Trusty Old Rock (Logic Problem)
- Thread starter Earthling
- Start date
Unless I have completely misunderstood the scenario, it would be idiotic to choose $2. If the aim is to maximise your potential monetary gain, without concern for beating the other contestants, then anything above $6, you are automatically assured to gain more financially than if you chose $2. Only if there is incentive in winning out over your competitors (e.g. the lowest bid gets what they bid, the next lowest bid gets half of what the first person wins, the next lowest bid gets half of that, etc.) does the situation become at all sensible.
Erik Mesoy
Core Tester / Intern
No you don't. Read the OP properly:
"If you choose different numbers, the pay out is as follows:
-The lower of the two numbers will be chosen
-The player who submitted the lower number gets that amount, plus $2.
-The player who submitted the higher number gets the amount minus $2."
If you pick 98, 99 or 100 and the other guy picks 50, you get 48 and the other guy gets 52.
The gain of 'cheating' in some way is $1. The loss is $98.
A game theory opponent would have to be odd to maximize gain by dropping all the way down to $2. That's not a maximal gain. His problem is that the cheating option that gives him more benefit than co-operation is not the cheating option that protects him from being undercut.
This, I believe, although I know extraordinarily little about the subject, is not usually how these games are set up.
Since beating your opponent is not part of the game there is no incentive to choose low numbers such as $2. There would be if only the winner got much more money. The benefit-maximizing opponent who thinks that you're the same can see that the gains to be had ($100) outweigh the losses from being just undercut ($3) or the gains from undercutting ($1) sufficiently for undercutting to be an unhelpful tactic.
The risk of being undercut remains every time one attempts to undercut oneself, so by aiming for $1 more one retains the chance to lose at least $3. Given that each undercut beyond the first reduces the possible gain by $1 the undercutting tactic is untenable. It depends on information which has not been given: the number that one's opponent has chosen.
The iteration of thoughts relies on taking the last thought (he chooses 96) and subtracting one, to give 95, then 94 and so on. But the last thought is entirely irrelevant: one needs to return to the beginning each time. What is relevant is that by choosing lower numbers the possible gain is automatically reduced.
So instead of thinking 'if he chooses 99 I need to choose 98' followed by 'if he chooses 98 I need to choose 97' I can rearrange all these thoughts in any order to give the same meaning, and end on 'if he chooses 68 I need to choose 67'. However, that's not the only concern. Against this need to gain $1 for any particular number my opponent has chosen I need to balance the loss of us choosing anything lower than 97, which is what I would get if I could be assumed to choose 100.
The potential gain of $1 is less than the loss of many dollars from choosing numbers below 97. Hence I, and a rational opponent, will choose in the high 90s.
If given the option of choosing from 2 to 7 I would call 7 the best choice. 2 to 5 or 6 would be trickier to call.
A game theory opponent would have to be odd to maximize gain by dropping all the way down to $2. That's not a maximal gain. His problem is that the cheating option that gives him more benefit than co-operation is not the cheating option that protects him from being undercut.
This, I believe, although I know extraordinarily little about the subject, is not usually how these games are set up.
Since beating your opponent is not part of the game there is no incentive to choose low numbers such as $2. There would be if only the winner got much more money. The benefit-maximizing opponent who thinks that you're the same can see that the gains to be had ($100) outweigh the losses from being just undercut ($3) or the gains from undercutting ($1) sufficiently for undercutting to be an unhelpful tactic.
The risk of being undercut remains every time one attempts to undercut oneself, so by aiming for $1 more one retains the chance to lose at least $3. Given that each undercut beyond the first reduces the possible gain by $1 the undercutting tactic is untenable. It depends on information which has not been given: the number that one's opponent has chosen.
The iteration of thoughts relies on taking the last thought (he chooses 96) and subtracting one, to give 95, then 94 and so on. But the last thought is entirely irrelevant: one needs to return to the beginning each time. What is relevant is that by choosing lower numbers the possible gain is automatically reduced.
So instead of thinking 'if he chooses 99 I need to choose 98' followed by 'if he chooses 98 I need to choose 97' I can rearrange all these thoughts in any order to give the same meaning, and end on 'if he chooses 68 I need to choose 67'. However, that's not the only concern. Against this need to gain $1 for any particular number my opponent has chosen I need to balance the loss of us choosing anything lower than 97, which is what I would get if I could be assumed to choose 100.
The potential gain of $1 is less than the loss of many dollars from choosing numbers below 97. Hence I, and a rational opponent, will choose in the high 90s.
