Talk:Traveler's dilemma

message: The article states that this is an extension of PD.  I don't see it in the SA article, but would it be inappropriate to say that this is a variant of the centipede game?  Smmurphy(Talk) 02:51, 23 May 2007 (UTC)
replies: text: It does bear a strong similarity to the centipede game, except the centipede game has more than one Nash equilbria (but only one subgame perfect equilibrium), while this game has only one NE.  This game also has some similarity to Guess 2/3 of the average in that it involves deep iterative deletion of dominated strategies in order to demonstrate the NE.  I think its okay to say it's similar to the PD.  The socially optimal strategy in this game is not an NE, while dominance reasoning leads individuals to a very inoptimal outcome (like the PD). --best, kevin [kzollman][talk] 20:29, 23 May 2007 (UTC)
thread_title:  centipede game 

message: The article stated that TD was an extension of PD if you are limited to 2 distinct numbers.  This is incorrect. In PD, the temptation to defect is greater than the reward for cooperation.  If you are limited to 2 and 100 as your only options, the temptation to defect is 4, and the reward for cooperation is 100.  This would result in 2 Nash Equilibrii, and thus not a prisoner's dilemma.  -Todd(Talk-Contribs) 00:46, 10 June 2007 (UTC)
replies: text: Indeed. Have re-formulated how this game is linked to PD. (see article) JocK 11:25, 10 June 2007 (UTC)
thread_title:  Not an extension of Prisoner's Dilemma 

message: What is the general consensus about when or how to capitalize Traveler's Dilemma, Prisoner's Dilemma, Nash's Equilibrium etc.  I see it both ways but usually capitalizing both words like any other proper noun.  I think all the WP game theory articles should use consistent style throughout on these names. Venado 23:36, 15 June 2007 (UTC)
thread_title: Style

message: I have never heard the term deep used for iterative elimination of dominated strategies.  Should it be explained or references identified?Venado 00:02, 16 June 2007 (UTC)
replies: text: A deep iteration is an iteration that is applied many times. Perhaps there is a better word, but can't think of any.JocK 06:53, 16 June 2007 (UTC)
replies: text: I think it better to leave off the word unless the term "deep iteration" is perhaps described through link etc.  Isn't the number of iterations to be performed simply determined by the number of choices available?  Classic TD is max $100 but Basu describes variant of TD played with only 2 choices.Venado 14:11, 16 June 2007 (UTC)
replies: text: The large number of iterations needed to reach the NE (in TD indeed related to the number of strategic choices) is key to the fact that in TD (and also in 'guess 2/3 of the average') real-life experiments (including real-money experiments) result in strong deviations from the game-theoretical solution. Would therefore not like to see the word 'deep' disappear. JocK 17:41, 16 June 2007 (UTC)
replies: text: Thank you. I think your article edit now explains its meaning much more clearly. Venado 15:04, 17 June 2007 (UTC)

text: A "deep" iteration can also indicate a degree of non-linearity, self-organized criticality, or insensitivity to (relatively large) perturbations of a system's internal or external parameters. -- TheLastWordSword (talk) 21:43, 1 December 2016 (UTC)
thread_title: Confused by word "deep" for IEDS

message: Why would anyone chooses 2 when they can benefit by choosing 98, 99, or 100. (97 and less are strictly dominated strategy, which means that it is nonsense to choose 2) 24.1.201.172 (talk) 00:42, 22 June 2010 (UTC)
replies: text: That's sort of the point of the game. (2,2) is the sole Nash Equilibrium, but in reality, it would probably never occur. If you're perfectly rational and assume the other player is perfectly rational and you are thinking about what number he is going to write down, you get caught in the vicious circle which leads you to (2,2).161.253.11.22 (talk) 16:57, 8 February 2013 (UTC)

text: This is a game.  Choosing $2 is the best move possible.  So by choosing $2 you would win the game, even though you would lose most of the value of your suitcase filled with antiques (that's why it's called a dilemma).—An Sealgair (talk) 03:14, 16 November 2014 (UTC)
thread_title: I don't get it

