Actually, here’s a better counter-example, one that actually exemplifies some of the claims of CDT optimality. Suppose that the universe consists of a bunch of agents (who do not know each others’ identities) playing one-off PDs against each other. Now 99% of these agents are UDT agents and 1% are CDT agents.
The CDT agents defect for the standard reason.
The UDT agents reason that my opponent will do the same thing that I do with 99% probability, therefore, I should cooperate.
CDT agents get 99% DC and 1% DD. UDT agents get 99% CC and 1% CD. The CDT agents in this universe do better than the UDT agents, yet they are facing a perfectly symmetrical scenario with no mind reading involved.
A version of this problem was discussed here previously. It was also brought up during the decision theory workshop hosted by MIRI in 2013 as an open problem. As far as I know there hasn’t been much progress on it since 2009.
I wonder if it’s even coherent to have a math intuition which wouldn’t be forcing UDT to cooperate (or defect) in certain conditions just to make 2*2 be 4 , figuratively speaking (as ultimately you could expand any calculation into an equivalent calculation involving a decision by UDT).
That shouldn’t be surprising. The CDT agents here are equivalent to DefectBot, and if they come into existence spontaneously, are no different than natural phenomena like rocks. Notice that the UDT agents in this situation do better than the alternative (if they defected, they would get 100% DD which is a way worse result). They don’t care that some DefectBots get to freeload.
Of course, if the defectbots are here because someone calculated that UDT agents would cooperate and therefore being defectbot is a good way to get free utilons… then the UDT agents are incentivized to defect, because this is now an ultimatum game.
And in the variant where bots do know each other’s identities, the UDT bots all get 99% CC / 1% DD and the CDT bots suck it.
The CDT agents here win because they do not believe that altering their strategy will change the way that their opponents behave. This is actually true in this case, and even true for the UDT agents depending on how you choose to construct your counterfactuals. If a UDT agent suffered a malfunction and defected, it too would do better. In any case, the theorem that UDT agents perform optimally in universes that can only read your mind by knowing what you would do in hypothetical situations is false as this example shows.
UDT bots win in some scenarios where the initial conditions of the scenario favor agents that behave sub-optimally in certain scenarios (and by sub-optimally, I mean where counterfactuals are constructed in the way implicit to CDT). The example above shows that sometimes they are punished for acting suboptimally.
Or how about this example, that simplifies things even further. The game is PD against CooperateBot, BUT before the game starts Omega announces “your opponent will make the same decision that UDT would if I told them this.” This announcement causes UDT to cooperate against CooperateBot. CDT on the other hand, correctly deduces that the opponent will cooperate no matter what it does (actually UDT comes to this conclusion too) and therefore decides to defect.
The game is PD against CooperateBot, BUT before the game starts Omega announces “your opponent will make the same decision that UDT would if I told them this.” This announcement causes UDT to cooperate against CooperateBot.
No. There is no obligation to do something just because Omega claims that you will.
First, if I know that my opponent is CooperateBot, then:-
It is known that Omega doesn’t lie.
Therefore Omega has simulated this situation and predicted that I (UDT) cooperate.
Hence, I can either cooperate, and collect the standard reward for CC.
Or I can defect, in order to access an alternative branch of the problem (where Omega finds that UDT defects and does “something else”).
This alternative branch is unspecified, so the problem is incomplete.
UDT cooperates or defects depending on the contents of the alternative branch. If the alternative branch is unknown then it must guess, and most likely cooperates to be on the safe side.
Now, the problem is different if a CDT agent is put in my place, because that CDT agent does not control (or only weakly controls) the action of the UDT simulation that Omega ran in order to make the assertion about UDT’s decision.
Fine. Your opponent actually simulates what UDT would do if Omega had told it that and returns the appropriate response (i.e. it is CooperateBot, although perhaps your finite prover is unable to verify that).
It’s not UDT. It’s the strategy that against any opponent does what UDT would do against it. In particular, it cooperates against any opponent. Therefore it is CooperateBot. It is just coded in a funny way.
