Traveler's Dilemma

[Yat]

"Restricted to [2,n]" simply means that any option other than those between (and including) 2 and n have been eliminated. It does not imply that every option between 2 and n is definitely valid.
In other words, the definition of "restricted" means that, if in fact Q can only play 2, then "Q is restricted to [2,x]" for any x is a true statement. We're not making subsequent mutually exclusive statements; we're making subsequent stronger statements.

I know, and it is the problem. The only thing which allows to restrict the range from [2,n+1] to [2,n-1] is the assumption that Q could actually play n, which is proved false in the next step.

No, it's the assumption that P could actually play n.
Maybe that realization will help.

I was talking about Q's range, but anyway this is exactly the same thing, because everything is you proof is based on a perfect symetry between P and Q.

P restricts his range from [2,n+1] to [2,n] because he knows Q's range is restricts [2,n+1], and when you are facing this range n is always better than n+1. But if P's range was not [2,n+1]n but [2,n-1]U{n+1}, he could not restrict his own range to [2,n-1], because if Q's range is [2,n+1], playing n+1 could actually be a better choice than playing n-1.
The problem is, a few steps later, P will restrict his range to [2,n-1], ruling out the n option.
But if n is not a rational option, restrictin P's range from [2,n+1] to [2,n] was based on an assumption that was proved wrong later :
As you said earlier, if 2 is P's only rational option, then restricting P's range to [2,x] is valid. But as you don't know which elements of this range are actually rational, there are some things which can't be done.

Imagine 43 is a forbidden choice. It is right to say that P's range is [2,100], because all rational options are contained in this set. It is also right to say that P's range is [2,44], and that Q's range is [2,44], at one point of the reasonning.
But then if you say that, as P knows Q's range is resticted to [2,44], then P can restrict his own range to [2,43], you make the exact same reasonning that the one you made in your actual proof. 43 is not a rational option, but it's not more of a problem than in your proof : you prove later that 43 is not a rational option too.

And in the situation where 43 is forbidden, the proof would be wrong, because the rational play is 44. So, if I take your definition

the definition of "restricted" means that, if in fact Q can only play 2, then "Q is restricted to [2,x]" for any x is a true statement

what is wrong with the proof where 43 is forbidden ?

[Chen]

Here is the argument restated a little more cleanly:

100 is dominated by 99 (also by 98). This means that any strategy that includes 100 can be made better, with no sacrifice, by assigning any probability of 100 to 99 (or 98). Therefore no strategy that includes 100 is optimal, therefore no rational player can play 100.

Lets expand the last statement. I'm sure this is where the confusion is coming in. "No strategy that includes 100 is optimal". This is true only because 99 (or 98) is playable. If our set was, for some reason [2, 97] U [100], its evident that 100 is not suboptimal over the whole range (if the opponent choses 100, the optimal response is also 100). The question that keeps coming up is why 100 can be removed from the range, if later on in the process the numbers that MAKE 100 suboptimal are themselves removed. Is there some sort of lack of dependence later on? Is this just a "common sense" argument that does not apply to a full logical argument? No one has answered this question. Its the same question Ender brought up and is being revisited now.

[Puck]

I think we have tried to answer this question, but the people raising it don't seem to understand the answer we're posing, and I'm not sure how to explain it further.

Even after we know 99 and 98 are ultimately not rational plays, it is still true that any strategy with a 100 in it can be improved by assigning that probability to 99 or 98. It's just that those strategies can be further improved, so they are not rational either. But they are still better than picking 100.

There is really no contradiction here. "Common sense" says there is, but there is not. 99 is only dominated in a universe where no one will play 100. Once we have determined that this is the universe we are in, the reasons for it do not matter.

[quintopia]

Here is the argument restated a little more cleanly:

100 is dominated by 99 (also by 98). This means that any strategy that includes 100 can be made better, with no sacrifice, by assigning any probability of 100 to 99 (or 98). Therefore no strategy that includes 100 is optimal, therefore no rational player can play 100.

