I'm crying inside.

If someone loses in chess its because they've made a mistake, no question about it. This doesn't disprove game theory; two game theory optimal actors would just be drawing against each other in the long run. The minute one of them deviates from that they'd now be losing.

If someone loses in chess its because they've made a mistake, no question about it. This doesn't disprove game theory; two game theory optimal actors would just be drawing against each other in the long run. — BitconnectCarlos

The larger truth here is that game theory, ought and should not be applied to reality, which indicates a disdain towards strategic outcomes derived from a rote logic of binary or higher order...

Analogously, think about chess for a moment. Given that chess is the oldest game in human history, and given that it is deterministic, then through enough iterations it can be demonstrated that both players, given a sufficiently long backlog of past historical games, are going to face situations where winning becomes... impossible.

What is left to entertain is simply a mistake committed by either player to ensure victory. Since both players, given enough iterations, become hyper-rational, then winning becomes impossible, and the game looses its "fun-factor".

I believe the analogy can be demonstrated for ANY deterministic game, and thus, game theory has been refuted for any deterministic game. — Shawn

You have a point. But your point holds up much more with tic-tac-toe than chess. For chess, a computer could possibly "memorize" the nearly infinite iterations of the game. But a person would never get there. Now to be fair, more experience with the game leads to MORE ability to create a stalemate, but unlike tic-tac-toe where 2 reasonably informed people will ALWAYS end in stalemate, chess would at best OCCASIONALLY end in stalemate, because no human could memorize every possibility. And if I remember correctly, the game "Go" has far more potential iterations than chess, so it would be even less possible to know enough to consistently reach a stalemate.

Given a sufficiently long enough interval to analyze all the potential iterations of a game, then a human being would become no different than a hyper-rational computer.

True or false?

Game theory is pretty evil stuff in my opinion.

Given a sufficiently long enough interval to analyze all the potential iterations of a game, then a human being would become no different than a hyper-rational computer.

True or false? — Shawn

I would say false. Let's say 50 years was enough time for me to view EVERY iteration possible in a chess game. Unfortunately, by the second year any normal human is already forgetting large chunks of the first year.

Am I missing something?

And a quick google search has the number of iterations in a chess game somewhere between 10 to the 111th power and 10 to the 123rd power, so 50 years is not going to be enough. (and "Go" is vastly more complex...more iterations than atoms in the universe).

I once memorized the periodic table for shits a giggles...yes, I am a real fun person, haha. After 3 MONTHS of totally ignoring the table, I was already forgetting portions. I am sure there are people with much better memories, but I have never heard of THAT much better.

Overall, I don't think you are wrong, just that, currently, humans are not capable of such feats. And our inability to memorize the totality of anything even a little complex (if we are counting chess or Go as complex, surely war or economics are vastly more complicated?), is why game theory would hold some applicability?

There is no question in my mind that you understand game theory better than I do, so if you feel my analysis is missing something related to the small details of game theory, feel free to let me know.

Yeah. I'm not quite sure if the point can be made with algorithms with infinite prior elasticity, manifesting in decisions that are strategically absolute, but I suppose the larger point that you sort of bring up is that despite however hard one might want to eliminate mistake making from human rationality, then mistakes will inevitably be made.

Kinda scary?

despite however hard one might want to eliminate mistake making from human rationality, then mistakes will inevitably be made. — Shawn

That is a much simpler way of saying it :up:

Kinda scary? — Shawn

It would be, if I wasn't so deeply conditioned to human fallibility :grimace:

In any sufficiently complex game, given enough iterations, it can be demonstrated that both players become hyper-rational, and thus a winning strategy cannot be entertained. — Shawn

Would you provide a source for this assertion, please. Thanks.

I actually came it up by myself as incredible as that sounds. I reached out to some mathematician friends to do a 3D analysis if possible.

Would you be able to help out?

I know very little about game theory other than Nash's work involved attracting fixed points (which I've dabbled with), but I'll be interested in what your friends have to say. Maybe there is a simple explanation. Fishfry? FDrake? Others?

What is game theory? How can we disprove game theory? Is there a single proposition that stands for game theory which can then be disproved?

I believe mathematical theories are not the same as scientific theories in that they can be disproved. :chin:

In any sufficiently complex game, given enough iterations, it can be demonstrated that both players become hyper-rational, and thus a winning strategy cannot be entertained. — Shawn

That's not true. There are games for which one side or the other has a winning strategy.

Analogously, think about chess for a moment. Given that chess is the oldest game in human history, and given that it is deterministic, then through enough iterations it can be demonstrated that both players, given a sufficiently long backlog of past historical games, are going to face situations where winning becomes... impossible. — Shawn

If you have such a demonstration it's publishable as the solution to an open problem. At present nobody knows whether white, with the first move, has a forced win or not.

That's not true — fishfry

Why not?

There are games for which one side or the other has a winning strategy. — fishfry

Oh, like chess? Have you played chess against a computer?

Why don't you test it out. Set a chess engine like Rybka, against Shredder, or Rybka vs Rybka, and see what happens?

Oh, like chess? Have you played chess against a computer?

Why don't you test it out. Set a chess engine like Rybka, against Shredder, or Rybka vs Rybka, and see what happens? — Shawn

This has nothing to do with the well-known fact that it is an open question as to whether white has a forced win with perfect play on both sides.

You have claimed a demonstration, your word, to the contrary. Have you got a link?

