Maurog
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Please discuss the Traveler's Dilemma Game here.

This is a game based on the following logic puzzle:

Lucy and Pete have identical antiques which are lost by the airline they took to get home. The manager will pay them back, but he assumes they'll try to inflate the price, so he decides to have them each (separately) write down the value of their item (an integer from 2 to 100 dollars). If they write the same number, he'll assume they're telling the truth and pay them each that. If they're different, he'll give them both the lower number, but also pay the person who wrote it a bonus of $2, and subtract $2 from the other one. What should Lucy write?

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[Last edited by Maurog at 08-29-2007 01:20 PM]

MeckMeck GRE
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Do I get a random opponent or the lowest opponent? So if... A says "8", B says "12" and C says "10" --> A will get "10" points and B and C "6" points. Is that correct?

Maurog
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Not exactly, it's everyone vs everyone in the end so it will be:
A gets 10 from (A vs C) and 10 from (A vs B) for a total of 20.
B will get 6 from (B vs A) and 8 from (B vs C) for a total of 14.
C will get 6 from (C vs A) and 12 from (C vs B) for a total of 18.

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Tahnan
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I've never seen this before; it's mildly interesting. I didn't do any searching for analyses, but I did run a few simulations and some calculation of expected payoff in the two-person version. (I hope that's allowed! Nothing in the rules seemed to forbid it. If I shouldn't have, Maurog, just withdraw my entry. Again, nothing that wasn't my own work--no Googling, say.)

It seems that the highest expected payoff is always achived by both n-3 and n-4, where n is the maximum value allowed.

Things are simpler here than in the Prisoner's Dilemma, since it's easy to know that you're hurting your opponent by "bidding" low, but you're hurting yourself by doing so as well. If your opponent bids x, your highest payoff is at x-1; x-2 has the same payoff as just bidding x, and anything lower and your payoff starts to drop. That means there's not much to be gained by bidding particularly low, which is why the highest expected payoff is just shy of the maximum bid.

Interestingly, if there are three participants all being compared against one another--that is, the airline executive pays them all x if they all write down x, and otherwise pays x+2 to the lowest bidder(s) and x-2 to the other(s)--then the bid with the highest payoff is the maximum - 8 or so.

Probably overanalyzing. Carry on.

Kwakstur
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It looks like Maurog reads Scientific American. Big 8-page or so article on Traveler's Delimma.

And, yeah, Prisoner's Delimma is, to me, a very easy choice. Rat him out! (or the number 2)

This is not as easy.
I tried a program once for this situation, and it would've given the perfect answer if there were a 99 people, all of whom had a different number from 2-100, thus occupying every number. Quite an obvious flaw that it assumes every number participates once, since that would be one HECK of a coincidental situation. But it's interesting to see how changes in the things (like range and amount of variance) affect the graph it makes (number on x and total profit on y).


Back on topic of this contest, I think this will be a very interesting event.

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[Last edited by Kwakstur at 08-30-2007 04:38 AM]

Maurog
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Actually, feel free to analyze it as deep as you want... that's the point of it. And if you think it's that much different from a 2-player Traveler's Dilemma, you are wrong. There still will be only one number you can choose that wins, and it will be based on the opponents' choices. Think of it that way - instead on n-1 opponents, 1 opponent with probability 1/(n-1) to choose either of the many options the other people wrote. Exact same situation, and I believe exact same calculations should hold.

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Rabscuttle
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Tahnan wrote:
Interestingly, if there are three participants all being compared against one another--that is, the airline executive pays them all x if they all write down x, and otherwise pays x+2 to the lowest bidder(s) and x-2 to the other(s)--then...

Be the executive. You pay out 3x-2 for things that you claim are worth 3x. You pocket the extra 2.

No wait, what if two people pick the lowest? Then you're out 2. Curses!

Meh, the executive is probably running some sort of scam anyway. He probably submits the higher value to their insurance company, and pockets double the difference between the two values.

I read an article on this recently too. I forget what most people would do (except that I think it depends on how long you think about it).

Tim
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Rabscuttle wrote:
No wait, what if two people pick the lowest? Then you're out 2. Curses!

I don't understand why anyone who choose the lowest. Because if everyone choose 100, everyone will get 100.

I think, in real life, you want to get as much as possible, not to "win" from everyone else.

But since we are playing a game to win instead of real life, you should realise that the strategies are different.

But to make this a bit more interesting, I declare that my entry was 100.

Disclaimer: Please also make sure to read my next post that says that the number I've mentioned here is merely a random one.

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[Last edited by Tim at 08-30-2007 10:46 AM]

Maurog
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That's what anyone would say if they picked 98. Please don't "reveal" you choice because it can't be trusted by definition. Not only it's unverifiable, it can be changed in the very last moment. Also, the original problem states that there is no communication between Lucy and Pete prior to writing the number, so let's not throw random numbers claiming we picked them, mmkay?