Lets expand the last statement. I'm sure this is where the confusion is coming in. "No strategy that includes 100 is optimal". This is true only because 99 (or 98) is playable. If our set was, for some reason [2, 97] U [100], its evident that 100 is not suboptimal over the whole range (if the opponent choses 100, the optimal response is also 100). The question that keeps coming up is why 100 can be removed from the range, if later on in the process the numbers that MAKE 100 suboptimal are themselves removed. Is there some sort of lack of dependence later on? Is this just a "common sense" argument that does not apply to a full logical argument? No one has answered this question. Its the same question Ender brought up and is being revisited now.

If your know your opponent can only play some thing in the set [2,97]U{100} then 100 is a suboptimal play for you because 99 is at least as good and sometimes better than 100 for every choice in that set. Thus, since you won't play 100 it would be irrational for your opponent to do so and hence you know that your opponent is actually restricted to [2,97]. Any time you start to say "Ah but we eliminated x as a rational choice so why shouldn't my opponent reconsider x+1?" then you can answer "because if he does, it would be in my best interest to reconsider x" and you'll come to the same conclusion you did before.

Also fyjham, thereason it seems like your objections from one argument are frequently countered with conclusions from the other is simply that the two are really just two different ways of stating the same argument. Method 2 states the induction proof in the usual form while Method 1 just spells out every step in the induction. They are the same argument and so if you have a problem with one way of writing it sometimes it helps to see it in another light: you find that your problem was just an illusion after all.

[Yat]

If your know your opponent can only play some thing in the set [2,97]U{100} then 100 is a suboptimal play for you because 99 is at least as good and sometimes better than 100 for every choice in that set.

That is true if 99 is still in our range. Chen said

If our set was, for some reason [2, 97] U [100]

I guess it needs to be specified that, the situation he is speaking about is the case where both players have restricted their range to [2,97]U{100}.

[Puck]

Yes, but that makes no difference. The fact remains that 99 is still a legal play, though not rational. The fact also remains that 100 is not rational.

How would I restrict my own range to [2, 97]U{100}? Would it be rational for me to reconsider 100, which was already eliminated, but not also reconsider 99, which has also been eliminated?

[quintopia]

Theorem: There is no situation in which it is irrational to play 99 but some player can rationally play 100.

Proof: Let both players have determined that 99 is an irrational play. Assume that player 1 can determine from the rules of the game that it is now rational to play 100 and therefore will play it. Then, by common knowledge of rationality and knowledge of those same rules, player 2 can determine that player 1 will play 100. In this situation, the play that will maximize player 2's gain is 99, so that is the rational move for player 2. But this is a contradiction, because player 2 cannot simultaneously know that is rational to play 99 and it is irrational to play 99. Thus, our assumption is false and 100 will not be played. QED.

[Chen]

Theorem: There is no situation in which it is irrational to play 99 but some player can rationally play 100.

Proof: Let both players have determined that 99 is an irrational play. Assume that player 1 can determine from the rules of the game that it is now rational to play 100 and therefore will play it. Then, by common knowledge of rationality and knowledge of those same rules, player 2 can determine that player 1 will play 100. In this situation, the play that will maximize player 2's gain is 99, so that is the rational move for player 2. But this is a contradiction, because player 2 cannot simultaneously know that is rational to play 99 and it is irrational to play 99. Thus, our assumption is false and 100 will not be played. QED.

I think this is where I am having an issue. In the original proof 100 is determined to be irrational because the choice of 99 results in more gain. The implication here is that 99 is a rational choice (since our players will not play irrational choices). Yet in the next step it becomes an irrational choice since 98 results in more gain when played against 99. The sequential elimination I guess is what is throwing me off. This duality of 99 being both rational (at one point) and irrational (afterwards) is where I saw the contradiction.

[Puck]

99 does not have to be a rational choice for it to be a better choice than 100. Even though you know that your final strategy will not contain a 99, it is still true that any strategy containing 100 can be improved by switching those 100's to 99's. You now know that the strategy can be improved further, but the first step is still valid.

