Main   Study 8R41 by Richard Pavlicek  

Tricks in a Strain

Have you ever tried to count the number of sticks in a train? Oops, hold that glass of wine! I mean tricks in a strain. On a bridge deal, if you add the number of tricks each player can win in a specified strain (suit or notrump) the most likely total is 26. For example, if North-South can make 2 S, the most likely apportionment per declarer is: North 8, South 8, West 5, East 5 — or 26 total tricks in spades.

Random DealsSymmetric DealsDeficient StrainsAbundant Strains

Variations in this total are common if not rampant, and usually lower than 26, because the defense has the advantage of the opening lead. Only in rare cases is the opening lead a disadvantage, allowing the strain total to exceed 26.

Random Deals

About 15 years ago I created a database of 10,485,760 random deals. Each deal was double-dummy solved 20 times (4 players × 5 strains) — a project that took almost two years to complete. This resource has proved beneficial toward a number of studies, including this one.

Each row in the following table shows the number of times that trick total (first column) occurred in each strain (suit or notrump) on each deal in the database. The last two columns show the number of times all four suits, or all five strains, produced identical totals on the same deal. The blue-tinted rows at the bottom show the total number of occurrences and average trick total for the respective column.

Results of 10,485,760 Random Deals
TricksClubsDiamondsHeartsSpadesNotrumpAll SuitsAll Strains
0000031400
10000000
2000093800
30000700
40000235600
500003800
60000475300
700009300
80000866800
9000026200
1022431701900
11121053100
12243827193746700
131410815122300
141911792152187980900
1564737377297600
16109410881130112814142100
17614551637590905800
18620560416144614625819820
1928242874280927933063000
2037765379623787738002515268620
21179501777917644178228823800
22291502291464291964291080103925359914
231458311462761458021456252705721005
24268246326832352682091268339926727418318623947
251196098119560611954991193908922305263307854
26610178061013126102551610364743803681626105880058
271338126712811287125330
280131100
29+0000000
Totals10485760104857601048576010485760104857601736387911878
Mean25.182825.182725.182825.183024.026325.887225.9388

The mode occurrence of all columns is the gold-tinted row, 26 total tricks, but note that suit strains achieve this far more often than notrump. Also note the plentiful dispersion of totals below 26, and the scarcity of totals above 26. Indeed, the highest total of all is only 28, and it was achieved by only six strains in 10,485,760 deals (52,428,800 strains). Not too likely!

Observe the much greater diversity in notrump than suits. In fact there were 314 deals on which no player could win a single trick in notrump! This phenomenon could never occur in a suit, of course, because high trumps are destined to win tricks no matter where they lie. The lowest total for any suit was 10 total tricks, occurring on 11 occasions.

Another curiosity is that even numbers of total tricks occur far more often than odd numbers (ignoring the fluke case of 28). This is because both players in a partnership usually win the same number of tricks, hence the sum will be even. Odd totals arise only when a strain is right-sided by one player to prevent a damaging opening lead, and even then the total may become even if the difference is an even number of tricks, or if a similar situation exists for the other partnership.

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Symmetric Deals

Symmetric deals consist of four hands that are identical except for suit rotation. While almost never occurring in real life, these deals often exhibit curious behavior — rarely spectacular (stay tuned and you will see). I created a database of every possible symmetric deal if the cards 5-4-3-2 in each suit are considered alike. Further, to eliminate redundant rotations, each ace has a fixed location: S A West, H A North, D A East, C A South. This streamlining reduced the size of the database to 2,293,760 deals.

Each symmetric deal was solved 20 times (4 players × 5 strains) at double-dummy, but this was greatly simplified by the symmetry: Whatever West wins in spades, North wins in hearts, East in diamonds, and South in clubs, etc., and all four players must win the same number of tricks in notrump. Effectively this dictated only five double-dummy solutions per deal.

Each row in the following table shows the number of times that trick total (first column) occurred in each strain (suit or notrump) on each deal in the database. The last two columns show the number of times all four suits, or all five strains, produced identical totals on the same deal. The blue-tinted rows at the bottom show the total number of occurrences and average trick total for the respective column.

