[Guest]

the questionable prisoners dilemma goes like this:

Two suspects are arrested by the police. The police have insufficient evidence for a conviction, and, having separated both prisoners, visit each of them to offer the same deal. If one testifies (defects from the other) for the prosecution against the other and the other remains silent (cooperates with the other), the betrayer goes free and the silent accomplice receives the full 10-year sentence. If both remain silent, both prisoners are sentenced to only six months in jail for a minor charge. If each betrays the other, each receives a five-year sentence. Each prisoner must choose to betray the other or to remain silent. Each one is assured that the other would not know about the betrayal before the end of the investigation. How should the prisoners act?

so we have the following table.

           p2    |   p2

          accuse | silent

          -------|--------

p1 accuse |5 yrs | 10 yrs

p1 silent |10 yrs| 6 mths

obviously both remaining silent is the best option, but remember, you don't necessaryly know what your partner is going to do. if you accuse, your guaranteed no more than 5 years and may not be charged at all, where as if you remain silent you may do some serious time, or hardly any time at all.

here's a fun one I came up with.

your boss brings you in and tells you the following. they are going to give out a raise to you and 9 other people, but with the following condition; if you cooperate, and all 9 other employees do as well, you all get a 4 dollar an hour raise. if you defect, they'll split a 10$ an hour raise amongst the defectors. ie. if you're the only defector you get a 10 dollar an hour raise, if only 2 people defect you get a 5 dollar an hour raise, etc. down to 1 dollar an hour for all defectors; while the employees who cooperated get zilch. which would you pick?

would the increase of cooperating to 7 dollars an hour influence your decision?

Edited January 5, 2010 by phillip1882

[Guest]

Well, if we assume that everyone involved is logical and rational (a stretch, I know ), then I would say that it's worth it to be silent. Of the nine people, only three need to "defect" to screw over everyone (including themselves). The only time anyone gets a better deal than the original offer is if only one or two people defect. If a third person defects, then the defectors all lose 66 cents while the rest of their coworkers will be really upset with them.

If you are considering defecting, chances are one or two others are as well. So it should be smarter for everyone involved to cooperate.

[unreality]

It really depends on what I know about the other person

[Izzy]

Depends on the person. If it's my best friend, I remain silent. Random stranger: bust 'em. I either get a free walk or five years, better than if they should have busted me.

Though, if there's no hard evidence, wouldn't neither of us go to jail regardless of a testimony?

[plasmid]

I think I read a Mathematical Recreations article on this a while back that got pretty sophisticated. Sure, if you're dealing with one other person (or set of people) in one instance and will never deal with them again, then it makes perfect sense to defect and be a traitor because it would give you the better outcome regardless of what the other person does.

A more interesting question is, what happens if you will deal with the same person again? Then it would be better if you can cooperate with the other person repeatedly to get a better outcome for all than if you were both constantly defecting. But you also have something to gain if you can cheat every once in a while without blowing the overall cooperation.

From there it delved into a game theory analysis which actually looks sort of like the rock-paper-scissors contest going on in the Games forum right now, except of course that it's a prisoner's dilemma contest between different algorithms instead. The exact values of how long each prisoner stays in prison for each possible outcomes are variables that can affect which strategies are optimal. I don't remember what the end result of the paper was anymore though. Maybe if any of us programmers get a significant amount of free time again we could try it out here after rock-paper-scissors. Unfortunately I'm once again kinda pressed for the forseeable future.

[Guest]

Are we to beieve everyone is capable of understanding the problem and, no errors occur, and wnat to achieve maximim possible return? And, that all are actually aware of each others status(ability) of understanding too - otherwise it goes to random depending on how few/many are able to appreciate the probem fully.

Edited January 6, 2010 by Lost in space

[Guest]

I think I read a Mathematical Recreations article on this a while back that got pretty sophisticated. Sure, if you're dealing with one other person (or set of people) in one instance and will never deal with them again, then it makes perfect sense to defect and be a traitor because it would give you the better outcome regardless of what the other person does.

A more interesting question is, what happens if you will deal with the same person again? Then it would be better if you can cooperate with the other person repeatedly to get a better outcome for all than if you were both constantly defecting. But you also have something to gain if you can cheat every once in a while without blowing the overall cooperation.

From there it delved into a game theory analysis which actually looks sort of like the rock-paper-scissors contest going on in the Games forum right now, except of course that it's a prisoner's dilemma contest between different algorithms instead. The exact values of how long each prisoner stays in prison for each possible outcomes are variables that can affect which strategies are optimal. I don't remember what the end result of the paper was anymore though. Maybe if any of us programmers get a significant amount of free time again we could try it out here after rock-paper-scissors. Unfortunately I'm once again kinda pressed for the forseeable future.

I'm pretty sure that when working with the same two agents, the optimal solution is to both stay silent every time until one defects. At that point, the Nash Equilibrium changes and it becomes optimal for both sides to defect every time after that.