To be clear letting Y(X) be what Y does against X we have that
BOT(X) = UDT(BOT) = C
This is different from UDT. UDT(X) is D for some values of X. The two functions agree when X=UDT and in relatively few other cases.
It’s clear that UDT can’t do better vs “BOT” than by cooperating, because if UDT defects against BOT then BOT defects against UDT. Given that dependency, you clearly can’t call it CooperateBot, and it’s clear that UDT makes the right decision by cooperating with it because CC is better than DD.
OK. Let me say this another way that involves more equations.
So let’s let
U(X,Y) be the utility that X gets when it plays prisoner’s dilemma against Y.
For a program X, let BOT^X be the program where BOT^X(Y) = X(BOT^X). Notice that BOT^X(Y) does not depend on Y. Therefore, depending upon what X is BOT^X is either equivalent CooperateBot or equivalent to DefectBot.
Now, you are claiming that UDT plays optimally against BOT_UDT because for any strategy X
U(X, BOT^X) ⇐ U(UDT, BOT^UDT)
This is true, because X(BOT^X) = BOT^X(X) by the definition of BOT^X. Therefore you cannot do better than CC.
On the other hand, it is also true that for any X and any Y that
U(X,BOT^Y) ⇐ U(CDT, BOT^Y)
This is because BOT^Y’s behavior does not depend on X, and therefore you do optimally by defecting against it (or you could just apply the Theorem that says that CDT wins if the universe cannot read your mind).
Our disagreement here stems from the fact that we are considering different counterfactuals here. You seem to claim that UDT behaves correctly because
U(UDT,BOT^UDT) > U(CDT,BOT^CDT)
While I claim that CDT does because
U(CDT, BOT^UDT) > U(UDT, BOT^UDT)
And in fact, given the way that I phrased the scenario, (which was that you play BOT^UDT not that you play BOT^{you} (i.e. the mirror matchup)) I happen to be right here. So justify it however you like, but UDT does lose this scenario.
Actually, you’ve oversimplified and missed something critical. In reality, the only way you can force BOT^UDT(X) = UDT(BOT^UDT) = C is if the universe does, in fact, read your mind. In general, UDT can map different epistemic states to different actions, so as long as BOT^UDT has no clue about the epistemic state of the UDT agent it has no way of guaranteeing that its output is the same as that of the UDT agent. Consequently, it’s possible for the UDT agent to get DC as well. The only way BOT^UDT would be able to guarantee that it gets the same output as a particular UDT agent is if the universe was able to read the UDT agent’s mind.
Actually, I think that you are misunderstanding me. UDT’s current epistemic state (at the start of the game) is encoded into BOT^UDT. No mind reading involved. Just a coincidence. [Really, your current epistemic state is part of your program]
Your argument is like saying that UDT usually gets $1001000 in Newcomb’s problem because whether or not the box was full depended on whether or not UDT one-boxed when in a different epistemic state.
Okay, you’re saying here that BOT has a perfect copy of the UDT player’s mind in its own code (otherwise how could it calculate UDT(BOT) and guarantee that the output is the same?). It’s hard to see how this doesn’t count as “reading your mind”.
Yes, sometimes its advantageous to not control the output of computations in the environment. In this case UDT is worse off because it is forced to control both its own decision and BOT’s decision; whereas CDT doesn’t have to worry about controlling BOT because they use different algorithms. But this isn’t due to any intrinsic advantage of CDT’s algorithm. It’s just because they happen to be numerically inequivalent.
An instance of UDT with literally any other epistemic state than the one contained in BOT would do just as well as CDT here.
It’s hard to see how this doesn’t count as “reading your mind”.
So… UDT’s source code is some mathematical constant, say 1893463. It turns out that UDT does worse against BOT^1893463. Note that it does worse against BOT^1893463 not BOT^{you}. The universe does not depend on the source code of the person playing the game (as it does in mirror PD). Furthermore, UDT does not control the output of its environment. BOT^1893463 always cooperates. It cooperates against UDT. It cooperates against CDT. It cooperates everything.