And by the way, if everyone choose 100, everyone will get 97 except for the guy who picked 99, who will get 101 and maximize his gain. Surely 101 is better than 100 in terms of personal gain? You *are* playing to get as much as possible here. The very dilemma in Traveler's Dilemma is that your opponent has too much control over what you're gonna get.

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Tim
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Maurog wrote:
That's what anyone would say if they picked 98. Please don't "reveal" you choice because it can't be trusted by definition. Not only it's unverifiable, it can be changed in the very last moment. Also, the original problem states that there is no communication between Lucy and Pete prior to writing the number, so let's not throw random numbers claiming we picked them, mmkay?

Oh, I can be very trustful. And if you don't believe me, you can always pick 94, or zero. While the game does assume everyone has no information about each other, in real life, Lucy will know about Pete, and she knows if she can trust him, or not.

But for the sake of this game, I am going to say that the previous 100 I've mentioned was merely a random number.

Surely 101 is better than 100 in terms of personal gain? You *are* playing to get as much as possible here.

In a game, yes. In real life, definitely not.
In real life I don't care if anyone gets a 101 while I get 100, because even 100 is a lot better than 0.

The very dilemma in Traveler's Dilemma is that your opponent has too much control over what you're gonna get.

I think the real dilemma is that there's absolutely no incentive to guess higher than 0, unlike the Prisoner's, where you actually get more if everyone cooperates.

Unless someone openly declares (s)he will guess 100.

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[Last edited by Tim at 08-30-2007 10:45 AM]

Maurog
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Fair enough. Actually, that doesn't change anything, seeing as there will be only one iteration. It's a wise choice to pick 100 only if you are pretty sure that the other guy picked a 100 too, because otherwise you only get what he picked minus 2. In a parallel between multiplayer dilemma game and two-player dilemma game, being (say) 50% sure that the other guy picked a 100, translates to believing that 50% of all the other people have a 100 under the unhidetime tags. I find that rather unlikely, but my opinion doesn't count. Dynamic thinking, revealing your number, discussing strategies aloud, do anything you like, but nobody will have any idea what the real situation is, until the unmasking.

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Tahnan
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By the way, Maurog, did you mean for whatever you have behind unhidetime tags in your original post to be revealed a day before the thing ends?

Maurog
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Yes, this is the deadline for submitting your entry. In that way, there will be one day in which no new submissions are accepted, but everything is still hidden. And then the unmasking.

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Casebier
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Hmm... should just write on the paper: "Pete's Amount Minus $1"


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Maurog
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That's it... no more entries will be accepted.

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Tim
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Hmmm. I'm surprised so many people didn't guess 0.

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Maurog
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0 was not a valid choice to begin with, there was one 0 and I ignored it, as the person clearly didn't read the rules.

The results are up in the original thread. Here is the full calculations log, in secret tags for length:



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[Last edited by Maurog at 09-06-2007 01:24 PM]

Maurog
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Second half:


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Jason
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Maurog wrote:
0 was not a valid choice to begin with, there was one 0 and I ignored it, as the person clearly didn't read the rules.

Ah... Oops. I knew that the rules said an integer (between 2 - 100)!

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Kwakstur
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[begin long explaination]

98 won. Wow. As someone who previously read the article on the Traveler's Delimma, I must say it is very boring that 98 won.
Well, you have to admit, the only reason the 98's won is because of this: The number of 98's, 99's, and 100's present (just one fewer of any of those numbers and 97 and 98 would've tied, as shown below).

First off, here's what I meant by the technique is different between 2 players and more than 2. You see, with two players, the lowest number always wins.

But with more than two players, you must factor these following things:
Where t is the highest legal number, l is the lowest legal number, and x is the amount you gain/lose over the winning predictor's price . . .

The statement that you will lose when guessing t is still true.
But you cannot pick too low a number any more. People will generally pick high, and that will make you lose for these reasons:
When each person goes against you and you pick a low number, they will not get much from the round against you.
However, those high numbers will go against eachother. And if those those higher numbers are more than x higher than you, you will lose when you go against them.

And some of you may be wondering, why did the 98's beat 97? I mean, 97 beat all three of those 98s!
True. 97 did beat all three of those 98s.
True. 97 has a total better score once all those 97s and 98s face off.
But look! Because x is only 2, those three matches only gave 97 a 6-point head start over each other competing number.
Let's continue. We can infer that:
When 97 and 98 compete with the values 2-96, neither will get any advantage, as 97 and 98 are both larger than any of those values.
We already know that when 97 and the 98s compete, 97 gets an advantage by 6.
But it's the numbers 99 and 100 that make a difference: When the 97 competes with either of those numbers, it gets 99. But the 98s get 100.
So for every entry of 99 or 100, the scores of the 98s catch up to that 97 by one point. And since there are 7 such scores, the 98s actually surpass 97 by one point.

With x only being 2, there is very little reason to not go with rediculously high numbers. I mean, if you lose a round, you hardly fall that much behind who you just competed.
I think the game would've been much more interesting if x had been a higher number, such as 8.

In that case, I think 95 would've won, and I would've been incapable of typing a post this length, because it would no longer be easy to explain the results.

[end long explaination]

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