[unreality]

plasmid: I remember reading something about that in a Martin Gardner article a while back. They had an algorithm contest similar to ours, and a t it-for-tat strategy won I believe; it did whatever the other prisoner did the previous round

[Guest]

If the prisoners testified against one another, wouldn't they both get 10 years instead of 5?

[Guest]

plasmid: I remember reading something about that in a Martin Gardner article a while back. They had an algorithm contest similar to ours, and a t it-for-tat strategy won I believe; it did whatever the other prisoner did the previous round

Consider a pair of perfectly rational players, playing repeated prisoner's dilemma with the sole intention of maximising their own reward. The number of rounds is not known but thought to be in the region of about 30. As far as I'm aware, tìt-for-tat is the best general strategy, and on that basis both players will take the cooperate (silent) option and maximise their shared rewards since under tìt-for-tat neither will be the first to defect (betray).

Ten rounds into the game, the players are made aware that the game will continue, not for about 30 rounds as thought, but for precisely 100 rounds.

What happens next? I'll spoiler it for added fun...

Both players know that on the last round, there will be no sense in continuing with tìt-for-tat and it would be best to defect on this round, since there will be no opportunity for the other player to retaliate. They also know that the other player, being similarly aware of the situation, is bound to do likewise.

Why not defect in the 99th round as well? The 100th round now offers no opportunities for reprisals since mutual defection in this round is certain in any case. So both players will also defect in the 99th round.

By induction, now that the length of the game is fixed, both players would choose to defect in all remaining rounds.

Can such a mutually destructive course of action really be the optimal strategy?

[Guest]

kinda reminds me of the hidden tiger paradox.

the solution in my opinion is to realize that while its true you can't retaliate after the last round this doesn't make the last round necessary for defection. even here, only doing what your opponent did the second to last round is still optimal.

let me put it to you this way. let's say you met God, and he said you would die tomorrow, guaranteed, but that he wouldn't punish any sins you had for today. would you go on a killing spree? take as many with you as you could? would you have sex with as many random beautiful women you could? me personally despite my "get out of jail free" card, I'd want to spend it with friends and family.

[Guest]

kinda reminds me of the hidden tiger paradox.

What's that?

the solution in my opinion is to realize that while its true you can't retaliate after the last round this doesn't make the last round necessary for defection. even here, only doing what your opponent did the second to last round is still optimal.

A strategy that works on the basis of "play tìt-for-tat until the last round and then betray on the last round" would clearly do better than a pure tìt-for-tat strategy. I'm considering this as a strategic question rather than a moral one. The moral thing to do in the repeated prisoners dilemma may be to cooperate no matter what the other player does. That's a pretty poor strategy though, as the other player can easily make a sucker out of you.

let me put it to you this way. let's say you met God, and he said you would die tomorrow, guaranteed, but that he wouldn't punish any sins you had for today. would you go on a killing spree? take as many with you as you could? would you have sex with as many random beautiful women you could? me personally despite my "get out of jail free" card, I'd want to spend it with friends and family.

As an atheist I am faced with that question every single day. Morality comes from within

[Guest]

Consider a pair of perfectly rational players, playing repeated prisoner's dilemma with the sole intention of maximising their own reward. The number of rounds is not known but thought to be in the region of about 30. As far as I'm aware, tìt-for-tat is the best general strategy, and on that basis both players will take the cooperate (silent) option and maximise their shared rewards since under tìt-for-tat neither will be the first to defect (betray).

Ten rounds into the game, the players are made aware that the game will continue, not for about 30 rounds as thought, but for precisely 100 rounds.

What happens next? I'll spoiler it for added fun...

Both players know that on the last round, there will be no sense in continuing with tìt-for-tat and it would be best to defect on this round, since there will be no opportunity for the other player to retaliate. They also know that the other player, being similarly aware of the situation, is bound to do likewise.

Why not defect in the 99th round as well? The 100th round now offers no opportunities for reprisals since mutual defection in this round is certain in any case. So both players will also defect in the 99th round.

By induction, now that the length of the game is fixed, both players would choose to defect in all remaining rounds.

Can such a mutually destructive course of action really be the optimal strategy?

What's that [Hidden Tiger paradox]?

The Hidden Tiger (a.k.a. the Unexpected Tiger) paradox involves a prisoner given a chance for freedom from the king.

The king calls the prisoner before him and tells the man that, being a just and fair king, he will give the prisoner a chance to escape. There are five doors before the prisoner. To escape, the prisoner must open the doors in order from left to right. Behind one of the doors will be an unexpected tiger, which the prisoner must kill to win his freedom.