If given the option of choosing from 2 to 7 I would call 7 the best choice. 2 to 5 or 6 would be trickier to call.
But regardless of what your opponent picks, choosing 99 or 98 gives you at least equal payoff. Choosing 100 is limiting your potential gain.
Consider this table showing your profit:
.................Opponents choice.........95........96..........97.........98...........99......100
Your choice
95..............................................95........97..........97.........97...........97.......97
96..............................................93........96..........98.........98...........98.......98
97..............................................93........94..........97.........99...........99.......99
98..............................................93........94..........95.........98...........100.......100
99..............................................93........94..........95.........96...........99.......101
100............................................93........94..........95.........96...........97.......100
I've put in green the situations where a move of 98 or 99 is superior to a move of 100. In all other possibilities, 98 or 99 is equally good as 100. It is never worse.
I'm not saying 98 or 99 is the answer necessarily, just that 100 certainly isn't.
Mise
isle of lucy
If the best choice was 98, then everyone would pick 98, and no-one would get more than 98. If the best choice was 100, then everyone would pick 100, and no-one would get less than 100. That's the rationale behind, e.g., Erik so dogmatically exhorting the virtue of picking 100 over any other number.
Yes, but you don't know that your opponent will pick any specific number be it 98, 100 or whatever. Regardless of what your opponent picks you are at least as well off choosing 98 or 99 as you are 100. You shouldn't pick 100 whether you think you can predict your opponent's move or not.
I don't follow.
The only information is the rules of the game, I can't say whether my opponent will pick 98. And so what if they do pick 98? 100 would still be an inferior choice for me to make.
Erik Mesoy
Core Tester / Intern
(Warning: Deliberately tongue-in-cheek post ahead in the style of St. Augustine, or rather of my probably ignorant impression of St. Augustine. 
)
Assuming arguendo that the heretic has eliminated 100 as an option, I have put in red where a move of 98 is superior to 99. In all other possibilities, 98 is equally good as 99. It is never worse.
.................Opponents choice......... 95........96..........97.........98...........99
Your choice
95.............................................. 95........97..........97.........97...........97
96.............................................. 93........96..........98.........98...........98
97.............................................. 93........94..........97.........99...........99
98.............................................. 93........94..........95.........98...........100
99.............................................. 93........94..........95.........96...........99
Thus the heresy, by the same aberrant argument which is used to condemn 100, must perforce abjure 99.
Now by the continuation of this argument, unless the heretic renounce one of his earlier propositions, he is forced to abjure 99 once 98 is shown superior, et cetera, until he must choose 2 because it is superior to 3.
But this is nonsense. Quod erat demonstrandum.
Objection. The heretic may find himself dealing with a man who wishes to be God-fearing and chooses 100 but nonetheless deals with heretics, in which case 99 is superior to 98.
Answer 1. In this case the heretic's strength is derived from the man choosing 100, which the heretic would have condemned if he were to do it himself.
Answer 2. Even then it is the case that 98 is superior to 99 by $2 in one instance, by $1 in another instance, and only inferior by $1 in a single instance.
)
Rebuttal of damned heresy by means of reductio ad absurdum:
Assuming arguendo that the heretic has eliminated 100 as an option, I have put in red where a move of 98 is superior to 99. In all other possibilities, 98 is equally good as 99. It is never worse.
.................Opponents choice......... 95........96..........97.........98...........99
Your choice
95.............................................. 95........97..........97.........97...........97
96.............................................. 93........96..........98.........98...........98
97.............................................. 93........94..........97.........99...........99
98.............................................. 93........94..........95.........98...........100
99.............................................. 93........94..........95.........96...........99
Thus the heresy, by the same aberrant argument which is used to condemn 100, must perforce abjure 99.
Now by the continuation of this argument, unless the heretic renounce one of his earlier propositions, he is forced to abjure 99 once 98 is shown superior, et cetera, until he must choose 2 because it is superior to 3.
But this is nonsense. Quod erat demonstrandum.
Objection. The heretic may find himself dealing with a man who wishes to be God-fearing and chooses 100 but nonetheless deals with heretics, in which case 99 is superior to 98.
Answer 1. In this case the heretic's strength is derived from the man choosing 100, which the heretic would have condemned if he were to do it himself.
Answer 2. Even then it is the case that 98 is superior to 99 by $2 in one instance, by $1 in another instance, and only inferior by $1 in a single instance.
