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Population dynamics

#1 User is online   hrothgar 

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Posted 2013-January-08, 08:22

I’ve long been interested in studying whether bidding systems exhibit transitivity. At the most basic level,

• If bidding system A is consistently better than bidding system B AND
• Bidding system B is consistently better than bidding system C
• Does it necessarily hold true that bidding system A is better than bidding system C?

It might be possible to run an experiment and gather some evidence, one way or another.

Consider the following hypothetical.

Seed a system with a large population of bidding systems in a roughly equal proportion. Hypothetically, your population might consist of 1000 robots, a hundred of which would be playing 2/1 GF. Another hundred would play Acol. Another 100 would play precision. So on down the line.

During the first round of the contest, the robots would be matched against one another at a 28 board team game with random hands. The IMP score for the match would determine the number of offspring for each of the competitors. If one team won a 30-0 blitz, the winning team would have 30 offspring while the losing team would have zero. If there were a tie, each team would have 15 offspring. At the end of the matches, we normalize the population down to a 1000 and run the second round of the Olympiad.

We continue the experiment until the system appears to converge on an equilibrium of some kind. (Either a monoculture or some optimal blend of populations)

Anyone else consider this problem at all interesting?
Alderaan delenda est
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#2 User is offline   helene_t 

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Posted 2013-January-08, 08:32

Doesn't it boil down to:
- what it the optimal defense (2nd seat opening scheme) against a Fantunes 1st seat pass?
- what it the optimal defense (2nd seat opening scheme) against a R/S 1st seat pass?
etc.

It may very well turn out that those are different.
The world would be such a happy place, if only everyone played Acol :) --- TramTicket
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#3 User is offline   gwnn 

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Posted 2013-January-08, 08:34

I think it's really interesting and I remember you mentioning your ideas about this a while ago. I can trivially think of three systems which are non-transitive, for example

system 1: strong club, stupid defence vs weak NT (e.g. bid 4S on any 13 cards)
system 2: strong NT, stupid defence vs strong club (e.g. bid 4S on any 13 cards)
system 3: weak NT, stupid defence vs strong NT (e.g. bid 4S on any 13 cards)

It would seem that each system would consistently lose to one system and consistently beat another.

Of course I assumed that competitive agreements constitute part of a given system but this is still a rather silly counterexample.
... and I can prove it with my usual, flawless logic.
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#4 User is offline   WellSpyder 

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Posted 2013-January-08, 09:34

View Postgwnn, on 2013-January-08, 08:34, said:

this is still a rather silly counterexample.

It may be silly, but it's still rather helpful! I thought instinctively on reading hrothgar's OP that the answer had to be non-transitive, but had no idea how to go about supporting this view.

It doesn't mean the problem isn't an interesting one for more typical systems, though. I wonder actually whether the experiment proposed would ever really converge? Maybe there are several local equilibria, and a particular set of hands could change the population enough to set off towards a different equilibrium, and so on?
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#5 User is online   mycroft 

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Posted 2013-January-08, 12:06

While it would be interesting to test, I think that evidence that "anti-X" systems work compared to "balanced" (otherwise why spend lots of time building defences?), when playing against X, combined with the pretty obvious (but unproven) "truth" that it is possible, even likely, that part of building an "anti-X" system will make it worse against "not X", would imply non-transitivity.

If I build a defence to FP that changes my second seat behaviour (not really a "defence", is it?), it will almost certainly be a net negative against "standard" against a non-forcing first-seat pass. Similarly, if I *don't* build a defence to FP, and play, say, K/S against it, I am going to lose badly. If K/S is better than EHAA, say, EHAA might be better than K/S against FP because it, too, doesn't make the "do your worst" bid very often, so second-seat FP doesn't get to open Pass; and EHAA-style preempts (especially 2!) will be better after an opening pass than K/S style (the 10-12 NT might not fare better than 12-14, though :-).

Also, the concept of "defence to system" tends to be orthogonal to "system"; my Precision defence is the same whether I'm playing Precision or K/S, for instance. And I have a separate "defence to system" to most systems, again orthogonal to what I play when I get to open.

But if you're discussing changes to system that include tweaking the openings (and therefore all the followups, at least slightly) to defend against system X which happens to be murdering you currently, I can't believe that that wouldn't influence how you do against system Z, say; and if you barely beat system Z before, and the change nets you +15 vs X, but converts vs. Z from +1 to -1, you're getting a net benefit if the metagame is 30% X, but a net loss if it's 99% Z.

I think if you ran your experiment to a monoculture or equilibrium, and investigated the end result, you could almost certainly build an "anti-metagame" system that would beat it. Once that settled into an equilibrium, you could repeat the process. Eventually after several of those, introduce a few of the eliminated "standard, balanced" systems into the mix, and see if they don't thrive against the systems tailored to beat the metagame.