Again, 99 does not have to be a rational choice for it to be better than 100 - at the end of the day, all choices except 2 are irrational anyway. But our rational actor, not originally knowing which play is the rational one, must have some way of eliminating the irrational ones.

[quintopia]

99 is only rational in certain situations. Like when your opponent is going to play 100. But a rational opponent will not play 100, so in the game we are discussing, 99 is irrational. It's not a matter of it being "rational before and irrational after" but simply a matter of being "rational in situations which can be ruled out and irrational in situations that actually occur."

[thc]

Any time you start to say "Ah but we eliminated x as a rational choice so why shouldn't my opponent reconsider x+1?" then you can answer "because if he does, it would be in my best interest to reconsider x" and you'll come to the same conclusion you did before.

Actually, if you reconsidered y>x, wouldn't you come to a completely different conclusion?

To reconsider, your range would be [2,x-1] U [x+1,100]. Following the same inductive reasoning starting back from 100, you'd arrive at x+1. Since you have eliminated x, there is no dominant strategy to x+1 and you stop at x+1. Contradiction?

[Puck]

First of all, that's incorrect. x-1 dominates x+1. Second, why are we choosing to reconsider numbers arbitrarily? "We earlier determined 100 is irrational, but we're going to reconsider it anyway; we also earlier determined x is irrational, but we're not going to reconsider that." Why?

[VMLM3]

I think I've got this one worked out.

Assume the antique's real value is n
Also assume neither of then know of the manager's little game.

Spoiler:

Lucy should seduce Peter and offer to sleep with him.

Lucy could try and agree with Peter to inflate n by s (so they'd both write down n+s). However assuming Peter's a cheap bastard and Lucy's a greedy bitch, Lucy will come to the conclusion that Peter will try and cheat her by writing down n+s-1. Obviously writing down n+s-2 won't solve this problem because Peter's inferred response will be to write down n+s-3, at which point Lucy realizes that no matter what she writes down Peter will always try to write one less. By offering to sleep with Peter she can later steal both antiques and the money Peter got out of the transaction and quickly hightail out of this hypothetical situation.

If Peter downright refuses, then she knows he's an honest idiot and writes down 99.

[thomblake]

I think I've got this one worked out.

Spoiler:

If Peter downright refuses, then she knows he's an honest idiot and writes down 99.

Spoiler:

I don't think it was in the problem statement that Lucy is hot.

[thomblake]

I think the reason intuitions are running counter to the mathematical solution is that we're encountering the sorites paradox. Any reasoning using mathematical induction on things people care about runs the risk of encountering this paradox. An example of the 'paradox':

Suppose we have n grains of sand, which are not enough to be called a 'heap'. Obviously, n+1 grains of sand also will not constitute a heap. Also, 1 grain of sand does not constitute a 'heap'. By mathematical induction, no amount of grains of sand constitutes a heap. But there are heaps constituted by grains of sand.

In the traveler's dilemma, it's obvious that playing x-1 is better than playing x for every x, and yet the outcome of $2 is obviously worse than the outcome of roughly $100 they'd both get if they'd just played high. If the game is zero-sum, then the play of 2 might still be justified; however, in real life we do not find ourselves primarily in zero-sum games.

As an aside, it's obvious to a virtuous person that the correct 'play' is the actual value of the item, and so that would also make the explanation counterintuitive.

[VMLM3]

Damn... my half hearted troll attempt foiled in one fell swoop.

Response:

Spoiler:

If both Peter and Lucy are willing to play the game "rationally" (as in economically rational), that means they'd probably find each other attractive anyway, each recognizing a geeky equal and/or equally cheap individual in the other.

Is it really that obvious to the virtuous person? The "honest" response to the question assumes that Lucy doesn't realize or doesn't care that Peter might get away with stealing a bit over the value of the item from her if she writes down the actual value of the item. If the person is virtuous, he/she will, probabilistically, also be righteous. So the problem becomes a moral dilemma:
Does Lucy, as a virtuous yet rational agent:
Allow Peter to get away with stealing from her?
Try to impose her moral principles on Peter by "teaching him a lesson"?