I don't follow.
The only information is the rules of the game, I can't say whether my opponent will pick 98. And so what if they do pick 98? 100 would still be an inferior choice for me to make.
You can assume that the other person knows everything that you do. And you know that picking 98 will most likely have the most beneficial outcome. So, the other person would know this too. Meaning that they would pick 98. Which would eliminate the benefit derived from picking that number.
Mise
isle of lucy
EDIT: Delete me pls!
(Warning: Deliberately tongue-in-cheek post ahead in the style of St. Augustine, or rather of my probably ignorant impression of St. Augustine.
Head asplode
This is going to go round and round! I don't for the record think that 98 "will most likely have the most beneficial outcome", just that it is better than choosing 100.
If you KNOW your opponent will pick 98 or indeed any number, then you will want to undercut them by 1 to get the greatest profit. But using your reasoning, your opponent is aware of this too and will want to undercut you by 1. Therefore you can't say that they will pick 98. Or something.
First of all I gotta say this - I love everything Erik has posted, showing some true human spirit there, and obviously humor 
As to a couple of other minor loose ends - someone mentioned the "social" factors of being on a game show or in an experiment. I really think it is best to disregard these, if only because in the first place I made up the game show situation. The original problem you'd read in any links in fact involves travelers who are being compensated by an airline. But at any rate, this scenario introduces some random social "reputational" factors as well which is why I tried to avoid it.
So I guess I'm glad to see the responses of everyone else agreeing with me, but then again people are right to critique a couple lines of reasoning - I don't think this problem is adequately addressed with trying to mathematically find probabilities against "random" or something - that's not the right approach. I think it is useful to look and see why "random" beats $2, which really makes playing $2 seem kinda ridiculous, but it can hardly be used to distinguish between the high number cases.
I've seen it argued but without any real proof that $2 must be rationally chosen because it is a Nash equilibrium, but again, that seems if anything like a misinterpretation of game theory to me. To be clear, I don't know what if any mixed strategy equilibrium this game has, however, I don't think it satisfies conditions that *forces* you to pick the Nash equilibrium, namely:
The game is NOT zero-sum
The is no explicit requirement for risk dominance as opposed to payoff dominance, furthermore, there is no strictly dominant strategy.
The game is not iterated and involves no chance at communication with the other player.
There's a good likelihood the game doesn't really even have complete information - your opponent is supposed to be maximizing his/her gain as well, and there's no way to know or assume that he/she will pursue a particular equilbrium strategy.
I would say I'm very glad to see the majority of the posters here agreeing that $2 is rather senseless, which leaves me still puzzled as to why that was the original conclusion. Again, I know that IF the goal was to "beat" the other player then $2 is optimal, hands down, one of those rather unhappy conclusions of game theory, but this is not the case. Likewise, as said above, if you KNOW what your opponent is going to play you want to undercut by one, but the game isn't set up in rounds where you and the other player go back and forth undercutting each other. And so I can only believe that some assumptions from alternate cases like that are being applied when they really shouldn't, since I'm not the type of person to try to argue that game theory in general is BS since it is rather useful, only that I think the original conclusion to this problem was mistaken applications of game theory.
So in the end I do like the conclusion of choosing $100 out of principle but if I really had to lay the chips down I think one could be better off with 97, 98, or 99 - if anyone tried out that similation from the university of Virginia earlier, you would see how you do can gain benefit from chumps trying to always play the max (though the bonus/penalty to payoffs were larger, which I imagine rather affects the game).
As to a couple of other minor loose ends - someone mentioned the "social" factors of being on a game show or in an experiment. I really think it is best to disregard these, if only because in the first place I made up the game show situation. The original problem you'd read in any links in fact involves travelers who are being compensated by an airline. But at any rate, this scenario introduces some random social "reputational" factors as well which is why I tried to avoid it.
So I guess I'm glad to see the responses of everyone else agreeing with me, but then again people are right to critique a couple lines of reasoning - I don't think this problem is adequately addressed with trying to mathematically find probabilities against "random" or something - that's not the right approach. I think it is useful to look and see why "random" beats $2, which really makes playing $2 seem kinda ridiculous, but it can hardly be used to distinguish between the high number cases.
I've seen it argued but without any real proof that $2 must be rationally chosen because it is a Nash equilibrium, but again, that seems if anything like a misinterpretation of game theory to me. To be clear, I don't know what if any mixed strategy equilibrium this game has, however, I don't think it satisfies conditions that *forces* you to pick the Nash equilibrium, namely:
The game is NOT zero-sum
The is no explicit requirement for risk dominance as opposed to payoff dominance, furthermore, there is no strictly dominant strategy.
