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September Symmetry


 by Richard Pavlicek

Three triads with an ‘e’ in the middle makes Sep-tem-ber a symmetric month, or so it is per-cei-ved by this mes-sen-ger in his hopes of being more res-pec-ted than neg-lec-ted. But never mind that. The symmetry this month is about bridge deals, so stick around and pay attention.

A symmetric deal requires all four suits to have the same sequential distribution with identical cards in each partition. Perforce this gives each hand one card of each rank (also known as a Margie) and the same distribution as each suit. Conversely, any hand with one card of each rank converts to a distinct symmetric deal. There are 413 = 67,108,864 different symmetric deals, assuming suit identity does not make a difference. If rotations are also considered the same symmetric deal, divide this number by four.

Indisputably, the most famous symmetric deal is the incredible fluke where all four players make 3 NT against any defense. Most bridge addicts have seen it, but as a ref-res-her it is rec-rea-ted below:

3 NT NorthS —
3 NT WestH A 8 7 6 5 4 3 2
D K J
C Q 10 9
S A 8 7 6 5 4 3 2TableS Q 10 9
H K JH —
D Q 10 9D A 8 7 6 5 4 3 2
C —C K J
S K J
H Q 10 9
3 NT EastD —
3 NT SouthC A 8 7 6 5 4 3 2

The key to its notoriety is that each long suit is blocked, but whoever is on lead must allow declarer to unblock one of his suits, else lead from K-J doubleton which is worse, yielding an overtrick. For example, if West leads a spade against South’s 3 NT, declarer pitches a club from dummy then ducks a club, freeing that suit to run. West’s spades are established but hopelessly blocked, so the defense can win only four tricks. Analogously, if West leads a diamond, declarer pitches a heart from hand then ducks a heart.

The above deal* is the only known case of this phenomenon, and based on my research, I believe it is unique. In fact there is no known deal where everyone makes two notrump, which I believe is impossible.

*actually 96 specific deals: the generic pattern 8-3-2-0 permutes 24 ways for South; for each, North is fixed and 4 arrangements exist for West-East.

One notrump, however, is a different story. Based on empirical evidence of one occurrence in 10,485,760 random deals, I guesstimate about five sextillion deals where everyone makes 1 NT. Among symmetric deals, there are over a million. All you need to do is find one!

Create a symmetric deal where everyone makes 1 NT against any defense.

Clever solvers have submitted their entries, with ties broken by the best poker hand.
Before reading further, venture your best guesses:

1. What is the generic hand pattern of the winning solution?
  4-3-3-3 5-4-3-1 5-5-2-1 5-5-3-0 6-5-1-1 6-6-1-0 7-6-0-0 8-3-2-0

2. What is the best possible poker hand?
  FL-AQT86 FL-AKJ97 FL-AKQJ9 SF-JT987 SF-QJT98 SF-KQJT9 SF-AKQJT

3. What is the worst possible poker hand?
  2P-3322A 2P-JJ22A ST-KQJT9 ST-AKQJT FL-75432 FL-K8642 FL-A9753

Quit Top September Symmetry

Cyrus Hettle Wins again!

For the month of September 2026, this puzzle was presented as a contest with 24 participants from 21 different locations. Thanks to all who entered, and congratulations to Cyrus Hettle (Kentucky) who was the only solver to find the optimal solution.* Cyrus is on a roll, winning five of the 10 contests he entered, including Half the Deck, Crooked Swiss, Most Cashing Tricks and Freeze Me! This extraordinary win percent of 50.00 far exceeds any other solver; next highest are Duncan Bell (England) 9/23 = 39.13, Andrew Spooner (Australia) 3/12 = 25.00, and Tim Broeken (Netherlands) 12/58 = 20.69.

*Solvers had to construct a deal for their solution. Multiple choice was only added for this write-up.

The object was to create a symmetric deal where all four players make 1 NT against any defense, and secondarily to have the best poker hand. Ranked below are the top 10 solvers, with ties broken by date and time of entry.

RankNameLocationTricks WonPoker Hand
1Cyrus HettleKentucky7SF-KQJT9
2Tina DenleeQuebec7SF-QJT98
3Nicholas GreerEngland7SF-T9876
4Chase LefevreGeorgia (US)7SF-98765
5Nick SmithEngland7SF-98765
6Lixin ZhengMaryland7SF-87654
7Richard SteinWashington7SF-87654
8Dan BakerTexas7FL-AKJ97
9Charles BlairIllinois7FL-AQT86
10Jim MundayNew Mexico7FL-AT876

For overall stats in all contests see the Puzzle Hall of Fame.

