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I should point out that I've already been into the answer to this in some detail in the "others" section so if you don't want it spoiled, stay away from there!

Two Masters of Logic sit down for a friendly or not so friendly game of Iterated Prisoners' Dilemma. The rules of the game are quite simple:

Each player has 2 cards, marked "COOPERATE" or "DEFECT".

In each round of the game, each player chooses one card and plays it face down, then both cards are revealed.

If both players played "COOPERATE", they are awarded a point each.

If they both played "DEFECT", they get nothing.

If one player played "DEFECT" and the other "COOPERATE", the defector gets 2 points, and the cooperator loses 1 point for being a sucker.

The objective of the game is to amass as many points as possible*, as these will be converted into beer tokens after the game, and paid to the players by the Grand Master who is hosting the game.

It doesn't matter whether the players score more or less than each other, their sole objective is to maximise their own score.

The players do not know each other and may not confer or agree on a combined strategy, but they are both Masters of Logic, so will both play the very best strategy possible for their own gain. Each player knows that the other is also a Master of Logic.

The length of the game is not decided at the beginning, but is announced after the tenth round. On this occasion it happens to be 50 rounds.

How many points will each player get in total?

The Grand Master has not brought any beer tokens to the game. He knows both players will get zero points. They will play "DEFECT" on every round.

Later edit: After debating this extensively with Neida I still can't decide if it's right or not. It seems to come down to a matter of opinion.

*For clarity, each player doesn't care what the other player gets, and is only concerned with their own points

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I have a question about the game. You state that the game is played for 10 rounds and then the actual length of the game is announced. Is it possible that the game could have been ended after the ten rounds, or were the players guaranteed of at least one more round (the 11th round)?

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I have a question about the game. You state that the game is played for 10 rounds and then the actual length of the game is announced. Is it possible that the game could have been ended after the ten rounds, or were the players guaranteed of at least one more round (the 11th round)?
Interesting... I'll say the game could have been ended after the ten rounds (no more rounds guaranteed).

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It won't effect the answer, but can the points go negative?
Yes. In this case they would have to buy beer for the Grand Master.

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Both will cooperate the 100% of the times, since it is the only way to maximize profits, that in the end will become beer for the players. The combination C/C will award them 2 beers for each round, C/D only wan, since it's +2 for one player and -1 for the other, and D/D will award zero points... no case using the latter. Hence, C/C is the winner combination since both of them know the logic behind the game AND both know that the other player is a Logic Master as well

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Thanks jagdmc, I won't comment on your answer just yet, I'd rather wait and see what others have to say...:mellow: (pokerface)

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By defecting, the worst case is that you get nothing. This is essentially a fallback position when cooperation fails. It is also a way of punishing the other player, because the best they can do is get nothing.

If you believe the other player is going to defect, then you should defect so you don't lose a token.

No master of logic would continue to allow the other player to defect while they cooperate. Therefore, if the other player defects, you should defect on the next round as punishment.

However, alternating is not a good strategy because both players would receive 1 token every two turns... and you can do better.

Lets look a round somewhere in the middle. If you defect on one round, you can expect the other player to defect on the next round. So your rewards would be 2, then -1. An average over the two rounds of .5.

If you cooperate, you get 1 (since the other player would want the sustained reward of 1 as well).

There is no punishment possible after the last round. If you defect on the last round, you could try to get a bonus of two at the end. The other person would defect too being the master of logic he is. This makes the last round amount to nothing. You could look at the second to last round as the last round and continue the pattern of mutual defection, but this lowers total reward since there will be a defection by the other player coming. So you can assume cooperation up to the last round.

So cooperation continues until the last round... where logic sort of breaks down.

You are both masters of logic, so you should both come to the same conclusion and perform the same action.

Assume you believe both players will play cooperate. You should defect since it is the last round and you get an extra token for defecting if the other will cooperate. If you are going to defect on the last round, you can assume the other player reached the same conclusion (both masters of logic). Therefore you should both cooperate to make it so you get something on the last round. But if you believe both players will cooperate, you should defect.... and the cycle continues.

You could try and break the cycle by knowing that since you both are masters of logic, and therefore will both choose the same action, you should cooperate so you get a token (as opposed to defecting so you both get nothing). But then you know the other player will cooperate, and you will do better to defect...

I'll assume both players will try and avoid the scowls of the other player after the game, and will therefore both cooperate on the last round.

Both players get 50 points.

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Yes. In this case they would have to buy beer for the Grand Master.

If one of the master's of logic is a teetotaler and a friend with the Grand Master, they could conspire and play defect every round. That way the worst case for the Grand Master is that he doesn't give away any free drinks, and will probably get one bought for him!