message: How can one traveler get paid $101 if the airline insurance limit is $100?
The article states if Alice quotes $99 and Babar quotes $100 then Alice gets $101. Since this is a distinct possibility, how then was the airline manager prepared to pay out this amount when, by airline policy, he could only award $100? I suppose you could say he would just pay the dollar out of his own pocket. But that would be a rather off-handed and clumsy explanation given this is a game-theory problem which usually has a very strict design. And besides, just out of principle, it seems awkward to suggest that someone would risk ANY amount of his own money just to do a job like this, particularly if there is no chance of any gain on that risk.
So then the more satisfactory explaination is: He's penalizing Babar for quoting $100 by subtracting $2 from the now agreed upon value of $99 for his luggage and then giving that $2 to Alice.
But for everything to be ligit he would still have indicate on his paperwork that he is awarding $99 to both passagers. So he would have to take $2 from Babar's awarded amount to leave him with $97 (in effect stealing from a customer for the airline) and then give Alice an extra $2 {in effect stealing from the airline to reward a customer). Babar could raise a big stink about the whole thing and the manager would lose his job when they look into the issue and find out how he's been falsifying baggage insurance records.
Considering this, if you assume that the airline manager cannot legally pay out any more than the $100 maximum, then this solves the dilemma. There would be little or no incentive for either passenger to quote $99 if they can't get the $101. And therefore we would never get into the reverse inductive reasoning thats leads them down to the $2 quote.
Dont get me wrong. I understand the point of this problem well and its very interesting. Just thought I'd point out what I believe to be a minor flaw.Racerx11 (talk) 05:03, 30 August 2010 (UTC)
replies: text: in none of the outcomes the airline has to pay more than $200 for both travelers combined. JocK (talk) 13:05, 2 September 2010 (UTC)
replies: text: True. In the example I give, the total is $198. But its not like he is able to just pull $198 in cash out of a register and divide it up however he sees fit. The airline will want a record of these two settlements. So he will have to fill out some kind of documentation and he wont be able to indicate in this documentation that he payed either passanger more than $100. As explained in the middle of my original post, this means he would have to falsify this documentation and the manipulation of the actual payoffs to the travelers could be viewed by the airline as stealing. Also I would expect an airline would require baggage loss settlements to be paid by check. Possibly mailed to the  recipients at a future date, further complicating the managers situation. I know all this is irrelevant to the intents and purposes of the problem. Its only a flaw as a real life example.Racerx11 (talk) 14:42, 2 September 2010 (UTC)
replies: text: Indeed, it's irrelevant for the arguments made in the article. However, if you consider it important enough to necessitate a modification of the text, I suggest you change the sentence "the airline is liable for a maximum of $100 per suitcase" into "the airline is liable for a maximum of $200 for both suitcases combined". JocK (talk) 16:58, 2 September 2010 (UTC)
replies: text: hehe. ok I don't want to put words in your mouth but Im kinda taking your point as: if I consider it important enough to elaborate on this talk page then I should think it important enough to fix in the article. Well taken and considered even if that wasn't your intent. I will think about this and your suggestion for the fix. I must say at first impression the condition "the airline is liable for a maximum of $200 for both suitcases combined" still doesn't sit too well for me as a satisfactory real life example. I guess Im unsure if I should "fix" it or not. But I will think about all of this some more later. Thanks.Racerx11 (talk) 18:02, 2 September 2010 (UTC)
thread_title:  $100 limit? 

