I always find it sad to see a thread downvoted with no comments or explanations, so I’m going to attempt to give my thoughts.
Newcomb’s problem seems absurdly easy to me. At least in the way it was presented by Eliezer, which is not necessarily a universal formulation. The way he expressed it, you observe Omega predicting correctly n times. (You could even add inaccurate observations if you wanted to consider the possibility that Omega is accurate, say, 90% of the time. We will do this in later steps, and call the number of inaccurate observations m.) If one box contains A dollars (or C dollars, if Omega predicted you to two box) and the other box contains B dollars, you can arrive at a pretty easy formulation of whether or not you should one box or two box. I almost wrote a MATLAB program to do it for arbitrary inputs that I was going to make into a post, but I figured most people wouldn’t find it very interesting, which was my conclusion after I got about halfway done with it.
First you arrive at a probability that Omega will predict you correctly, assuming that you are no different from anyone else with whom Omega has played the game. To do this, you estimate the accuracy, p, over a range of values from 0 to 1. 1.0 means Omega is perfectly accurate, 0.0 would mean ey is always wrong. The probability of obtaining the results you observed (n accurate predictions by Omega and m inaccurate ones) given any probability of em being accurate (p) is then p^n (1-p)^m. This gives us a distribution that represents the probability that Omega has a certain accuracy. We will call this distribution *D(p).
We then need to consider our two alternatives and select the one that maximizes expected utility. The utility of two boxing is:
U(two box) = p(Omega is wrong) (Value of box B + Value of Box A) + p(Omega is right) (Value of box B + Lesser value Omega puts in A)
U(one box) = p(Omega is wrong) (Value of box B + Lesser value Omega puts in A) + p(Omega is wrong) (Value of box B + Value of box A)
(Remember that we are considering the possibility that instead of replacing A with 0 dollars, Omega puts some value C dollars in A. All that really matters is the difference in these two values, though.)
With the variables we used, p(Omega is right) is the probability that ey has a certain accuracy, our distribution D(p) times that accuracy, p. p(Omega is wrong) is one minus this. The value of box A is obviously A, the value of box B is B, and the lesser value that Omega puts in box A is C
So our expected utilities are then a function of p as follows:
U(two box) = (1 - D(p)p)(B+A) + D(p)p(B+C) = [1 - (p^n (1-p)^m)] p (B+A) + (p^n (1-p)^m) p (B+C)
U(two box) = (1 - D(p)p)(B+C) + D(p)p(B+A) = [1 - (p^n (1-p)^m)] p (B+C) + (p^n (1-p)^m) p (B+A)
All that needs to be done is then to integrate the expected utilities over p from zero to one. Whichever value is greater is the correct choice.
Note that this analysis has a number of (fairly obvious and somewhat trivial) assumptions. One, the probability of Omega being right is constant over both one boxers and two boxers. Two, one’s utility function in money is linear (although compensating for that would not be very difficult). Three, Omega has no more or less information about you than anyone else about whom ey made this prediction.
(Here’s a slightly less mind-killing variant: let’s say that regularly taking aspirin is correlated with risk of a heart attack, but not because it causes them; in fact, aspirin (in this hypothetical) is good for anyone’s heart. Instead, there’s an additional risk factor for heart attacks, which also causes discomfort beneath the threshold of full consciousness. People with this risk factor end up being more likely to take aspirin regularly, though they’re not able to pinpoint why, and the effect is large enough that the correlation points the “wrong” way. Now if you know all of this and are wondering whether to take aspirin regularly, the calculation you did above would tell you not to take it!)
We can get down to a discussion of evidential vs casual decision theory if you want, certainly, but I think that’s a bit off topic.
I have a couple of reactions to your point. My initial reaction is that evidential decision theory is superior in the case of Omega because nothing is known about em. Since Omega is a black box, the only thing that can really be done is gather evidence and respond to it.
But more generally, I think your example is somewhat strawman-ish. Just like in the smoking problem, there is other evidence suggesting that Asprin has the opposite effect. Saying that evidential decision theory has to ignore this is pretty unfair to it. Furthermore, you know that you can’t really rely on the evidence you have (that Asprin is correlated with heart attack, because you know you don’t have a random sample. Moreover, the evidence that Asprin is actually good for your heart was supposedly generated with some kind of statistical controls. It’s the same reason I mentioned in my analysis the assumption that Omega knows nothing more or less about you than ey did about any of the other people. The second you don’t have a representative sample, all of your statistics can be thrown out of the window.
Let average_one_box_value = the average value received by people who chose one box. Let average_two_box_value = the average value received by people who chose two boxes.
If average_one_box_value > average_two_box_value, then pick one box, else pick two.
As a bonus, this eliminates the need to assume Omega being right is constant over both one boxers and two boxers.
[Edit—just plain wrong, see Misha’s comment below]
Minor quibble; It’s also not necessary to assume linear utility for dollars, just continuous.