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If one of the master's of logic is a teetotaler and a friend with the Grand Master, they could conspire and play defect every round. That way the worst case for the Grand Master is that he doesn't give away any free drinks, and will probably get one bought for him!
It's a good plan, but I must stress that no conspiracies of any sort are taking place. That includes conspiratorial "understandings" between the players. Neither will care if the other is left scowling at the end. I predict much controversy from this topic. It's far from over...:ph34r:

And with that, I bid you good night.

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Interesting... I'll say the game could have been ended after the ten rounds (no more rounds guaranteed).

The players need to consider the 10 round game

In a 10 round game the max any one player could score is 20 (he defects 10 times and the other player cooperates all 10) Now that is not going to happen because as soon as a player defects the trust is broken and the subsequent player would more than likely start to defect in an act of self preservation.

So both players in the interest of maximizing their own points will cooperate for some period of time but for how long?

consider always cooperating: if they both cooperated for all 10 rounds they each would be awarded 10 pts. (with a possibility of more to come)

consider defecting: If one defects at round 10 he could get either 9 points or 11 points depending if the other player also defects. (but virtually eliminates any chance for more points because the trust is broken. now in our case where 40 more rounds are announced I'm betting a fragile trust would be reestablished with both playing a game of chicken on when it would be broken again)

if one defects before round 10 the points get lower still, so this option is eliminated.

my guess the is they would cooperate for the first 10, because the value of a lot more possible rounds it much greater than the 11 points you could get by defecting.

So would someone then defect at round 50 to get 51 points instead of just 50?

if both are masters I say they would reason that if one were to defect, then both would and then they would end at 49, so both would reason the best they could hope for is both cooperating the entire game and both would have 50 points.

Edited by preflop

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Simplest Case: Deterministic Symmetric Logicians:

Assuming that both are equally logical in every aspect, then no matter what one decides, the other will also reach the same conclusion and both will be aware of this fact.

This symmetry indicates that only a cooperative strategy would work. Since there are no descriptions in the problem to distinguish one player from another it also makes sense that each player will make the same move the first round (because of the prior assumption that they are exactly equal in logical skill). Realizing this, the two outcomes are: cooperate gives 1 point, defect gives 0 points. Both players will choose to cooperate. The same thing will happen for all subsequent rounds.

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Both players will always choose to defect. Imagine that you are player A. No matter what the other player does, you will always be better off if you defect. If you defect and B cooperates, you get two beers. If you defect and B defects, you get nothing. However, if you cooperate and B cooperates, you only get 1 beer (where you would get 2 had you defected). And worst of all, if you cooperate and B defects you get minus 1 beer, where you would have gotten a net of 0 had you defected. So, even though this case means nobody gets any beer, it is the most logical in this situation.

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It's simple and makes sense. If two identical players have identical decisions to make every round, and both players know it, then defecting will never be the correct play. If it was, both players would pick it and neither would be awarded any points.

Edited by Tuckleton

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OK, I guess it's time to give some input. For anyone who's stuck, this spoiler gives about half the answer:

Typically with Prisoners' Dilemma games, the optimal strategy depends heavily on circumstances.

As Classicstang pointed out, for a single round, defection gives each player the highest score possible given that you cannot control what the other player does. Paradoxically it makes both players worse off than if they had both chosen to cooperate, but since you can only control what you do and not what the other player does, neither player has it in their power to ensure mutual cooperation.

Tuckleton and mmiguel1 have pointed out that since the players are both Masters of Logic, they will both make the same decisions, so asymmetric results can be ignored and both can choose to cooperate confident in the knowledge that the other will do the same. This may prove to be a bone of contention. I disagree on the basis that Masters of Logic can be expected to make the same decisions as each other as a consequence of their reasoning, but to use that as a starting assumption of such reasoning is circular logic. It's no more valid than a Master of Logic thinking they are right about something just because Masters of Logic are never wrong. A player who thinks the other will cooperate would still have a choice about her own move. To think "if I choose to defect, he will too" is to depend on a line of cause and effect which does not exist. Neither player's choice can directly influence the other's choice. Therefore, in a single round, both would independently choose to defect.

However, when Prisoners' Dilemma is repeated over an indeterminate number of rounds, things get more interesting. Defection isn't necessarily the best strategy when the other player may retaliate in subsequent rounds. When the other player's strategy is not known, "tіt-for-tat"-style strategies work well. Basic tіt-for-tat is a strategy that cooperates on the first round then always repeats the other player's last move. When faced with another player playing tіt-for-tat, both will cooperate on all rounds. When faced with a defecting strategy, tіt-for-tat loses out on the first defection only. If the other player is likely to play something like tіt-for-tat, it would be wise to do similar, since an "always defect" strategy would win you only 2 beer tokens rather than the many you get by playing tіt-for-tat. Tіt-for-tat is a robust and essentially cooperative strategy.

In this case, however, the number of rounds is known...

