law to pasture. I’m sure the contest entries you choose to show will confirm that idea.

As this contest was running I received a note from Larry Cohen, who was amused by the title. When I invited him to submit an entry, he confided that the best he could come up with is a deal off by one. I’m kidding of course, as Larry is sharp as a tack, to which I can attest first-hand from my playing days. More importantly, however, he is one of the most good-natured guys in the game.

Thirteen, anyone?

Considering the wide array of winning solutions, there are probably other deal patterns as well, so I am not convinced that 12 is the greatest possible negative deviation. For the time being I’ll consider the contest still open. Show me 13 and I’ll star you in my next documentary “The Beerhunter.” A sure hit, eh?

As you exit, please put your empty bottles in the trash can.

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© 2011 Richard Pavlicek