J 4 3 2
K 7 6
A Q 5
Q 8 7
10 6 5
A Q 5
J 9 8
J 4 3 2
Q 8 7
10 4 3 2
K 7 6
A K 9
A K 9
J 9 8
10 4 3 2
10 6 5
There are no symmetric deals where everybody can make the same suit contract, but in this deal, we find that every declarer can make
1 NT,
1 ♠, and
1 ♥. No higher contracts make.
The contract
1 ♦ makes when declared by North or South.
The contract
1 ♣ makes when declared by East or West.
I have yet to write up a analysis of why
1 ♠ makes from each seat, so I leave it as an exercise for the reader.
Richard Pavlicek has
previously found a deal where everybody makes
1 ♠, but in that deal, some declarers can make more than seven tricks in other suits.
I found this deal by combining two symmetric deals.
In this first symmetric deal, West can only make six tricks in spades, but every other seat can make seven tricks in spades.
J 4 3 2
Q 8 7
A K 9
10 6 5
10 6 5
J 4 3 2
Q 8 7
A K 9
Q 8 7
A K 9
10 6 5
J 4 3 2
A K 9
10 6 5
J 4 3 2
Q 8 7
In this second symmetric deal, East can only make six tricks in spades, but every other seat can make seven tricks.
10 4 3 2
J 9 8
A Q 5
K 7 6
K 7 6
10 4 3 2
J 9 8
A Q 5
J 9 8
A Q 5
K 7 6
10 4 3 2
A Q 5
K 7 6
10 4 3 2
J 9 8
On an interesting side note, in both of these symmetric deals, there are four one-level suit contracts that do not make. In both examples, each seat fails to make seven tricks in one suit.
But in our example where
1 ♠ makes from every seat, there are also exactly four one-level contracts which fail, namely,
1 ♦ by East and West, and
1 ♣ by North and South.
Is this possibly the mathematical limit, or are there deals where fewer one-level contracts fail? Ideally, is it possible for every hand to make every one-level suit contract?