P.S. I reckon EventHorizon has come closest to getting to the bottom of all this

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Probablistic Logicians:

Even if the masters of logic are not identical in thinking, they will both surely consider the cooperative case as one of their considerations. Even if defecting the first round will give 2 points over one point it lowers the probabality of the other player cooperating in future rounds. Every time the other player does not cooperate, you will either gain nothing or lose one point. As a consequence of broken trust and the loss of potential gains in the future, I believe that this will decrease the expected value of tokens won for the original player. There is more value in the promise of SUSTAINED GROWTH than in a 2 to 1 point increase for a single turn. Two people do not have to have identical thoughts to adopt the logically sound strategy of choosing such to maximize the probablistic expected value of points. There is always a chance of the other player defecting but the probability is lower when there is more trust.

If one player began the game trying to cooperate every round and the other player began the game with plans to defect every round. The cooperative player would with high probability ultimately decide to defect for the rest of the game. There is no more potential to earn tokens for either player and this was not the best decision for the player who originally decided to defect (because he ends up with less tokens than 50 due to his choice).

There can be more complicated plays: for example, to lure the other player back into the game, the original defector may cooperate a few times, hoping that these losses will be paid for by future plays of defect-cooperate. In some cases they may get more points than 50, but based on human nature I think this is not very probable because on every single round the first player has a chance to "quit" and defect for the rest of the game or to punish their opponent by a string of defections (which prevents SUSTAINED growth for both parties). I have not done calculations but I think that the action of defecting has the result of lowering the expected value of won tokens. It is therefore logical for both players to cooperate in order to maintain (maximize) the expected value of winning.

Edited by mmiguel1

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Probablistic Logicians:

Even if the masters of logic are not identical in thinking, they will both surely consider the cooperative case as one of their considerations. Even if defecting the first round will give 2 points over one point it lowers the probabality of the other player cooperating in future rounds. Every time the other player does not cooperate, you will either gain nothing or lose one point. As a consequence of broken trust and the loss of potential gains in the future, I believe that this will decrease the expected value of tokens won for the original player. There is more value in the promise of SUSTAINED GROWTH than in a 2 to 1 point increase for a single turn. Two people do not have to have identical thoughts to adopt the logically sound strategy of choosing such to maximize the probablistic expected value of points. There is always a chance of the other player defecting but the probability is lower when there is more trust.

I'm not sure if you've read the spoiler in my last post, if not you may as well. I have criticised your approach of depending on the symmetry between players to remove asymmetrical possibilities from their reasoning, since I wanted to establish that in a single round game, we'd expect to see mutual defection. However, what you say here about multiple round games is quite valid. Cooperation may be the best way to start (and if reciprocated, continue) in these cases. The tіt-for-tat strategy I mentioned works like that (it's never the first to defect).

That said, there's more to the answer...

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cooperate until the tenth round when told the number of rounds, then defect every round after that because of the unexpecting-hanging (or "surprise tiger") paradox. No matter what they've been doing before, playing defect on the very last round (when known) will do them better (2/0 instead of 1/-1). However, then going back a round they would defect then as well since that has become the last round before defection. Except for every round that they both defect, they get 1 less point each than if they had both cooperated (assuming they both go back the same level of thought nesting)... anyway, I realized it wouldn't matter before round 10 cuz they do know that it has to end SOMETIME (right?) so that means their chain of logic will go infinitely back (being Masters) and they will defect every time. But even if one starts the defect chain earlier, he doesn't make up for it and they both get sub-optimal.

So I conclude that they both Cooperate every single round. Even on the last round, because they both get 1 point then too, instead of 0 points (if they both defect), and since they know the other person is also a perfect logic master, conspiring to defect means the other is also considering defecting. They will mutually realize the value of cooperation then because even after infinitely going back chains of logic, the conclusion is the same spread out over any number of rounds: both cooperating gets more than both defecting.

That being said, your actions don't affect theirs, so still defect on that last round. But the round before it will still be subject to tít-for-tat because defect then and both of you will defect on the final round and get no points.

So

they'll cooperate up until the final round where either both or one of them will defect, unless they are so masterful they realize that the other person is thinking the same and follows the same paths - in which case they would cooperate 100%

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I'm not sure if you've read the spoiler in my last post, if not you may as well. I have criticised your approach of depending on the symmetry between players to remove asymmetrical possibilities from their reasoning, since I wanted to establish that in a single round game, we'd expect to see mutual defection. However, what you say here about multiple round games is quite valid. Cooperation may be the best way to start (and if reciprocated, continue) in these cases. The tіt-for-tat strategy I mentioned works like that (it's never the first to defect).

That said, there's more to the answer...