But this isn’t due to any intrinsic advantage of CDT’s algorithm. It’s just because they happen to be numerically inequivalent.
No. CDT does at least as well as UDT against BOT^CDT. UDT does worse when there is this numerical equivalence, but CDT does not suffer from this issue. CDT does at least as well as UDT against BOT^X for all X, and sometimes does better. In fact, if you only construct counterfactuals this way, CDT does at least as well as anything else.
An instance of UDT with literally any other epistemic state than the one contained in BOT would do just as well as CDT here.
This is silly. A UDT that believes that it is in a mirror matchup also loses. A UDT that believes it is facing Newcomb’s problem does something incoherent. If you are claiming that you want a UDT that differs from the encoding in BOT because, of some irrelevant details in its memory… well then it might depend upon implementation, but I think that most attempted implementations of UDT would conclude that these irrelevant details are irrelevant and cooperate anyway. If you don’t believe this then you should also think that UDT will defect in a mirror matchup if it and its clone are painted different colors.
I take it back, the scenario isn’t that weird. But your argument doesn’t prove what you think it does:
Consider the analogous scenario, where CDT plays against BOT = CDT(BOT). CDT clearly does the wrong thing here—it defects. If it cooperated, it would get CC instead of DD. Note that if CDT did cooperate, UDT would be able to freeload by defecting (against BOT = CDT(BOT)). But CDT doesn’t care about that because the prisoner’s dilemma is defined such that we don’t care about freeloaders. Nevertheless CDT defects and gets a worse result than it could.
CDT does better than UDT against BOT = UDT(BOT) because UDT (correctly) doesn’t care that CDT can freeload, and correctly cooperates to gain CC.
If you are claiming that you want a UDT that differs from the encoding in BOT because, of some irrelevant details in its memory...
Depending on the exact setup, “irrelevant details in memory” are actually vital information that allow you to distinguish whether you are “actually playing” or are being simulated in BOT’s mind.
No. BOT^CDT = DefectBot. It defects against any opponent. CDT could not cause it to cooperate by changing what it does.
If it cooperated, it would get CC instead of DD.
Actually if CDT cooperated against BOT^CDT it would get $3^^^3. You can prove all sorts of wonderful things once you assume a statement that is false.
Depending on the exact setup, “irrelevant details in memory” are actually vital information that allow you to distinguish whether you are “actually playing” or are being simulated in BOT’s mind.
OK… So UDT^Red and UDT^Blue are two instantiations of UDT that differ only in irrelevant details. In fact the scenario is a mirror matchup, only after instantiation one of the copies was painted red and the other was painted blue. According to what you seem to be saying UDT^Red will reason:
Well I can map different epistemic states to different outputs, I can implement the strategy cooperate if you are painted blue and defect if you are painted red.
Of course UDT^Blue will reason the same way and they will fail to cooperate with each other.
No. BOT^CDT = DefectBot. It defects against any opponent. CDT could not cause it to cooperate by changing what it does.
Maybe I’ve misread you, but this sounds like an assertion that your counterfactual question is the right one by definition, rather than a meaningful objection.
Well, yes. Then again, the game was specified as PD against BOT^CDT not as PD against BOT^{you}. It seems pretty clear that for X not equal to CDT that it is not the case that X could achieve the result CC in this game. Are you saying that it is reasonable to say that CDT could achieve a result that no other strategy could just because it’s code happens to appear in the opponent’s program?
I think that there is perhaps a distinction to be made between things that happen to be simulating your code and this that are causally simulating your code.
Well I can map different epistemic states to different outputs, I can implement the strategy cooperate if you are painted blue and defect if you are painted red. Of course UDT^Blue will reason the same way and they will fail to cooperate with each other.
No, because that’s a silly thing to do in this scenario. For one thing, UDT will see that they are reasoning the same way (because they are selfish and only consider “my color” vs “other color”), and therefore will both do the same thing. But also, depending on the setup, UDT^Red’s prior should give equal probability to being painted red and painted blue anyway, which means trying to make the outcome favour red is silly.