The prisoner does not know which door the tiger is behind, but as the location of the tiger is given to be unexpected, the prisoner reasons that the tiger cannot be behind door 5 since he would be expecting it by that time. Since door 5 is out, similarly door 4 is out and following the induction, he reasons that there is no tiger since there is no door behind which it could be a surprise. So he opens the doors one by one and the tiger jumps out at him when he opens door 3, quite unexpectedly.

His reasoning seems sound, but leads to an illogical conclusion. Your statement in the spoiler above fits into that situation nicely. The reasoning to immediately defect since t it-for-tat won't work on round 100 leads to the same suboptimal solution that the prisoner reached in dealing with the tiger.

[Guest]

...His reasoning seems sound, but leads to an illogical conclusion. Your statement in the spoiler above fits into that situation nicely. The reasoning to immediately defect since t it-for-tat won't work on round 100 leads to the same suboptimal solution that the prisoner reached in dealing with the tiger.

Ahh, thanks for that. Intuitively, you'd think there's something wrong with my inductive reasoning too, though I wonder where exactly it breaks, and what the optimal solution would be.

I couldn't resist doing a 10-round experiment on a spreadsheet to get a rough idea...

When you know the number of rounds, "tìt-for-tat until last round, then betray" (TT-1B for short) is certainly a superior strategy to "tìt-for-tat" (TT). Whatever the other player does, TT-1B gives you at least as good a result as TT, and sometimes a better one.

After that, it gets a bit more complicated. TT-2B (betray on last 2 rounds) doesn't beat TT-1B in all cases. Playing against TT, TT-1B yields a better result than TT-2B, but otherwise TT-2B is better. If the other player is equally likely to play any of the 3, TT-2B is marginally the best, based on the negative reward table in the OP. In fact, if we consider the other player to be equally likely to play any of TT, TT-1B, TT-2B, TT-3B .... TT-10B, then TT-2B is the best bet.

But here's the paradox. The other player is not equally likely to play any of those 11. As I've defined them, the players will choose the best strategy they can, so you can count out TT completely since it is always inferior to TT-1B. If the remaining 10 are equally likely, TT-3B has now become the best bet, but TT-1B is inferior to TT-2B in all cases so you can count that out... and we're back to the tiger paradox.

The difference between this and the tiger paradox is that there is an answer to this, namely that the optimal strategy is just to betray all the time. If two Masters of Logic were playing, they may be forced to do exactly that. Two less well-informed players might innocently score a much better result by cooperating, blissfully unaware that their more successful strategy is not the "optimal" one

[Guest]

Both players know that on the last round, there will be no sense in continuing with tìt-for-tat and it would be best to defect on this round, since there will be no opportunity for the other player to retaliate. They also know that the other player, being similarly aware of the situation, is bound to do likewise.

Why not defect in the 99th round as well? The 100th round now offers no opportunities for reprisals since mutual defection in this round is certain in any case. So both players will also defect in the 99th round.

By induction, now that the length of the game is fixed, both players would choose to defect in all remaining rounds.

Can such a mutually destructive course of action really be the optimal strategy?

Reminds me of the blue-eye problem.

Although... I don't understand why anybody would ever even consider choosing anything but stay-silent all the time. Unless you know the other guy is guilty and deserves greater jail time, or unless you're scared that the other guy might have it in for you, then staying silent is always optimal. Of course, those are two big assumptions...

The prisoner does not know which door the tiger is behind, but as the location of the tiger is given to be unexpected, the prisoner reasons that the tiger cannot be behind door 5 since he would be expecting it by that time. Since door 5 is out, similarly door 4 is out and following the induction, he reasons that there is no tiger since there is no door behind which it could be a surprise. So he opens the doors one by one and the tiger jumps out at him when he opens door 3, quite unexpectedly.

I've encountered this same paradox with different wording before. Here's my view:

Once you're told that there's going to be a lion behind a door (and you assume he's telling the truth, at least somewhat), then it's impossible for a lion to be _unexpected_. The first door: you expect a fifth of a lion (lol, but that's expected value/quantum superposition for you... but really, it just means there's a 20% chance). If no lion, then on the second door, you expect two fifths of a lion, and so on until you expect one full lion - IF you didn't encounter lions before hand.

So it's just a semantic misunderstanding of "expectation" on the part of either the king or the prisoner.

However, if you word the problem differently, then I'm not sure if my circumvention of the paradox still holds...

The way I heard it was basically this: there's a prisoner that's gonna be killed on some day Mon through Fri at noon, and he won't know what day that is ahead of time.

When I first analyzed the paradox, I think I broke it down into three statements, any two of which will work together just fine, but all three together don't work. So I think I concluded that the judge had to have been lying about one of any of the three statements (or at least lying in the sense that he knew one had to be false, even if HE wasn't sure which would eventually become false). I think. I could figure it out again if I felt like it...

Edited January 7, 2010 by DarthNoob

[Guest]

Seeing as this is a puzzle forum, I've reworded my previous ramblings as a puzzle. If you've been reading this topic, leave it alone!