Maybe I'm too corrupted by Magic analysis, where it's quite obvious that there are dominating strategies, but there is always a deckbeater; the problem is that that anti-X deck fares poorly against anything but X. So Anti-X strategies only work when the metagame stabilizes in a state that is "dominating and a few outliers". So that works once or twice, and then when enough people start running Anti-X, people come in with strategies that will kill Anti-X, and have a "good game" against X - X is better, but not enough better that skill or luck will overcome the deck frequently. And then... But the metagame changes in Magic much more frequently than Bridge, because:
- there's no partnership element, so the only person that has to know how to use the strategy is the one player;
- the "language" is a lot less constrained; and, of course
- Wizards keeps changing the card list, and one of their goals is to kill any dominating situation and restore balance. And the players know that.
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#6 User is offline   straube 

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Posted 2013-January-08, 16:05

I would think that each system would carry its own (optimal) defense. In this case, I would think that if A is better than B and B is better than C, then A is better than C.
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#7 User is offline   dake50 

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Posted 2013-January-10, 14:43

How about changing your question into
How near is system A to matching "what is in the cards"?
Then nearness of B and nearness of C.
The bids that most nearly reflect what the play can do
semm to concretely compare the systems thus transitive.
Trying to imagine some "perfect" defense to each is a
mirage - can never be sure "best" is found.
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#8 User is offline   Cthulhu D 

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Posted 2013-January-10, 22:45

You could easily test this with Jack or GIB - they both support enough systems - ACOL, 2/1, Moscito to make it fun.
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#9 User is offline   FM75 

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Posted 2013-January-11, 18:24

I find the question interesting. Your method of testing the concept seems flawed.

I think if you want to determine whether a bidding systems have a transitive relationship, you need to test the bidding systems only. To do this, you need a measurement of the system's bidding performance. Perhaps this means that you really should do something like pass off the hands to a double dummy analyzer after they have been bid in A vs B, B vs C, and C vs A "contests".

It is not clear that rounds and "population dynamics" addresses the question at all. For example, if you started in the extreme case with one hand rounds, and two robot teams per contest, it is entirely possible that A, or B, or C was "eliminated" in the first round (assuming elimination instead of proliferation). So a random event that was a weakness of a system could rule it as the weakest system. Suppose in the most extreme imaginable case - system A could win 95% of all possible bridge hands against B and C (or 96 and 93). (I will concede that the example might be computationally impossible.) Then there is a 5% chance that what you probably would agree is the strongest system would be eliminated as the weakest at the outset.

Substituting in multiplication instead of elimination does not seem to me to alter the approach materially. It front loads extreme randomness problems. So if you were to repeat the exercise serially, you would have a reasonable expectation of different outcomes. (Kind of the butterfly wing boundary condition effect.)

You might first try picking 3 truly trivial systems - such as always pass, always open 1N, always open 1M, with each never competing and each making at most one bid. Then see if you can show that transitivity was violated. It only takes one violation to disprove the conjecture.
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#10 User is offline   awm 

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Posted 2013-January-12, 06:19

It seems like a completely specified system would indicate what my call should be given any hand and any auction up to this point (including full specification of the meanings of all prior calls). This gives a finite (but ridiculously large) number of systems. With such a construction, I don't think my first seat bidding algorithm can be effected by your first-seat algorithm, since only one of us will be in first seat on any given auction. My second-seat algorithm obviously should depend on your first-seat algorithm (i.e. it's unlikely that I should play the same system in second seat after Pass if your pass was 0-7 playing magic diamond, 0-13 playing Roth-Stone, or 14+ playing some strong pass system); you might wish your first-seat algorithm to depend on my second-seat algorithm, but the laws of bridge do not really permit this.

The upshot is that if you view the specification of the Nth seat call as a distinct problem, you won't get these sorts of non-transitivities.

Of course, the reality is that all real bidding systems use certain techniques to reduce the number of specifications, such that it's common (for example) to play the same first-seat system and second-seat system after Pass (at least assuming the initial Pass shows weakness and not strength). It's only these techniques that open the door to your non-transitive relationships, and Gwnn already gave an easy construction where this happens.

Maybe a more interesting question is whether the optimum bidding system depends on what's going on at the other table(s). It seems like it actually does -- if we assume that given all four hands we can compute a "score" for each contract then the optimum bidding system must uniquely exist (given my definitions above). But such a score in real bridge depends on what goes on at the other table(s). For an easy example, suppose that I know the pair sitting my direction at the other table never bids slams. Then the payoff for a grand slam by my side is much less than if I can expect the other table to reach at least a small slam when it's "obvious" to do so. This substantially skews the odds, and might even mean that I want to avoid playing a relay system (which helps my grand slam bidding significantly but also gives opponents more chances to i.e. double for the lead).
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#11 User is offline   FM75 

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Posted 2013-January-12, 09:42

I should have been far clearer and more succinct.

There is no point in trying to establish whether bidding systems have a transitive property until you can define the ">" operator.

Until you can establish a metric that will order 2 systems, the transitive concept has no meaning.

It should be apparent, but sometimes it it important to state the obvious:
If the greater than operator is to establish that a > b, then it must always be true that a > b.

Though this might seem obvious, it is an extremely important restriction on the definition of "greater than" in the context of bridge bidding systems. Presumably this makes it clear that Monte Carlo style simulations can't serve as a metric.
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#12 User is online   hrothgar 

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Posted 2013-January-12, 10:36

View PostFM75, on 2013-January-12, 09:42, said:

It should be apparent, but sometimes it it important to state the obvious:
If the greater than operator is to establish that a > b, then it must always be true that a > b.

Though this might seem obvious, it is an extremely important restriction on the definition of "greater than" in the context of bridge bidding systems. Presumably this makes it clear that Monte Carlo style simulations can't serve as a metric.


I think that you are obsessing on minutia.

People use sampling methods all the time...

Moreover, given the amount of luck involved in bridge, I doubt that any (reasonable) system can said to be inferior on every possible set of 28 boards.
Alderaan delenda est
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