I’m wondering from a mathematical perspective, what the best theoretically possible bridge system is.
I know this is a bit of a muddy question, such as what systems are allowed, and what criteria best is, so I’ll try to rigorously define it.
In human terms, “Bidding must be deterministic, but may be defensive. Try to maximize IMPs.” (To keep things simple, assume hands are scored using doubly dummy analysis, but this could be extended to include information given during the bidding in gameplay.)

To be mathematically precise:
Find a system S*, so that for all systems S, and summing over all possible bridge hands, the total sum of IMP(Table(S*, S, S*, S), Table(S, S*, S, S*)) >= 0.

Wherein a system is defined as a function that takes as input:

• The current game state. (Specifically, the vulnerability, the bid, the declarer, and how many passes until end of bidding.)
• A bridge hand.
• For each of the four players, the set of bridge hands that player is representing.

And gives as output:

• A legal bid, including pass.

For instance, a player playing Standard American is first to bid, and they pick up AQJ6c 9d K632h Q984s.
Then, we apply the function
SA(No sides vulnerable, no bid, 4 more passes to end auction, hand = AQJ6c 9d K632h Q984s, {all possible hands}, {all possible hands}, {all possible hands}, {all possible hands})
This evaluates to 1c.

Then, the next player playing the optimal system S*, and picking up hand H, would apply to get their bid:
S*(No sides vulnerable, 1c by E, 3 more passes to end auction, hand = H, {all possible hands SA would bid 1c with}, {all possible hands}, {all possible hands}, {all possible hands})

Note that if the first person was instead playing precision, their bid would instead be:
S*(No sides vulnerable, 1c by E, 3 more passes to end auction, hand = H, {all possible hands Precision would bid 1c with}, {all possible hands}, {all possible hands}, {all possible hands})
which may or may not be the same bid.

This problem is difficult: there are 52C13 possible bridge hands. That means there are 2^(52C13) sets of bridge hands. The number of systems, which takes sets of bridge hands as input, is much larger than that. It’s likely that even if S* were found, it would be too difficult to use in practice.