The "rational" answer is already somewhere in this thread, I've already read it. You might consider this more of a philosophical question really.

[Silas]

I think the reason intuitions are running counter to the mathematical solution is that we're encountering the [S]orites paradox.

I disagree. I don't really see the Sorites paradox coming into play here- it's just an observation about the categories people use to think about the world, and doesn't have much to do with behavior. The paradox that I think underlies the contention here is Newcomb's paradox. I would bet a significant sum of money that there's a strong correlation between supporting the dominance answer in Newcomb's case and supporting the (2, 2) case in the TD.

[Ghona]

Here's my question:

Given the same information and motivations, will a perfectly rational person always come to the same decision making strategy?

If the answer is yes, it seems obvious that a perfectly rational person would be aware of this, and pick 100. Your opponent will always end up at the same decision as you.

I don't see any probabilistic strategy that increases the average return beyond 100-100, and since what you're doing is exactly equivalent to writing down a strategy and then having two robots follow it, getting the return from robot A instead of robot 1.

If the answer is no, it's pointless to argue about which strategy is the "correct" one, since you've already stated that there are multiple possible strategies. Since this thread has how many pages? I think we can discard that answer.

[Puck]

Ghona:

Rationality does not imply superrationality. See the previous 18 pages of discussion. It seems obvious that both players can use the idea that they should arrive at the same answer, but it's not true. A logical fallacy is introduced.

Specifically, it is not valid to make the leap from "we must both arrive at the same answer" to "all choices where we arrive at the same answer are possible results for rational actors".

It is provable (for any reasonable definition of a rational actor) that a rational actor will never play 100.

If a rational actor can predict with certainty that his opponent will play 100, then he will play 99 in response.

[Yakk]

Ghona, more accurately, you won't get Puck to agree that anything you or Puck labels as rationality implies super rationality. You won't get good definitions of rationality -- just hand-waving ones. The actual axiom of these definitions is "it plays a Nash equilibrium when played against itself", as any deduction from the definition that doesn't lead to that result is taken as solid proof that it is an invalid argument.

See the previous 18 pages of discussion.

This thread isn't engaging in formal logic, it is hand-waving around using the forms of logic. With only hand-waving "logic", disagreements at this level cannot be dealt with via the "logic". A chain of argument can be dismissed because it doesn't lead to a conclusion one party likes (and not because a step was shown to be invalid) -- it must be wrong, instead of saying why it is wrong.

Another chain of argument cannot be shown to be correct because each step is full of assumptions that aren't stated, and cannot be checked by even an approximation of an automaton.

Don't be fooled: if someone isn't even willing to try to make a formal logic statement out of their 'logical' arguments, their logical arguments aren't worth the bits they are transmitted on.

[Puck]

Yakk, quit with the ad-hom attacks and misrepresentations of what I've said. You seem to have missed about the last two pages.

Superrationality and rationality directly contradict each other. They are not compatible. It's very confusing if you want to define superrationality as being rational, because it's not a concept you can add to the "normal", accepted definition of rationality without producing a contradiction.

By the normal, accepted definition, I mean a player who will always play 99 if he can predict with certainty that his opponent is going to play 100. You will have a hard time convincing me that an actor who does something different in that situation is rational.

I have shown you the invalid step in your argument several times, including my previous post. I have stated the axioms upon which I base my definition of "rationality", and shown that superrationality is not implied by them. Stop building straw men.

[Yakk]

As of the last time I posted to the thread, your proof that my argument was wrong consists of "I don't like the conclusion".

You haven't used the word axiom since, not that I noticed (or at least defined them). You aren't engaging in "real" logic: I like my logic as close to the bone as possible. I'm just letting others know that the level of detail that the problem(s) and issues brought up occurs at isn't happening at the level of logic that is being argued in this thread. You need a much stronger definition of what rational is to reason about this problem in a consistent way than is being used.