Compare to the version of newcomb’s where the bot in the room is UDT^Red, while Omega simulates UDT^Blue. UDT can implement the conditional strategy {Red ⇒ two-box, Blue ⇒ one-box}. This is obviously unlikely, because the point of the Newcomb thought experiment is that Omega simulates (or predicts) you. So he would clearly try to avoid adding such information that “gives the game away”.
However in this scenario you say that BOT simulates UDT “by coincidence”, not by mind reading. So it is far more likely that BOT simulates (the equivalent) of UDT^Blue, while the UDT actually playing is UDT^Red. And you are passed the code of BOT as input, so UDT can simply implement the conditional strategy {cooperate iff the color inside BOT is the same as my color}.
OK. Fine. Point taken. There is a simple fix though.
MBOT^X(Y) = X’(MBOT^X) where X’ is X but with randomized irrelevant experiences.
In order to produce this properly, MBOT only needs to have your prior (or a sufficiently similar probability distribution) over irrelevant experiences hardcoded. And while your actual experiences might be complicated and hard to predict, your priors are not.
No. BOT(X) is cooperate for all X. It behaves in exactly the same way that CooperateBot does, it just runs different though equivalent code.
And my point was that CDT does better against BOT than UDT does. I was asked for an example where CDT does better than UDT where the universe cannot read your mind except via through your actions in counterfactuals. This is an example of such. In fact, in this example, the universe doesn’t read your mind at all.
Also your argument that UDT cannot possibly do better against BOT than it does in analogous to the argument that CDT cannot do better in the mirror matchup than it does. Namely that CDT’s outcome against CDT is at least as good as anything else’s outcome against CDT. You aren’t defining your counterfactuals correctly. You can do better against BOT than UDT does. You just have to not be UDT.
Actually, this is a somewhat general phenomenon. Consider for example, the version of Newcomb’s problem where the box is full “if and only if UDT one-boxes in this scenario”.
UDT’s optimality theorem requires the in the counterfactual where it is replaced by a different decision theory that all of the “you”’s referenced in the scenario remain “you” rather than “UDT”. In the latter counterfactual CDT provably wins. The fact that UDT wins these scenarios is an artifact of how you are constructing your scenarios.
This is a good example. Thank you. A population of 100% CDT, though, would get 100% DD, which is terrible. It’s a point in UDT’s favor that “everyone running UDT” leads to a better outcome for everyone than “everyone running CDT.”
Actually, here’s a better counter-example, one that actually exemplifies some of the claims of CDT optimality. Suppose that the universe consists of a bunch of agents (who do not know each others’ identities) playing one-off PDs against each other. Now 99% of these agents are UDT agents and 1% are CDT agents.
The CDT agents defect for the standard reason. The UDT agents reason that my opponent will do the same thing that I do with 99% probability, therefore, I should cooperate.
CDT agents get 99% DC and 1% DD. UDT agents get 99% CC and 1% CD. The CDT agents in this universe do better than the UDT agents, yet they are facing a perfectly symmetrical scenario with no mind reading involved.
A version of this problem was discussed here previously. It was also brought up during the decision theory workshop hosted by MIRI in 2013 as an open problem. As far as I know there hasn’t been much progress on it since 2009.
I wonder if it’s even coherent to have a math intuition which wouldn’t be forcing UDT to cooperate (or defect) in certain conditions just to make 2*2 be 4 , figuratively speaking (as ultimately you could expand any calculation into an equivalent calculation involving a decision by UDT).
That shouldn’t be surprising. The CDT agents here are equivalent to DefectBot, and if they come into existence spontaneously, are no different than natural phenomena like rocks. Notice that the UDT agents in this situation do better than the alternative (if they defected, they would get 100% DD which is a way worse result). They don’t care that some DefectBots get to freeload.
Of course, if the defectbots are here because someone calculated that UDT agents would cooperate and therefore being defectbot is a good way to get free utilons… then the UDT agents are incentivized to defect, because this is now an ultimatum game.