message: Although there is nothing wrong with the inductive reasoning used by passengers to conclude that $2 is the best quote, it occurs to me that the travelers could each use a very similar line of reasoning and yet come to a very different conclusion.
Each passenger reasons: If I quote $99, I could get $101. But if the other is also thinking the same and quotes $99, THEN WE BOTH GET $99. So I should quote $98 in order to get $100, but then the other is going to think the same and quote $98. WE BOTH GET $98. So I quote $97 and so does he and we BOTH GET $97...and so on until we are both down to each of us quoting $2. Well that sucks! The other person seems reasonable and should think the same way as me. So I will quote $100 and so shall he.
To me this is very similar to the reasoning used in the "equilibrium strategy" which yields the $2 conclusion, but with just a minor tweak, it is changed to the $100 conclusion. The "purely optimal strategy" is still $100 but for different reasons. Its more to do with expected gain. Quoting $2 yields a 2 or 4 dollar gain, while quoting $100 yields a range from 0 to 100 dollars. You should only quote $2 if you are absolutely sure the other will quote $2. In the problem there is nothing to suggest that the other is certain to quote $2, so its safe to assume it is likely he WONT quote $2. Its the gambler's equivalent to making $2 bet on a good chance of winning $100. Thats a bet we should take all day long. So thats the optimum strategy. The equilibrium strategy in the problem is a little like an insurance policy that guarantees at least a $2 payoff, but no more than $4. The price of that policy is giving up the chance at a much higher payoff up to and including $100. Not a very attractive deal.
I don't agree that the equilibrium necessarily be set at $2 since the inductive reasoning used can just as easily (as shown in 1st paragraph) conclude a strategy of quoting $100. Which happens to be the purely optimal strategy anyway. Racerx11 (talk) 17:54, 30 August 2010 (UTC)
replies: text: Indeed. Give it a few more years and maybe you'll see some professor coming up with this idea as an epiphany, although it seemed fairly obvious to me. Could be that someone else has already made the point - I haven't looked into it much. II | (t - c) 08:39, 25 November 2010 (UTC)
thread_title:  A different strategy for the travelers 

message: Just a random reader tab-surfing here, I have no wiki account to tie this to. I'm far from an expert on game theory but this seems to me to rely on a premature termination of predictive strategy and response.
The entire paradox is defined by prediction of the second traveler's strategy by the first. Since it is a hypothetical situation you in practice assign a behaviour to the second traveler, writing them a course of logic comparable to a computer program.
Suppose we assign the second traveler a "snap decision" or "stupid" (to be less than PC) strategy. The second traveler makes a decision on the spot without iterating through responsive theory whatsoever. They will pick $100 because this is the highest value, and not think about the first traveler's/"player"'s response. In this situation the optimal strategy is picking $99, thus netting $101.
On another tact, we assign the second traveler an "infinite contemplation" or "genius" behaviour. This is the behaviour seemingly employed in the article, where one follows prediction and response in a leap-frog down to the minimum value. This relies on the ability of the second traveler to apply logic perfectly to the first traveler under the unwavering assumption the first traveler will act accordingly - it does not for example consider the possibility the first traveler would be snap decision/stupid, in which case the optimum strategy would be to pick $99 as given above. The logic for the second traveler in this scenario seems to be "I operate under perfect logic and assume that the first traveler does as well". The second traveler believes the two travelers can "read" each other's "program" and thus even though they cannot watch one another run the program, they know infallibly the outcome. Now, if there were some clause that the travelers could not agree upon a price, that they had to select 2 separate values, and there was some kind of turn taking mechanism, such as a man turning to each to ask what they pick, then the other, until both decide to settle, in this scenario picking $2 instantly would be logical as by getting to it first you prevent the other traveler from beating you to the punch, limiting them to only higher values and thus securing the $2 bonus. But there is no such mechanism in play, and both travelers can pick the same value. The second traveler's assumption that "I operate under perfect logic and assume that the first traveler does as well", can thus be condensed first to "The first traveler will arrive at the same conclusion as me" and then "the first traveler will pick the same value as me". The second traveler thus believes that whatever value it chooses, the first traveler - the player - will also choose. This means both first that attempting to pick less than the first traveler is logically futile, and that if the second traveler picks $100, the first traveler will also pick $100, while if the second traveler attempts to undercut the first traveler and pick $99, the first traveler will also have attempted the exact same thing and picked $99, thus eliminating the possibility of earning the $2 bonus. This could be repeated down to the base minimum of $2, as it has originally, but it seems to me premature to stop the chain of prediction there. Unless there is a flaw in my logic, the second traveler's behaviour is perfectly equivalent to "The first traveler will pick the same value as me". Reaching the $2 minimum should draw the conclusion "It is impossible to bid less than the first traveler and win the additional $2", not the given "This is the best solution". The bonus/malus is thus effectively unobtainable and removed from the game logic, having as much weight on the matter as the statement "The airline manager is a sexist and will always believe a man over a woman. Both travelers are male/female" - both are beyond reach and thus irrelevant. Having removed the bonus/malus factor from the equation, the optimal choice is $100, since the second traveler believes whatever it picks the first traveler will as well, and in this scenario the best choice is $100. The first traveler, the player, has a choice of either following this logic as well, picking $100 and getting $100, or following the logic exhibited on the page, picking $2 and getting $4, or believing they can trick the second traveler and picking $99 and getting $101.
In summary, either $100 or $99 are the optimal choice for the first traveler if the second traveler operates logically, the better of which I won't attempt to conclude here, but $2 is not the optimal choice unless you believe/know the second traveler to operate under a flawed logic, and not the optimal game theory of picking $100 under the belief the player will pick whatever value they do. It is entirely possible for the logic to break under other behaviours. For example if dealing with a traveler using "mean" logic, you know they will choose $51 as the middle of the possible range, and $50 becomes the optimal strategy. As listed above $99 is the best choice with a "snap decision" traveler, while $2 is optimal if dealing with a "paranoid"/"malicious" or similar traveler, who will place priority on how certain a pay-off is over how much it is or who will place priority on inflicting the malus on the other traveler respectively. As I stated in the beginning the problem is defined by prediction and what you believe the other traveler will do - however, if you believe the other traveler to employ optimum game theory then $100 seems demonstrably the choice they will make and either $100 or $99 the best choice "you" can make. Kinda long-winded, but there you go.
PS: Got my terminologies a bit mixed up. I'm referring to Nash equilibrium. It seems to me Nash equilibrium would never arrive at $2, or more accurately not stop at it for the same reason it doesn't stop at $99, $98 or any other value. The point of termination seems arbitary, to me.  — Preceding unsigned comment added by 81.178.243.190 (talk) 15:10, 9 July 2011 (UTC)
thread_title:  Definition Of Prediction 