I do hold that there are models of "rationality" that do result in "playing 2".

I put forward one -- where the rational player plays against a strictly less knowledgeable rational player. Then you take the leap of faith that the strategies that evolve in the limit are the strategies that a perfectly rational player would play. This leads to Nash Equalibirum -- but at no time does the rational player of order N know that they are playing a rational player of order N. But that doesn't seem to be what people are using. Nor have I seen any replacement for it.

The problem is that the thread lacks sufficient rigour to be as confident in the conclusions as people claim they are.

I fully suspect that the definitions of rational player that people are holding in their head are inconsistent ones. Making a self-referential definition like the one being thrown around in this thread without being inconsistent usually takes a lot of care.

I left the thread because nobody was interested in formalising the problem. I'm just pointing out that, barring a willingness to head towards formalisation, what is going on is rhetoric, not logic.

Let X be the statement: it can be deduced what my strategy will be from the axioms.
Let A be the statement: Let P_1 be player 1's strategy, and P_2 be player 2's strategy. Then P_1 = P_2.
Let G` be the game G, minus any axioms about how your opponent decides to play. All of them. Whatever 'rational' is defined as, and that you are rational, is left in. You just don't know what your opponent is.
Let O = G \ G`. Then G = G` and O.

In logic, when you remove axioms, anything you can deduce can also be deduced from your original set of axioms. If you generate a contradiction, you just proved that your axioms are inconsistent. This is not a problem, other than to your axioms.

Assuming ~X:
* If you manage to produce a strategy, all you have done is proven a contradiction. Because ~X assumes that you cannot produce a strategy.
Assuming X:
* You can prove A. (or do you disagree?)
* G` and A -- can you see how this proves "Play 100"? (or do you disagree?) -- remember, G` is G minus statements about how your opponent decides to play.
* G and A may or may not lead to a contradiction.

If both branches lead to a contradiction, your axioms are inconsistent. The ability to prove that (2,2) is a rational move doesn't matter if the system you are proving it is inconsistent. It is utterly irrelevant.

It doesn't matter if you can deduce (2,2) without thinking about the statement X. It could matter if you could generate a axiomatic system that allows the (2,2) deduction, but not the X deduction -- but that requires smithing at the axioms and deduction rules, which isn't what is going on in this thread.

Now, you can get around this a number of ways. You can make X or A not a statement in your logical system -- and that would be interesting. It isn't a given, not by any stretch of the imagination.

You can remove the axioms that tell you your opponent is identical to you (the "rational player of order n" solution[1]). But that becomes the fundamental problem -- that the self-referential nature of the game either leads to super-rationality from rationality, or a contradiction, unless you are very careful about what you mean by rational.

The existence of a contradiction doesn't mean "don't follow that chain of logic", it means that any argument from those axioms is invalid. This isn't how you think, Puck, so that is why arguing with you about consistency of your assumptions is relatively pointless: if you see that a chain of logic leads to inconsistency, you simply say "clearly that means the chain of logic is invalid" (I can cite where you said this in an earlier post, which I'm assuming is a position you still hold.)

Self reference is hard. As I have said before, I'd be interested in seeing the formal system in which you can describe the "We both know everything about my opponent, and both are rational" that is both consistent and doesn't lead to playing 100 at 100% chance. I doubt that this thread will produce it.

Footnote [1]: The "rational player of order n" model.
R_n := A rational player of order n with seed s
R_0 := s
R_n is a rational player who thinks it is playing against R_{n-1}.

R_infinity with seed s is the limit, if it exists, of R_n. Note that R_{infinity+1} could be distinct from R_infinity in theory (the player who thinks it is playing against the rational player of order infinity with seed s). Ie, I'm using + in the Peano arithmetic/ordinal sort of sense, not in the arithmetic sense ( inc(foo) := {{foo}, foo} ). The plus 1 just means "I'm thinking I'm playing the previous player".