Solutions

Of the 67+ million symmetric deals, there are over 1.6 million where everyone makes 1 NT, so finding one of them was not a formidable task. The real challenge was to find the best poker hand, which by symmetric definition cannot have any pairs, trips or quads, and perforce must have at least an ace-high straight. Therefore, the goal was to beat that with a flush, or the elusive straight flush.

Tie me kangaroo down

In pursuit of a straight flush, this solver got lost in the outback:

1 NT AllS —
7 tricksH 2
D A Q 10 8 6 4
C K J 9 7 5 3
S 2 TableS K J 9 7 5 3
H A Q 10 8 6 4H —
D K J 9 7 5 3D 2
C —C A Q 10 8 6 4
S A Q 10 8 6 4
H K J 9 7 5 3
PokerD —
FL-AQT86C 2

Charles Blair: The return of the kangaroo straight flush. Hey…Abb-b-o-t-tt!

Would Bud Abbott accept A-Q-10-8-6? Not by his definition of a kangaroo straight.

Set me kangaroo loose

1 NT AllS —
7 tricksH —
D A K J 9 7 5 3
C Q 10 8 6 4 2
S — TableS Q 10 8 6 4 2
H A K J 9 7 5 3H —
D Q 10 8 6 4 2D —
C —C A K J 9 7 5 3
S A K J 9 7 5 3
H Q 10 8 6 4 2
PokerD —
FL-AKJ97C —

Dan Baker: I started with the idea of spiraling cards in each suit (S A-10-6-2 H K-9-5 D Q-8-4 C J-7-3) to freeze every suit, but that has the worst possible poker hand (a straight). Trying something similar in two suits at least gets a fairly strong flush.

If you played in Bud Abbott’s game, you could claim a true kangaroo straight flush!

From kangaroo to hamster

This solver managed to find a real straight flush, albeit a low one:

1 NT AllS —
7 tricksH 8 7 6 5 4 2
D K J 9 3
C A Q 10
S 8 7 6 5 4 2 TableS A Q 10
H K J 9 3H —
D A Q 10D 8 7 6 5 4 2
C —C K J 9 3
S K J 9 3
H A Q 10
PokerD —
SF-87654C 8 7 6 5 4 2

Richard Stein: After pressing the Make Deal button like a hamster, about the 30th try I found an eight-high straight flush.

Our top two solvers added fuel to my theme of symmetry:

Queen for the queen!

Our solving queen created a queen-high straight flush:

1 NT AllS A 3 2
7 tricksH —
D Q J 10 9 8
C K 7 6 5 4
S — TableS K 7 6 5 4
H Q J 10 9 8H A 3 2
D K 7 6 5 4D —
C A 3 2C Q J 10 9 8
S Q J 10 9 8
H K 7 6 5 4
PokerD A 3 2
SF-QJT98C —

Tina Denlee: After a low diamond lead, North wins cheaply, allowing declarer to set up spades with the D A entry; then at some point West or East must lead a club, heart or diamond, blowing the odd trick. After a heart lead, the H K scores early; spades are played, and after West scores his hearts he must blow the odd trick in a minor. In some sense, West’s only safe lead is his void, then North plays to his doom… Di-rec-tor!

King for the king!

Not to be outdone, our solving king created a king-high straight flush:

1 NT AllS 7
7 tricksH A 8 5 4 3 2
D K Q J 10 9 6
C —
S A 8 5 4 3 2 TableS —
H K Q J 10 9 6H 7
D —D A 8 5 4 3 2
C 7C K Q J 10 9 6
S K Q J 10 9 6
H —
PokerD 7
SF-KQJT9C A 8 5 4 3 2

Cyrus Hettle: I believe it’s impossible for all to make 1NT with a royal flush, though I don’t have a rigorous proof. Therefore, this deal has the highest possible straight flush; and as a bonus, the lowest one too.

Cyrus is correct. Solutions are possible with every straight flush except a royal flush.

Symmetry Mimicry

Did you know: (1) The number of divisors of 635,013,559,600 (total bridge hands) = the highest par score possible at all vulnerabilities = 1440. (2) The number of generic high-card holdings = the number of suit holdings without touching cards = the only Fibonacci bridge plus score = 610. (3) The number of suit holdings with 3 HCP = the number of suit holdings with 4 HCP = the number of suit holdings with 5 HCP = the number of suit holdings with 6 HCP = the number of suit holdings with 7 HCP = 1024. (4) And if you really dare, witness The Devil’s Product.

Symmetry has to end somewhere, and a few words from the out-of-sync will ensure it:

Richard Stein: I discovered the basic pattern by flinging cards around like a monkey.

Jim Munday: I only need seven tricks as declarer? Sounds like something right in my wheelhouse!

Jim underrates himself. I can assure you he regularly wins eight tricks in 3 NT, nine in 4 S, and a whopping 11 in slam contracts!

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© 2026 Richard Pavlicek