A more effective robot only looks for the most probable digit, it doesn’t need to know how probable that digit is (edit: that is, in this specific problem where utilities are equal. There is nothing ad-hoc about use of algebra). E.g. if it would figure out that for some reason the large primes are most likely to end in 3 (or rather, that no digit is more likely than 3), it does not even need to know that 0,2,4,5,6,8 are not an option.
Furthermore, as you lower the prime, there has to be a seamless transition to the robot actually calculating the last digit in an efficient manner.
A bounded agent has to rationally allocate it’s computing time. Calculating direct probabilities or huge sets of relational probabilities—rather than minimum necessary—is among the least rational things that can be done with the computing time. (Even worse thing to do could be a comparison between two estimates of utility which have different errors, such as two incomplete sums). Outside very simple toy problems, probabilistic reasoning is incredibly computationally expensive, as often every possible combination of variables has to be considered.
A more effective robot only looks for the most probable digit
I agree with the rest of your comment, but this seems too ad hoc. It runs into trouble if outcomes differ in utility, so that you can’t just look for high probability. And storing a number seems like a much better way of integrating lots of independent pieces of information than storing a list.
Then you look for largest probability*utility , which you generally do by trying to find a way to demonstrate A>B which you can do in many cases where you can’t actually evaluate either A or B (and many cases where you can only evaluate A and B so inaccurately that outcome of comparison of evaluations of A and B is primarily dependent on inaccuracies).
Furthermore, a “probability” is a list due to loss of statistical independence with other variables. edit: the pieces of information are very rarely independent, too. Some reasoning that 3 is more likely than other digits would not be independent from 2 being a bad choice.
edit: also, holy hell, trillionth prime does end with 3.
A more effective robot only looks for the most probable digit, it doesn’t need to know how probable that digit is (edit: that is, in this specific problem where utilities are equal. There is nothing ad-hoc about use of algebra). E.g. if it would figure out that for some reason the large primes are most likely to end in 3 (or rather, that no digit is more likely than 3), it does not even need to know that 0,2,4,5,6,8 are not an option.
Furthermore, as you lower the prime, there has to be a seamless transition to the robot actually calculating the last digit in an efficient manner.
A bounded agent has to rationally allocate it’s computing time. Calculating direct probabilities or huge sets of relational probabilities—rather than minimum necessary—is among the least rational things that can be done with the computing time. (Even worse thing to do could be a comparison between two estimates of utility which have different errors, such as two incomplete sums). Outside very simple toy problems, probabilistic reasoning is incredibly computationally expensive, as often every possible combination of variables has to be considered.
I agree with the rest of your comment, but this seems too ad hoc. It runs into trouble if outcomes differ in utility, so that you can’t just look for high probability. And storing a number seems like a much better way of integrating lots of independent pieces of information than storing a list.
Then you look for largest probability*utility , which you generally do by trying to find a way to demonstrate A>B which you can do in many cases where you can’t actually evaluate either A or B (and many cases where you can only evaluate A and B so inaccurately that outcome of comparison of evaluations of A and B is primarily dependent on inaccuracies).
Furthermore, a “probability” is a list due to loss of statistical independence with other variables. edit: the pieces of information are very rarely independent, too. Some reasoning that 3 is more likely than other digits would not be independent from 2 being a bad choice.
edit: also, holy hell, trillionth prime does end with 3.