Results of 2,293,760 Symmetric Deals
TricksClubsDiamondsHeartsSpadesNotrumpAll SuitsAll Strains
0000012845700
10000000
20000000
30000000
4000015817900
50000000
60000000
70000000
80000139400
90000000
100000000
110000000
12000020856600
130000000
142113211321132113021130
15181818180180
16131271312713127131276089413127190
1761861861861806180
18305523055230552305520305520
192720272027202720027200
2068541685416854168541324887685412287
216788678867886788067880
2219612719612719612719612701961270
23385513855138551385510385510
247833267833267833267833261357293783326451223
2522159522159522159522159502215950
2692564992564992564992564909256490
271122112211221122011220
2829132913291329135408929131756
290000000
300000000
310000000
320000000
330000000
340000000
350000000
360000100
37+0000000
Totals229376022937602293760229376022937602293760455456
Mean24.451224.451224.451224.451219.491324.451223.9920

Observe that the columns for each suit and for “all suits” are identical, which is forced by the symmetry.

As with random deals, the mode occurrence for each suit is 26 total tricks. But look what happened in notrump: never 26 but a mode of 24. This is obvious once you realize that each player must win the same number of tricks in notrump, hence the total must always be divisible by four. Scanning the notrump column confirms this.

Here again notice the plentiful dispersion of totals below the mode, and the scarcity above. It seems like 28 is still the highest possible total… or is it? Cast your eyes downward to the red-tinted cell. Could this really be? One and only one deal producing 36 total tricks in notrump? Or is this a glitch or a typo? Read on and find out.

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Deficient Strains

Strains that do not yield the normal 26 total tricks are almost always deficient, producing fewer than 26, sometimes a lot fewer. This is because the defense strikes the first blow with the opening lead, which is sometimes a killer, depriving declarer of many tricks that he could have won on his own. Indeed, the total tricks in notrump can be zero, and it’s not as rare as you might think.

Zero tricks in notrump

The following deal occurred in a 1999 match against Bobby Goldman and Paul Soloway, where I was East:

1. 5 H
South
S A J 10 9 8 7 5
H 9 8 4 3
D 10 4
C
Makes
North
South
West
East
Deal
NT
0
0
0
0
0
 S
11
11
0
0
22
 H
11
11
0
0
22
 D
0
0
10
10
20
 C
0
0
11
11
22
S 6
H 6
D Q 9 8
C A K Q 10 9 8 6 2
Table S 2
H J 10
D A K J 6 5 2
C J 7 4 3
S K Q 4 3
H A K Q 7 5 2
D 7 3
C 5

Of course nobody was crazy enough to play in notrump. After a competitive auction, Soloway, South, bought the contract in 5 H, making seven when my partner led his (our) singleton spade. Oh well; better in five than against a slam.

While every strain is deficient, note the complete wipeout in notrump: zeros down-the-column, because the defense always has two long suits to run. Amusingly, this total might even be considered negative, as East-West could win the first 14 tricks if allowed.

Fourteen tricks in spades

The preceding anomaly can only exist in notrump. Suit strains must always allow tricks to be won, no matter who declares, because high trumps cannot be shut out. The lowest suit total I’ve ever noticed in actual play occurred on BBO in 2018:

2. 1 NT
South
S K J 10 2
H Q 10 9
D K 6 5
C 10 5 4
Makes
North
South
West
East
Deal
NT
6
6
3
3
18
 S
4
4
3
3
14
 H
7
7
3
3
20
 D
6
6
6
6
24
 C
6
6
7
7
26
S 9 8 6 4
H 5 3
D J 9 8 7 3 2
C K
Table S Q 3
H K 8 6 2
D A
C A Q 8 7 6 3
S A 7 5
H A J 7 4
D Q 10 4
C J 9 2

This deal seemed innocuous if not boring at the time. South became declarer in 1 NT after opening 1 C, which kept East from entering the bidding. The friendly diamond lead and club blockage allowed eight easy tricks with the heart finesse, and the S Q dropping added two more.