The game is not iterated and involves no chance at communication with the other player.
There's a good likelihood the game doesn't really even have complete information - your opponent is supposed to be maximizing his/her gain as well, and there's no way to know or assume that he/she will pursue a particular equilbrium strategy.
I would say I'm very glad to see the majority of the posters here agreeing that $2 is rather senseless, which leaves me still puzzled as to why that was the original conclusion. Again, I know that IF the goal was to "beat" the other player then $2 is optimal, hands down, one of those rather unhappy conclusions of game theory, but this is not the case. Likewise, as said above, if you KNOW what your opponent is going to play you want to undercut by one, but the game isn't set up in rounds where you and the other player go back and forth undercutting each other. And so I can only believe that some assumptions from alternate cases like that are being applied when they really shouldn't, since I'm not the type of person to try to argue that game theory in general is BS since it is rather useful, only that I think the original conclusion to this problem was mistaken applications of game theory.
So in the end I do like the conclusion of choosing $100 out of principle but if I really had to lay the chips down I think one could be better off with 97, 98, or 99 - if anyone tried out that similation from the university of Virginia earlier, you would see how you do can gain benefit from chumps trying to always play the max (though the bonus/penalty to payoffs were larger, which I imagine rather affects the game).
I see no reason why $2 being the Nash Equilibrium would propose actually choosing it. The game is about getting free money and by choosing $2 you're minimizing you're potential gain. So yes, I'd call that a misinterpretation as well.
I still don't see the logic in choosing less than 99. If you choose 98 you can get $100 at most so it would be better for both players to choose 100 instead. 99 has slightly better potential than 100 so it's the only alternative worth considering but IMO in the long run players would benefit from always choosing 100.
Mise
isle of lucy
Because if the other person chooses 99 and you choose 98 then you get 100, whereas if the other person chooses 99 and you choose 100 then you only get 97.
MagisterCultuum
Great Sage
Under these rules, going for a low number doesn't make much sense. If you changed it so that the person who bids highest gets nothing (or maybe added that you have an irrational hatred of your opponent) then things would change dramatically.
Catharsis
catch u on the flip scythe
Curse you and your interesting problem, Earthling; this has been bothering me for almost a day straight now. 
Admittedly I don't know much about game theory, but isn't there a contradiction in the question? It says that there's no incentive to defeat the other player, but it also says that both I and the other player are trying to maximise our gain, and the only way to do that is to undercut in the manner that leads us to this $2 figure; in effect, competing with the other player. So there is an incentive to defeat our opponent.
As for the problem, then assuming that both I and my opponent are super-logical gain-maximisation robots, and not just, like, regular joes who would quite like $100 each (I would, by the way) then I can't not choose $2. But, you know, those of you who would choose a high-90s number, I admire your optimism, really I do. I'll wave my $4 in your face.
(EDIT: Actually, I didn't consider mixed-strategies, did I. Hmm. That's probably a better option than just picking $2. Too complex for me, though. I'm joining Perfection in the chair-throwing.)
Admittedly I don't know much about game theory, but isn't there a contradiction in the question? It says that there's no incentive to defeat the other player, but it also says that both I and the other player are trying to maximise our gain, and the only way to do that is to undercut in the manner that leads us to this $2 figure; in effect, competing with the other player. So there is an incentive to defeat our opponent.
As for the problem, then assuming that both I and my opponent are super-logical gain-maximisation robots, and not just, like, regular joes who would quite like $100 each (I would, by the way) then I can't not choose $2. But, you know, those of you who would choose a high-90s number, I admire your optimism, really I do. I'll wave my $4 in your face.
(EDIT: Actually, I didn't consider mixed-strategies, did I. Hmm. That's probably a better option than just picking $2. Too complex for me, though. I'm joining Perfection in the chair-throwing.)
Tell me how does $2 maximize your gain? You're not investing a dime so the ratio of amount chosen and money gained is meaningless. Maximum possible gain is $101 but chasing it has a risk of lowering your gain. By choosing $100 you do your part in ensuring the maximum mutual gain (100 + 100 > 101 + 97) of which statistically 50% will come to you.
Yeah, and continuing that road leads to damnation - or at least to each of us ending up with just $2. Chasing the lower number than your opponent is a losing road as it diminishes the mutual gain.
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