The problem here is that in your original post, you state that the goal is not to have more points than the other person, just to maximize your own points. Meaning that it is not a competition, and so there is no intrinsic value in defecting unless it is going to increase your points. given the rules of the game it dictates mutual cooperation is the only way to ensure a highest possible score. Because if you decide to defect after 49 rounds you may end up with 51, but you may end up with 49. (if after 48 rounds its either 48 or 50). two masters would come to the conclusion that if they go for the 51st point it is equally likely the other would as well and then end up with 49 points. So the solution is to cooperate and be guaranteed of 50, instead of defecting and averaging to 50 if you are lucky (which you wont be, because the other guy is a master).

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The problem here is that in your original post, you state that the goal is not to have more points than the other person, just to maximize your own points. Meaning that it is not a competition, and so there is no intrinsic value in defecting unless it is going to increase your points. given the rules of the game it dictates mutual cooperation is the only way to ensure a highest possible score. Because if you decide to defect after 49 rounds you may end up with 51, but you may end up with 49. (if after 48 rounds its either 48 or 50). two masters would come to the conclusion that if they go for the 51st point it is equally likely the other would as well and then end up with 49 points. So the solution is to cooperate and be guaranteed of 50, instead of defecting and averaging to 50 if you are lucky (which you wont be, because the other guy is a master).
If you cooperate on the last round you are not guaranteed 50. The other player could defect and reduce your score to 48 :(

cooperate until the tenth round when told the number of rounds, then defect every round after that because of the unexpecting-hanging (or "surprise tiger") paradox. No matter what they've been doing before, playing defect on the very last round (when known) will do them better (2/0 instead of 1/-1). However, then going back a round they would defect then as well since that has become the last round before defection. Except for every round that they both defect, they get 1 less point each than if they had both cooperated (assuming they both go back the same level of thought nesting)... anyway, I realized it wouldn't matter before round 10 cuz they do know that it has to end SOMETIME (right?) so that means their chain of logic will go infinitely back (being Masters) and they will defect every time. But even if one starts the defect chain earlier, he doesn't make up for it and they both get sub-optimal.

So I conclude that they both Cooperate every single round. Even on the last round, because they both get 1 point then too, instead of 0 points (if they both defect), and since they know the other person is also a perfect logic master, conspiring to defect means the other is also considering defecting. They will mutually realize the value of cooperation then because even after infinitely going back chains of logic, the conclusion is the same spread out over any number of rounds: both cooperating gets more than both defecting.

That being said, your actions don't affect theirs, so still defect on that last round. But the round before it will still be subject to tít-for-tat because defect then and both of you will defect on the final round and get no points.

So

they'll cooperate up until the final round where either both or one of them will defect, unless they are so masterful they realize that the other person is thinking the same and follows the same paths - in which case they would cooperate 100%

Now this is more like it!

Woo yay! You even got past the red-herring 10th round! But then...

"so that means their chain of logic will go infinitely back (being Masters) and they will defect every time. But even if one starts the defect chain earlier, he doesn't make up for it and they both get sub-optimal."

I honestly don't know if that's nonsense or I just don't understand it.:D But I'm loving your answer anyway. Why wouldn't they just defect on all 50 rounds?

"So they'll cooperate up until the final round where either both or one of them will defect"

If one will defect on the final round it follows that they both will, since a MoL would not be the one to cooperate when the other must defect.

Given that, why would either player cooperate on the 49th round?

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If you cooperate on the last round you are not guaranteed 50. The other player could defect and reduce your score to 48 :(

Now this is more like it!

Woo yay! You even got past the red-herring 10th round! But then...

"so that means their chain of logic will go infinitely back (being Masters) and they will defect every time. But even if one starts the defect chain earlier, he doesn't make up for it and they both get sub-optimal."

I honestly don't know if that's nonsense or I just don't understand it.:D But I'm loving your answer anyway. Why wouldn't they just defect on all 50 rounds?

"So they'll cooperate up until the final round where either both or one of them will defect"

If one will defect on the final round it follows that they both will, since a MoL would not be the one to cooperate when the other must defect.

Given that, why would either player cooperate on the 49th round?

if they don't, they'll get less than if they do. And if they don't, they won't the 50th round, and they'll get even less.

As has been said, the most optimal is getting all 50 in cooperation for 50 points

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if they don't, they'll get less than if they do. And if they don't, they won't the 50th round, and they'll get even less.

As has been said, the most optimal is getting all 50 in cooperation for 50 points

If one player plays cooperate on all 50 rounds, the other can get 51 points for betraying on the last round. Where's the downside in that?

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If one player plays cooperate on all 50 rounds, the other can get 51 points for betraying on the last round. Where's the downside in that?

but if they both defect, they both get 49. 100 may be maximum, but 50 is optimal

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but if they both defect, they both get 49. 100 may be maximum, but 50 is optimal
You can't control what the other player does on the last round. If he defects and you dont you only get 48. Either way you're better off defecting yourself.

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