A subsequent look revealed the remarkably low spade makes. With North or South declarer, the defense can wreak havoc with a club lead, allowing a crossruff in the red suits to hold declarer to just four tricks. But if West or East is declarer, the defense can draw trumps, finesse hearts, and establish two diamond tricks while clubs are blocked, holding declarer to just three tricks.

Curious enough, but how low can you go? Is there an absolute limit for the fewest total tricks in a suit?

Eight tricks in spades

Yes there is, as the following construction illustrates:

3. S
South (WNE)
S 2
H A K Q J 10 9 8 7 6
D 3 2
C 2
Makes
North
South
West
East
Deal
NT
4
4
0
0
8
 S
4
4
0
0
8
 H
10
10
0
0
20
 D
4
4
8
8
24
 C
4
4
8
8
24
S 10 9 8 7
H 5 4 3
D 10 9 8
C A K 7
Table S 6 5 4 3
H
D A K Q J
C Q J 10 9 8
S A K Q J
H 2
D 7 6 5 4
C 6 5 4 3

In spades, if North or South declares, the defense scores a heart ruff then leads a spade, holding South to his four natural trump tricks. But if West or East is declarer, they get shut out completely! South draws trumps, then North runs hearts for 13 tricks. So the trick total in spades around the table is only eight, which I am confident cannot be reduced any further.

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Abundant Strains

We now arrive at the most interesting part of this study, at least for this writer. Only in extremely rare cases are the total tricks in a strain above 26. Indeed, attempts to create such a deal can be an exercise in futility. Trust me. I’ve wasted many hours on this.

Three spades both ways

To emphasize the frustration in obtaining abundant totals, consider this deal I constructed as a puzzle in 1998. The objective was to determine the optimum contract for each side, which amazingly happens to be 3 S for both.

4. 3 S
South (E)
S 9
H J 10 9
D A Q 4 3 2
C A K Q 2
Makes
North
South
West
East
Deal
NT
5
5
8
8
26
 S
4
9
4
9
26
 H
5
5
8
8
26
 D
4
5
8
9
26
 C
5
5
8
8
26
S
H
D J 10 9 8 7 6 5
C 8 7 6 5 4 3
Table S A K Q 8
H A K Q 8 7
D K
C J 10 9
S J 10 7 6 5 4 3 2
H 6 5 4 3 2
D
C

That’s right. South can make 3 S, and East can make 3 S — both against any defense — and plus 140 is the highest score either side can obtain. (Play it out to appreciate.) This delivers 18 tricks for just two players, so surely the total will exceed 26 for all four players. Right?

Wrong. Despite the bizarre apportionment in spades, the overall total is still 26 — not only in spades but in every other strain as well. Sigh. Just another ordinary deal!

Four spades both ways

Another incredible deal, constructed many years ago by Moses Ma and handed down to me via Bart Bramley:

5. 4 S
South (W)
S
H A K Q J 10 9 8 7 6 5 4 3
D
C 7
Makes
North
South
West
East
Deal
NT
0
12
1
0
13
 S
3
11
10
2
26
 H
12
12
1
0
25
 D
0
4
13
9
26
 C
0
5
13
8
26
S J 10 9 8
H
D A Q J 10 9 8 7 6 5
C
Table S 7 6 5 4
H
D 4 3
C A K Q J 10 9 8
S A K Q 3 2
H 2
D K 2
C 6 5 4 3 2

While the optimum contracts are far beyond (7 C or 7 D West, 6 NT South) consider the play in spades. If South is declarer, West can win only the D A and a trump trick, as declarer immediately clears trumps and has beaucoup hearts to win the rest — not only making 4 S but an overtrick.

Now consider the play by West. North can’t lead a spade, so assume a club, giving declarer five pitches as South follows suit. Then two rounds of diamonds leaves S J-10-9-8 D J-10 opposite S 7-6-5-4 C 9-8. A diamond is ruffed, and South is helpless to win more than his three top trumps — makingS. The play is slightly different with a heart lead (ruffed in dummy) but 10 tricks are easy. Is bridge a crazy game or what?

Nonetheless, despite 21 tricks being makable by two players, the trick total in spades is still only 26. Argh! Even a crazed construction like this does not exceed the norm.