If for all s, R_infinity is identical, then this is defined to be the rational player of order infinity.

[Puck]

If you can prove A using our set of axioms, then A is a deduction, not an axiom itself.

Further, you cannot prove A for arbitrary values of P_1 and P_2. You can prove A only when P_1 and P_2 are determined using (or: are consistent with) X and G.

If you wish to use A to determine P_1 and P_2, then you have exactly the tricky self-referential problem you're worrying about.

Can you please formally flesh out your proof that G' + A implies "Play 100"? I disagree - I think there is hand-waving going on here.

[Ghona]

My definition of a rational actor is one that always comes to the strategy with the highest expected return for him/herself, given the knowledge available.

Is there any other one?

[Puck]

I think we can agree on that. Do you disagree that such a rational actor, if he were able to determine with 100% certainty that his opponent would play 100, would respond by playing 99?

[Puck]

Yakk, this is a somewhat limited analogy, but I think it might explain the problem I'm having:


G: Both players want to choose the highest number they can between 1 and 10.
H: Players may choose only prime numbers.
A: Both players must come to the same conclusion; that is, P_1 = P_2.

G + A => both players will play 10?

In other words, there are other axioms that restrict which choices are allowed. In TD, the axioms of rationality prohibit rational players from choosing 100, so [100, 100] is not possible (just as impossible as [100, 99]), even though it is better than other choices.

[Oculus Vespertilionis]

* G` and A -- can you see how this proves "Play 100"? (or do you disagree?)

Maybe it's just me, but you said all those things about how you like formal argument, and dislike hand-waving -- and then you hand-waved the core of your argument. I certainly don't see it.

[Yakk]

Ah. Interesting! G+H+A, as an axiomatic system, doesn't generate an answer, as it doesn't rule out playing 5. There is no axiom that says "you can play 7", so you cannot prove (from the axioms given) that 7 is a legal move. You can prove from G+H+A that the strategy of the players won't include (8,9,10).

If you word G as "Player may play any number from 1 to 10, and will seek to play the highest number", then H is inconsistent with G.

G+A => nothing, because there are no way to prove that any move is legal.

Interesting.

A less toy example:
You can play moves 1 through 100 in this game.
In this game, you earn 100+X points if you get the same number X as your opponent, and 0 otherwise.
#1 My opponent always plays the same number. Say 7.
I am a "rational player".
#2 I know everything about my opponent.

---
Thus I will play the same as my opponent (P_1 == P_2) -- namely we will both play 7.

Remove axiom #1 and #2, add in P_1 == P_2 as an axiom. While I know that I will play the same as my opponent, I cannot determine what that move will be from the collection of axioms.
---

However, this still leaves me at a loss on how to prove that playing 100 is dominated against a true mirror player. The proof that 99 dominates 100 still relies on the possibility that your strategy will differ from your opponents, which you know (and they know) not to be the case. Then again, maybe I'm assuming completeness?

Barring X (that I can prove what I'm going to play) from being expressed is a possibility, but then you run into problems of other kinds of comprehensions -- you want perfect knowledge of what your opponent is, and that you are the same as your opponent.

[Puck]

Alright, let me rephrase slightly:

G1: Players want to play the highest (natural) number they can between 1 and 10 (inclusive).
G2: All moves between 1 and 10 are legal.
H: Players following decision-making process Z can only choose prime numbers.
J: Both players follow decision-making process Z.

I'm still awaiting your formal proof that (in your axiomatization of TD) G' + A implies "play 100".

[thc]

Sorry for the interruption, but my mind really refuses to accept 2,2 for some reason. :p

Can you tell me what is wrong with this line of thought? Assume you are playing a rational player capable of deducing your strategy. Your pay-off matrix looks like this:

Code: Select all

Guess     Pay-off
100       97
99        96
98        95
...
4         1
3         0
2         2


Doesn't that imply that the dominant strategy is to play 100 under these conditions? If so, how is this different than the conditions set forth that concludes 2,2?