And in the variant where bots do know each other’s identities, the UDT bots all get 99% CC / 1% DD and the CDT bots suck it.
And the UDT agents are equivalent to CooperateBot. What’s your point?
The CDT agents here win because they do not believe that altering their strategy will change the way that their opponents behave. This is actually true in this case, and even true for the UDT agents depending on how you choose to construct your counterfactuals. If a UDT agent suffered a malfunction and defected, it too would do better. In any case, the theorem that UDT agents perform optimally in universes that can only read your mind by knowing what you would do in hypothetical situations is false as this example shows.
UDT bots win in some scenarios where the initial conditions of the scenario favor agents that behave sub-optimally in certain scenarios (and by sub-optimally, I mean where counterfactuals are constructed in the way implicit to CDT). The example above shows that sometimes they are punished for acting suboptimally.
Or how about this example, that simplifies things even further. The game is PD against CooperateBot, BUT before the game starts Omega announces “your opponent will make the same decision that UDT would if I told them this.” This announcement causes UDT to cooperate against CooperateBot. CDT on the other hand, correctly deduces that the opponent will cooperate no matter what it does (actually UDT comes to this conclusion too) and therefore decides to defect.
No. There is no obligation to do something just because Omega claims that you will.
First, if I know that my opponent is CooperateBot, then:-
It is known that Omega doesn’t lie.
Therefore Omega has simulated this situation and predicted that I (UDT) cooperate.
Hence, I can either cooperate, and collect the standard reward for CC.
Or I can defect, in order to access an alternative branch of the problem (where Omega finds that UDT defects and does “something else”).
This alternative branch is unspecified, so the problem is incomplete.
UDT cooperates or defects depending on the contents of the alternative branch. If the alternative branch is unknown then it must guess, and most likely cooperates to be on the safe side.
Now, the problem is different if a CDT agent is put in my place, because that CDT agent does not control (or only weakly controls) the action of the UDT simulation that Omega ran in order to make the assertion about UDT’s decision.
Fine. Your opponent actually simulates what UDT would do if Omega had told it that and returns the appropriate response (i.e. it is CooperateBot, although perhaps your finite prover is unable to verify that).
Err, that’s not CooperateBot, that’s UDT. Yes, UDT cooperates with itself. That’s the point. (Notice that if UDT defects here, the outcome is DD.)
It’s not UDT. It’s the strategy that against any opponent does what UDT would do against it. In particular, it cooperates against any opponent. Therefore it is CooperateBot. It is just coded in a funny way.
To be clear letting Y(X) be what Y does against X we have that BOT(X) = UDT(BOT) = C This is different from UDT. UDT(X) is D for some values of X. The two functions agree when X=UDT and in relatively few other cases.
What is your point, exactly?
It’s clear that UDT can’t do better vs “BOT” than by cooperating, because if UDT defects against BOT then BOT defects against UDT. Given that dependency, you clearly can’t call it CooperateBot, and it’s clear that UDT makes the right decision by cooperating with it because CC is better than DD.
OK. Let me say this another way that involves more equations.
So let’s let U(X,Y) be the utility that X gets when it plays prisoner’s dilemma against Y. For a program X, let BOT^X be the program where BOT^X(Y) = X(BOT^X). Notice that BOT^X(Y) does not depend on Y. Therefore, depending upon what X is BOT^X is either equivalent CooperateBot or equivalent to DefectBot.
Now, you are claiming that UDT plays optimally against BOT_UDT because for any strategy X U(X, BOT^X) ⇐ U(UDT, BOT^UDT) This is true, because X(BOT^X) = BOT^X(X) by the definition of BOT^X. Therefore you cannot do better than CC. On the other hand, it is also true that for any X and any Y that U(X,BOT^Y) ⇐ U(CDT, BOT^Y) This is because BOT^Y’s behavior does not depend on X, and therefore you do optimally by defecting against it (or you could just apply the Theorem that says that CDT wins if the universe cannot read your mind).