Twenty-eight tricks in spades

Okay, enough misses already; it’s time to get abundant. (I hope I’ve made this abundantly clear.) In my database of 10,485,760 random deals there are only six instances — out of 52,428,800 strains — yielding 28 total tricks. Here’s one of those rare birds:

6. S
South (WNE)
S A 9
H 10 8 7 2
D A 10 4 2
C J 8 4
Makes
North
South
West
East
Deal
NT
6
6
6
6
24
 S
8
8
6
6
28
 H
6
6
7
7
26
 D
8
8
5
5
26
 C
5
5
8
8
26
S 10 4
H A K 9 5 4
D K 7 6
C K 6 5
Table S J 6 5 2
H J 3
D J 3
C A 10 9 3 2
S K Q 8 7 3
H Q 6
D Q 9 8 5
C Q 7

While three suits are normal (26 tricks) and notrump is deficient (24 tricks), note the spades. North or South can win 8 tricks, and West or East can win 6 tricks. For this abundance to occur, it means that whoever is on lead against a spade contract must sacrifice a trick. Though difficult to analyze completely, this becomes evident by the frozen nature of every suit. Like the zugzwang in chess, whoever moves first loses.

One of my projects about 10 years ago was to create a deal in which the 28 total tricks in spades were divided 7:7:7:7, i.e., so everyone makes 1 S. It wasn’t easy, but I persisted and succeeded. The pursuit is documented in Chasing Rainbows.

Twenty-eight tricks in notrump

Random deals with 28 tricks in notrump are extremely rare — just one in my database of 10,485,760 — but they’re easy to construct symmetrically. In fact my symmetric deal database has over 58,000! Here’s a cutie:

7. 1 NT
South (WNE)
S 9 8 6
H A 5 4 3
D K Q 2
C J 10 7
Makes
North
South
West
East
Deal
NT
7
7
7
7
28
 S
6
6
7
7
26
 H
7
7
6
6
26
 D
6
6
7
7
26
 C
7
7
6
6
26
S A 5 4 3
H K Q 2
D J 10 7
C 9 8 6
Table S J 10 7
H 9 8 6
D A 5 4 3
C K Q 2
S K Q 2
H J 10 7
D 9 8 6
C A 5 4 3

All the suit totals are completely normal, but everyone can make 1 NT! Whoever is on lead must begin the trail of decadence, like the toppling of dominoes, to surrender a trick. For example, West leads a low spade (nine, 10, queen); South leads a low club (nine, 10, queen); East leads a low diamond (nine, 10, queen); North leads a low heart (nine, 10, queen); then it’s back to West, who must now lose a trick outright. The play continues with back-and-forth exchanges, but declarer nets the odd trick because the defense had to commit first.

Thirty-six tricks in notrump

The next deal is the “one of a kind” alluded to previously. As incredible as it may be to accept, all four players can make 3 NT against any defense. Does this make a mockery of our game or what?

8. 3 NT
South (WNE)
S
H A 8 7 6 5 4 3 2
D K J
C Q 10 9
Makes
North
South
West
East
Deal
NT
9
9
9
9
36
 S
2
2
11
11
26
 H
11
11
2
2
26
 D
2
2
11
11
26
 C
11
11
2
2
26
S A 8 7 6 5 4 3 2
H K J
D Q 10 9
C
Table S Q 10 9
H
D A 8 7 6 5 4 3 2
C K J
S K J
H Q 10 9
D
C A 8 7 6 5 4 3 2

The gist of this symmetric layout is that every long suit is blocked. Whoever is on lead must allow declarer to unblock his or dummy’s suit, else lead fatally from K-J doubleton. For example, if West leads a low spade, dummy pitches a club to unblock that suit, then declarer ducks a club to establish it. West’s spade suit, while established, remains hopelessly blocked.

Note that all suit totals are exactly normal — 26 tricks — yet hidden in the abyss is the notrump phenomenon. Go figure.

A bold prediction

Except for the extraordinary 36-trick monster shown above, based on research and failed attempts, I believe it is impossible to exceed 28 total tricks in any strain (suit or notrump). Prove me wrong, somebody! I’d love to see it.

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© 2019 Richard Pavlicek