message: Someone has added a paragraph at the bottom of the introduction, outlining a vague strategy involving completely arbitrary "estimates" of the probability that the other player will pick a certain value. I think the very specific meaning given to the term "rational" in game theory was missed by whoever posted that. I do not have an account so I don't want to make any edits, but if you think I am correct please delete that last paragraph. If I am missing something I'd be grateful if someone could explain.  — Preceding unsigned comment added by 217.165.114.132 (talk) 03:40, 3 April 2012 (UTC)
thread_title:  Changes to the introduction 

message: Dr. Morone has reviewed this Wikipedia page, and provided us with the following comments to improve its quality:
We hope Wikipedians on this talk page can take advantage of these comments and improve the quality of the article accordingly.
Dr. Morone has published scholarly research which seems to be relevant to this Wikipedia article:
*Reference : Andrea Morone & Piergiuseppe Morone, 2012. "Individual and Group Behaviours in the Traveller's Dilemma: An Experimental Study," Working Papers 2012/09, Economics Department, Universitat Jaume I, Castellon (Spain).
ExpertIdeasBot (talk) 18:56, 27 June 2016 (UTC)
thread_title:  Dr. Morone's comment on this article 

message: This article is of extremely low quality and may cause more confusion than it does good. 
- The "solution" of the game talks about asymmetric information, which hints at the airline being a player in the game, which it is not. 
- It also makes no sense to even mention Backwards Induction as a solution algorithm, since it only applies to dynamic games and this is a static game. 
- If the goal of the section "Analysis" is to prove that (2,2) is a Nash Equilibrium, then the section should be written up in this way: it should just prove that the best response to the opponent reporting 2 is to also report 2. Then one could go on to also prove that there can be no other Nash Equilibrium (x,x) where x > 2 (this can be proven simply by showing that the best response to x is x-1 for any x > 2. 
- There is a sentence saying that "The traveler's dilemma can be framed as a finitely repeated prisoner's dilemma". It cites two references, where the first does not exist online (there are dead links to the paper online), and the second does not support the claim (it only references the word "repeat" once and does not support the sentence). This claim does not make much sense: the game is a static game, I don't see what it has to do with repeated prisoners' dilemma at all.  
Superpronker (talk) 13:11, 22 April 2024 (UTC)
thread_title:  Low quality article