[Ghona]

I think we can agree on that. Do you disagree that such a rational actor, if he were able to determine with 100% certainty that his opponent would play 100, would respond by playing 99?

Assuming that playing 99 would not result in his opponent playing anything other than 100, no.

[Puck]

THC, a rational player cannot deduce your strategy unless he knows that you are playing rationally. If he could, it would constitute a violation of the "no collusion" rule of the game.

Ghona, your choice cannot "result" in your opponent playing anything. One of the basic rules of the game is that the two choices must be made independently; thus, whatever I ultimately choose, it will not change my opponent's play. However, what my opponent can deduce about what I should choose can certainly affect his play.

Regardless of any other conditions, if P1 can predict with certainty that P2 will play 100, P1 will play 99. If P2 can predict this play, then P2 (if he is rational) will not play 100 in the first place, meaning that P1 could not have predicted with certainty P2's play of 100. (It is possible that P1 could have made a mistake, but our perfectly rational actors don't make mistakes.)

Yakk, I realize that this proof does not rule out the possibility that we have inconsistency in our axioms. I would really like to see a formal derivation of the proof that rational actors must play 100.

[Ghona]

I think the issue here is the halting problem, actually.

No, really.



Let's ignore probabilistic strategies for the moment (assume they don't have dice or something)

As I am defining perfectly rational to only come to one conclusion, I am making it an axiom that there is only one possible strategy. Thus, it is axiomatic that it is an autostrategy. Thus, there is no option to defect from 100 to 99 - your opponent's moves are automatically the same as your own. If they were not, one of you would not be perfectly rational.

To use a computer analogy, you're both running the same "perfectly rational" program, and fed in the same "conditions" data. Your opponent's strategy isn't changed by any of your decisions, but it will always be identical anyway.

In other words, it is 100% probable that whatever price you say is the price your opponent will say. Just like it's 100% probable that the same button inputs at exactly the same milliseconds into a game loaded from the same initial read only memory will always result in victory in 23:58.

[Puck]

As I am defining perfectly rational to only come to one conclusion, I am making it an axiom that there is only one possible strategy. Thus, it is axiomatic that it is an autostrategy. Thus, there is no option to defect from 100 to 99 - your opponent's moves are automatically the same as your own.

Your last sentence doesn't follow from the first two. Yes, the end result must be an autostrategy. However, that doesn't restrict your freedom of choice. The very fact that the rules of the game don't prevent you from defecting to 99 is the reason why rational players can't possibly play 100, 100.

There is a proof earlier in this thread that if there exists a unique rational solution for this game, then it is (2, 2). So far, no one has found a fatal flaw in this proof (although I had to adjust the wording somewhat), so I'm pretty convinced that the only possibility other than it being correct is an inconsistency in our initial assumptions. I have yet to see a proof of this inconsistency that appears correct to me.

[Ghona]

Perhaps I should have said, any transition from 100 to 99 cannot be a defection, since your opponent will also make the same transition.

You can't move from 100-100 to 99-100, only from 100-100 to 99-99. 99-99 is a worse strategy that 100-100.

If you know that all strategies must be autostrategies, than considering a case where you have a different strategy than your opponent is in contradiction of the previously made axiom. You might as well say "well, maybe my opponent is a numerologist, and will always bid $66.66"

If there is only one possible strategy, and that strategy is an autostrategy, let's compare the possible strategies to see which one has the highest expected return. Since all strategies are autostrategies, I've taken the liberty of having filled in what your opponent will have chosen as well.

2-2 is really sucky
3-3 is slightly less sucky
...
99-99 is really quite good
100-100 is the best, return wise.

[Nitrodon]

The fact that both players will play the same thing is a direct consequence of the fact that there is only one play a rational player will make in this game when both players' rationality is common knowledge. Any argument which presumes that both players will do the same thing and uses that condition to show an optimal move is begging the question (or is assuming superrationality instead of rationality).