In any sufficiently complex game, given enough iterations, it can be demonstrated that both players become hyper-rational, and thus a winning strategy cannot be entertained. — Shawn

It's certainly the case that some games with perfect information (both sides know the current state of the game at all times) do have winning strategies. The example I thought of is a little technical but it's an interesting story I happen to know so I'll talk a little about it.

To be fair this is an infinitary game, it consists of a countably infinite sequence of moves. When mathematicians talk about game theory this is one class of games that's studied.

Choose a subset of the closed unit interval, $X \subset [0,1]$. There are two players, Alice and Bob as they're often called in these scenarios. Alice goes first, choosing a zero or a one. Bob then chooses a zero or a one, and so forth. They continue for a countably infinite sequence of moves, defining an infinite sequence of zeros and ones.

If you put an implied binary point in front of the resulting string, you have the binary representation of some real number in the unit interval. If that real number is in $X$ Alice wins; otherwise Bob wins.

Is there a winning strategy for Alice or Bob? Clearly it depends on which set $X$ we choose. For example suppose $X = [0, \frac{1}{2})$. That's the half-interval that includes 0 but excludes 1/2. Then Alice chooses 0 as her first move, and no matter what Bob does, Alice has already won the game.

I hope this example is clear. Any binary number .0xxxxxanything is automatically in the left half of the unit interval.

So it's clear that at least for SOME sets, there is a winning strategy for one player or the other.

The statement For every set there is a winning strategy is known as the axiom of determinacy. It's one of the alternative axioms set theorists like to play with. It's inconsistent with the axiom of choice which is why it's not generally adopted. It's been proven true for various interesting classes of set of reals.

[The description of the game in the Wiki article is a little different than how I described it but it's equivalent].

I admit this is the example that popped into my head when you claimed that with perfect play all games must be draws. This is not true.

Even for finite games, which can in principle be completely analyzed, some games have winning strategies and others don't. And as far as chess is concerned, it's unknown whether there is a winning strategy. The problem space is so vast that even though it's finite, our current computers can't evaluate the entire game tree.

ps -- Here is a reference for the solvability of chess.

https://en.wikipedia.org/wiki/Solving_chess

Are you trying to point out that natural advantageous positions exist? Yeah, sure. But, even in chess, where white get's the first move, or utilizes the strongest opening, being the Italian, it still is tantamount to having a player that is super-rational as black deciding that a stalemate is the only winning strategy.

This is basic game theory, 101. Should I go on?

This is basic game theory, 101. Should I go on? — Shawn

By all means, since you claim a solution to an open problem.

Winning strategies is something that can only exist for participants of a deterministic game where mistakes can be made. Once you have a super-rational player that is immune from making mistakes via forward and backward induction, along with no asymmetrical information problems, then winning becomes next to impossible.

Agree?

Winning strategies is something that can only exist for participants of a deterministic game where mistakes can be made. Once you have a super-rational player that is immune from making mistakes via forward and backward induction, along with no asymmetrical information problems, then winning becomes next to impossible.

Agree? — Shawn

No, that's simply wrong. There are games in which one player or the other has a forced win with perfect play on both sides. I gave you an example of one earlier.

Doesn't count. Otherwise it wouldn't be much of a game if one had to memorize a certain causal chain in a deterministic tree to ensure victory at all times.

Does that make sense?

Doesn't count. Otherwise it wouldn't be much of a game if one had to memorize a certain causal chain in a deterministic tree to ensure victory at all times.

Does that make sense? — Shawn

No, that's not what game theory's about. You seem more interested in the practical aspects of playing games. In that light your comments make more sense. When I think of game theory I think of the formal mathematical discipline of that name.

A game theoretic scenario entails that both players have an equal chance at winning... But, once you introduce the notion of a developing and advancement, through numerous iterations, and thus, hyper-rational players, then in some sense any notions of winning in a deterministic game becomes obsolete.

A game theoretic scenario entails that both players have an equal chance at winning... — Shawn

There are mathematical games in which one or the other player has a forced win with perfect play. You're redefining what game theory means. I'm curious. Where are you getting your definition from? You're correct that memorizing perfect play wouldn't be fun; but the idea of game theory is abstract, it's not about you and Uncle Fred sitting down for a friendly game of checkers.

https://en.wikipedia.org/wiki/Solved_game

For what it's worth, Hex is a game with a winning strategy for the first-player, but which is both mathematically interesting (apparently the existence of the strategy is equivalent to Brouwer's fixed point theorem) and fun to play with your Uncle Fred.

Hello Nagase,

What are your thoughts about the first sentiment proposed in this thread, about games being unitary or zero-sum, after an exhaustive method of rote analysis of winning and counter-winning strategies?

Well, it is wrong, if it is implying that no deterministic game of perfect information can have a winning strategy for one of the players; indeed, I just gave you a counter-example (Hex). You seem to be supposing that (e.g.) the second-player can, by rote analysis, always find a counter-move to the first player, but this is highly non-trivial, and, in fact, false, as the example of Hex illustrates. The first-player may have a strategy that (i) either forces the second player to make a series of moves or else (ii) makes the moves of the second player irrelevant.

Well, it is wrong, if it is implying that no deterministic game of perfect information can have a winning strategy for one of the players — Nagase

Perfect information seems to be irrelevant here. The point is that stochastically it would become deterministic after enough iterations of game playing, assuming that learning is possible.

If there is a deterministic game of perfect information with a winning strategy for one of the players, then, a fortiori, there is a deterministic game with a winning strategy for one of the players. So I don't understand your point.