Our disagreement here stems from the fact that we are considering different counterfactuals here. You seem to claim that UDT behaves correctly because U(UDT,BOT^UDT) > U(CDT,BOT^CDT) While I claim that CDT does because U(CDT, BOT^UDT) > U(UDT, BOT^UDT)
And in fact, given the way that I phrased the scenario, (which was that you play BOT^UDT not that you play BOT^{you} (i.e. the mirror matchup)) I happen to be right here. So justify it however you like, but UDT does lose this scenario.
Actually, you’ve oversimplified and missed something critical. In reality, the only way you can force BOT^UDT(X) = UDT(BOT^UDT) = C is if the universe does, in fact, read your mind. In general, UDT can map different epistemic states to different actions, so as long as BOT^UDT has no clue about the epistemic state of the UDT agent it has no way of guaranteeing that its output is the same as that of the UDT agent. Consequently, it’s possible for the UDT agent to get DC as well. The only way BOT^UDT would be able to guarantee that it gets the same output as a particular UDT agent is if the universe was able to read the UDT agent’s mind.
Actually, I think that you are misunderstanding me. UDT’s current epistemic state (at the start of the game) is encoded into BOT^UDT. No mind reading involved. Just a coincidence. [Really, your current epistemic state is part of your program]
Your argument is like saying that UDT usually gets $1001000 in Newcomb’s problem because whether or not the box was full depended on whether or not UDT one-boxed when in a different epistemic state.
Okay, you’re saying here that BOT has a perfect copy of the UDT player’s mind in its own code (otherwise how could it calculate UDT(BOT) and guarantee that the output is the same?). It’s hard to see how this doesn’t count as “reading your mind”.
Yes, sometimes its advantageous to not control the output of computations in the environment. In this case UDT is worse off because it is forced to control both its own decision and BOT’s decision; whereas CDT doesn’t have to worry about controlling BOT because they use different algorithms. But this isn’t due to any intrinsic advantage of CDT’s algorithm. It’s just because they happen to be numerically inequivalent.
An instance of UDT with literally any other epistemic state than the one contained in BOT would do just as well as CDT here.
So… UDT’s source code is some mathematical constant, say 1893463. It turns out that UDT does worse against BOT^1893463. Note that it does worse against BOT^1893463 not BOT^{you}. The universe does not depend on the source code of the person playing the game (as it does in mirror PD). Furthermore, UDT does not control the output of its environment. BOT^1893463 always cooperates. It cooperates against UDT. It cooperates against CDT. It cooperates everything.
No. CDT does at least as well as UDT against BOT^CDT. UDT does worse when there is this numerical equivalence, but CDT does not suffer from this issue. CDT does at least as well as UDT against BOT^X for all X, and sometimes does better. In fact, if you only construct counterfactuals this way, CDT does at least as well as anything else.
This is silly. A UDT that believes that it is in a mirror matchup also loses. A UDT that believes it is facing Newcomb’s problem does something incoherent. If you are claiming that you want a UDT that differs from the encoding in BOT because, of some irrelevant details in its memory… well then it might depend upon implementation, but I think that most attempted implementations of UDT would conclude that these irrelevant details are irrelevant and cooperate anyway. If you don’t believe this then you should also think that UDT will defect in a mirror matchup if it and its clone are painted different colors.
I take it back, the scenario isn’t that weird. But your argument doesn’t prove what you think it does:
Consider the analogous scenario, where CDT plays against BOT = CDT(BOT). CDT clearly does the wrong thing here—it defects. If it cooperated, it would get CC instead of DD. Note that if CDT did cooperate, UDT would be able to freeload by defecting (against BOT = CDT(BOT)). But CDT doesn’t care about that because the prisoner’s dilemma is defined such that we don’t care about freeloaders. Nevertheless CDT defects and gets a worse result than it could.
CDT does better than UDT against BOT = UDT(BOT) because UDT (correctly) doesn’t care that CDT can freeload, and correctly cooperates to gain CC.
Depending on the exact setup, “irrelevant details in memory” are actually vital information that allow you to distinguish whether you are “actually playing” or are being simulated in BOT’s mind.