[thomblake]

thc, your payoff matrix is wrong. You're assuming your opponent always plays just a little less than yourself. If your opponent plays 2, then 3 is just as bad as 100.

[thc]

thc, your payoff matrix is wrong. You're assuming your opponent always plays just a little less than yourself. If your opponent plays 2, then 3 is just as bad as 100.

No it is not wrong. Re-read my post. The payoff matrix is under the assumption that your opponent can deduce your strategy.

THC, a rational player cannot deduce your strategy unless he knows that you are playing rationally. If he could, it would constitute a violation of the "no collusion" rule of the game.

Really? I don't see how the second statement follows from the first.

I.e., if you know that your opponent can deduce your strategy, then it is best (and therefore most rational) to play the number that maximizes your returns, which is 100. (If you played 99, you know your opponent will play 98, which will return you less.)
Accordingly, I fail to see how there is any collusion going on. Collusion implies that there is some some sort of conspiracy/secret agreement, and for all it matters, your opponent can be an automaton incapable of such.

[Ansain]

It's been a while since I've posted in here so here's my thoughts on the issue (slightly changed from last time).

-You can't know what your opponent picked until after the game is over, so we have some unknown X between 2 and 100.
-The optimal pick is X-1, in which case you get x+1 dollars.
-If you pick exactly X you will get X, or 2 below optimal.
-If you pick a number above X, you will get x-2 dollars, or 3 below optimal.
-If you pick a number below X you will get Y+2 dollars, or X-Y-1 below optimal. (Y is your pick)

Now, it seems to me that you have 2 logical choices here, pick as high a number as possible and accept 3 less than the optimal amount, or try to undercut your opponent and get the optimal amount.

Two arguments now come to mind. One is that if we know that the logical choice to make is picking a higher number then undercutting at 99 will give more money. Thus undercutting is actually a more logical choice meaning that it is contradictory to say that picking a higher number is logical. The other argument is that if undercutting is logical, then nobody would ever pick 100, and if nobody would ever pick 100 then nobody would ever pick 99, ect... $2. Clearly $2 is less than the expected yield with picking a higher number, so its also contradictory to say that undercutting is the more logical choice. Therefor I come to the conclusion that neither choice is more logical than the other.

Given that picking a number above X (ie 100) is just as viable an option as trying to undercut X, I look again at what happens when you undercut your opponent. Now 99 seems like a very good option if there is a chance that your opponent will play 100. 98 seems like a slightly better option because of how good 99 is, but if your opponent plays 100 you would rather have picked 99. If you graphed how good each choice looked it (starting at 100 and decreasing to 2) it would increase a bit as you went down, then it would max out (somewhere between 98 and 95) then decrease asymptotically to 0 as you approached 2.

I am currently working on a program that takes an initial graph like that and makes a new graph based on the expected yields if your opponent plays a strategy that uses the old graph. I think eventually it will stop changing between each loop and I shall claim that picking a number based on those percentages is the most logical choice.

Edit: upon looking at the end result I must take back that claim.

[Ansain]

well I can't quite figure out how to post the graph, but I can describe it to you. It looks like a graph of ln(x) from 2 to 96. it crosses .5% at 19 and 1% at 44. it peaks at 96 (1.49%) and decreases slightly till it reaches 100. 100 is higher than 93 but lower than 94. It seems to make this graph regardless of what the starting values are (except if you start with 2 at 100 percent in which case you always get 2 at 100 percent). The expected return is $40.8965.

This expected return of ~$41 is what happens if you accept that undercutting is a logical choice. No matter what you start with, if you make new values based on the expected return of each value you will get to this graph after enough iterations.

-If both you and your opponent come to the conclusion that undercutting is the best strategy you get $2

-If both you and your opponent come to the conclusion that undercutting and bidding $100 are viable options you get ~$41

-if both you and your opponent come to the conclusion that bidding $100 is the best option you get $100.

Yes it's true that my solution can easily be taken advantage of, but if that happens i really do not care about the small amount of money I lose and would not risk $60 just to get that extra $3.