As a function of M, |P| is very likely to be exponential and so it will take O(M) symbols to specify a member of P. Under many encodings, there isn’t one that can even check whether the inputs are equal before running out of time.
That aside, why are you assuming that program b “wants” anything? Essentially all of P won’t be programs that have any sort of “want”. If it is a precondition of the problem that b is such a program, what selection procedure is assumed between those that do “want” money from this scenario? Note that being selected for running is also a precondition for getting any money at all, so this selection procedure is critically important—far more so than anything the program might output!
As a function of M, |P| is very likely to be exponential and so it will take O(M) symbols to specify a member of P.
O-ops, I didn’t think about it, thanks! Maybe it would be better to change it so input is “a=b” or “a!=b”, and a always gets “a=b”.
That aside, why are you assuming that program b “wants” anything? Essentially all of P won’t be programs that have any sort of “want”. If it is a precondition of the problem that b is such a program, what selection procedure is assumed between those that do “want” money from this scenario? Note that being selected for running is also a precondition for getting any money at all, so this selection procedure is critically important—far more so than anything the program might output!
Programmer who wrote b decided that it should be consequentialist agent who wants to get money. (Or, if this program is actually, a, it wants to maximize the payment for b just because such a program was chosen by Omega by pure luck)
Suppose b “knows” that Omega runs this experiment for all programs b. Then the optimal behaviour for a competent b (by a ridiculously small margin) is to 1-box.
Suppose b suspects that box-choosing programs are slightly less likely to be run if they 1-box on equal inputs. Then the optimal behaviour for b is to 2-box, because the average extra payoff for 1-boxing on equal inputs is utterly insignificant while the average penalty for not being chosen to run is very much greater. Anything that affects probability of being run as box-chooser with probability greater than 1000/|P| (which is on the order of 1/10^10^10^10^100) matters far more than what the program actually does.
In the original Newcombe problem, you know that you are going to get money based on your decision. In this problem, a running program does not know this. It doesn’t know whether it’s a or b or both, and every method for selecting a box-chooser is a different problem with different optimal strategies.
As a function of M, |P| is very likely to be exponential and so it will take O(M) symbols to specify a member of P. Under many encodings, there isn’t one that can even check whether the inputs are equal before running out of time.
That aside, why are you assuming that program b “wants” anything? Essentially all of P won’t be programs that have any sort of “want”. If it is a precondition of the problem that b is such a program, what selection procedure is assumed between those that do “want” money from this scenario? Note that being selected for running is also a precondition for getting any money at all, so this selection procedure is critically important—far more so than anything the program might output!
O-ops, I didn’t think about it, thanks! Maybe it would be better to change it so input is “a=b” or “a!=b”, and a always gets “a=b”.
Programmer who wrote b decided that it should be consequentialist agent who wants to get money. (Or, if this program is actually, a, it wants to maximize the payment for b just because such a program was chosen by Omega by pure luck)
I’ll try to make it clearer:
Suppose b “knows” that Omega runs this experiment for all programs b. Then the optimal behaviour for a competent b (by a ridiculously small margin) is to 1-box.
Suppose b suspects that box-choosing programs are slightly less likely to be run if they 1-box on equal inputs. Then the optimal behaviour for b is to 2-box, because the average extra payoff for 1-boxing on equal inputs is utterly insignificant while the average penalty for not being chosen to run is very much greater. Anything that affects probability of being run as box-chooser with probability greater than 1000/|P| (which is on the order of 1/10^10^10^10^100) matters far more than what the program actually does.
In the original Newcombe problem, you know that you are going to get money based on your decision. In this problem, a running program does not know this. It doesn’t know whether it’s a or b or both, and every method for selecting a box-chooser is a different problem with different optimal strategies.