No. BOT^CDT = DefectBot. It defects against any opponent. CDT could not cause it to cooperate by changing what it does.
Actually if CDT cooperated against BOT^CDT it would get $3^^^3. You can prove all sorts of wonderful things once you assume a statement that is false.
OK… So UDT^Red and UDT^Blue are two instantiations of UDT that differ only in irrelevant details. In fact the scenario is a mirror matchup, only after instantiation one of the copies was painted red and the other was painted blue. According to what you seem to be saying UDT^Red will reason:
Well I can map different epistemic states to different outputs, I can implement the strategy cooperate if you are painted blue and defect if you are painted red.
Of course UDT^Blue will reason the same way and they will fail to cooperate with each other.
Maybe I’ve misread you, but this sounds like an assertion that your counterfactual question is the right one by definition, rather than a meaningful objection.
Well, yes. Then again, the game was specified as PD against BOT^CDT not as PD against BOT^{you}. It seems pretty clear that for X not equal to CDT that it is not the case that X could achieve the result CC in this game. Are you saying that it is reasonable to say that CDT could achieve a result that no other strategy could just because it’s code happens to appear in the opponent’s program?
I think that there is perhaps a distinction to be made between things that happen to be simulating your code and this that are causally simulating your code.
No, because that’s a silly thing to do in this scenario. For one thing, UDT will see that they are reasoning the same way (because they are selfish and only consider “my color” vs “other color”), and therefore will both do the same thing. But also, depending on the setup, UDT^Red’s prior should give equal probability to being painted red and painted blue anyway, which means trying to make the outcome favour red is silly.
Compare to the version of newcomb’s where the bot in the room is UDT^Red, while Omega simulates UDT^Blue. UDT can implement the conditional strategy {Red ⇒ two-box, Blue ⇒ one-box}. This is obviously unlikely, because the point of the Newcomb thought experiment is that Omega simulates (or predicts) you. So he would clearly try to avoid adding such information that “gives the game away”.
However in this scenario you say that BOT simulates UDT “by coincidence”, not by mind reading. So it is far more likely that BOT simulates (the equivalent) of UDT^Blue, while the UDT actually playing is UDT^Red. And you are passed the code of BOT as input, so UDT can simply implement the conditional strategy {cooperate iff the color inside BOT is the same as my color}.
OK. Fine. Point taken. There is a simple fix though.
MBOT^X(Y) = X’(MBOT^X) where X’ is X but with randomized irrelevant experiences.
In order to produce this properly, MBOT only needs to have your prior (or a sufficiently similar probability distribution) over irrelevant experiences hardcoded. And while your actual experiences might be complicated and hard to predict, your priors are not.
No. BOT(X) is cooperate for all X. It behaves in exactly the same way that CooperateBot does, it just runs different though equivalent code.
And my point was that CDT does better against BOT than UDT does. I was asked for an example where CDT does better than UDT where the universe cannot read your mind except via through your actions in counterfactuals. This is an example of such. In fact, in this example, the universe doesn’t read your mind at all.
Also your argument that UDT cannot possibly do better against BOT than it does in analogous to the argument that CDT cannot do better in the mirror matchup than it does. Namely that CDT’s outcome against CDT is at least as good as anything else’s outcome against CDT. You aren’t defining your counterfactuals correctly. You can do better against BOT than UDT does. You just have to not be UDT.
Actually, this is a somewhat general phenomenon. Consider for example, the version of Newcomb’s problem where the box is full “if and only if UDT one-boxes in this scenario”.
UDT’s optimality theorem requires the in the counterfactual where it is replaced by a different decision theory that all of the “you”’s referenced in the scenario remain “you” rather than “UDT”. In the latter counterfactual CDT provably wins. The fact that UDT wins these scenarios is an artifact of how you are constructing your scenarios.
This is a good example. Thank you. A population of 100% CDT, though, would get 100% DD, which is terrible. It’s a point in UDT’s favor that “everyone running UDT” leads to a better outcome for everyone than “everyone running CDT.”