That is, more money is always better. However, I’m pretty sure that’s true in your example as well.
It is definitely necessary to assume linear utility for dollars. For example: suppose your (marginal) utility function for money is U($0) = 0, U($1000) = 1, U($1000000) = 2 (where $1000 and $1000000 are the amounts of money that could be in the two boxes, respectively). Furthermore, suppose Omega always correctly predicts two-boxers, so they always get $1000. However, Omega is very pessimistic about one-boxers, so only 0.2% of them get $1000000, and the average one-box value ends up being $2000.
It is then not correct to say that you should one-box. For you, the expected utility of two-boxing is exactly 1, but the expected utility of one-boxing is 0.2% x 2 = 0.004, and so one-boxing is a really stupid strategy even though the expected monetary gain is twice as high.
Edit: of course, there’s an obvious fix: compute the average utility received by people, according to your utility function, and optimize over that.
That was my goal, the same but less verbose. and without needing to factor out probabilities that are later factored back in.
My question was unclear, let me try again;
(Why) is it necessary to go through all the work to arrive at a probability that Omega will predict you correctly?
[edit question: is there anyway to do strike-through text in markdown? Or embed html tags?]
I always find it sad to see a thread downvoted with no comments or explanations, so I’m going to attempt to give my thoughts.
Newcomb’s problem seems absurdly easy to me. At least in the way it was presented by Eliezer, which is not necessarily a universal formulation. The way he expressed it, you observe Omega predicting correctly n times. (You could even add inaccurate observations if you wanted to consider the possibility that Omega is accurate, say, 90% of the time. We will do this in later steps, and call the number of inaccurate observations m.) If one box contains A dollars (or C dollars, if Omega predicted you to two box) and the other box contains B dollars, you can arrive at a pretty easy formulation of whether or not you should one box or two box. I almost wrote a MATLAB program to do it for arbitrary inputs that I was going to make into a post, but I figured most people wouldn’t find it very interesting, which was my conclusion after I got about halfway done with it.
First you arrive at a probability that Omega will predict you correctly, assuming that you are no different from anyone else with whom Omega has played the game. To do this, you estimate the accuracy, p, over a range of values from 0 to 1. 1.0 means Omega is perfectly accurate, 0.0 would mean ey is always wrong. The probability of obtaining the results you observed (n accurate predictions by Omega and m inaccurate ones) given any probability of em being accurate (p) is then p^n (1-p)^m. This gives us a distribution that represents the probability that Omega has a certain accuracy. We will call this distribution *D(p).
We then need to consider our two alternatives and select the one that maximizes expected utility. The utility of two boxing is:
U(two box) = p(Omega is wrong) (Value of box B + Value of Box A) + p(Omega is right) (Value of box B + Lesser value Omega puts in A)
U(one box) = p(Omega is wrong) (Value of box B + Lesser value Omega puts in A) + p(Omega is wrong) (Value of box B + Value of box A)
(Remember that we are considering the possibility that instead of replacing A with 0 dollars, Omega puts some value C dollars in A. All that really matters is the difference in these two values, though.)
With the variables we used, p(Omega is right) is the probability that ey has a certain accuracy, our distribution D(p) times that accuracy, p. p(Omega is wrong) is one minus this. The value of box A is obviously A, the value of box B is B, and the lesser value that Omega puts in box A is C
So our expected utilities are then a function of p as follows:
U(two box) = (1 - D(p)p)(B+A) + D(p)p(B+C) = [1 - (p^n (1-p)^m)] p (B+A) + (p^n (1-p)^m) p (B+C)
U(two box) = (1 - D(p)p)(B+C) + D(p)p(B+A) = [1 - (p^n (1-p)^m)] p (B+C) + (p^n (1-p)^m) p (B+A)
All that needs to be done is then to integrate the expected utilities over p from zero to one. Whichever value is greater is the correct choice.
Note that this analysis has a number of (fairly obvious and somewhat trivial) assumptions. One, the probability of Omega being right is constant over both one boxers and two boxers. Two, one’s utility function in money is linear (although compensating for that would not be very difficult). Three, Omega has no more or less information about you than anyone else about whom ey made this prediction.
This looks like evidential decision theory, which gives the wrong answer in the Smoking Lesion problem.
(Here’s a slightly less mind-killing variant: let’s say that regularly taking aspirin is correlated with risk of a heart attack, but not because it causes them; in fact, aspirin (in this hypothetical) is good for anyone’s heart. Instead, there’s an additional risk factor for heart attacks, which also causes discomfort beneath the threshold of full consciousness. People with this risk factor end up being more likely to take aspirin regularly, though they’re not able to pinpoint why, and the effect is large enough that the correlation points the “wrong” way. Now if you know all of this and are wondering whether to take aspirin regularly, the calculation you did above would tell you not to take it!)
We can get down to a discussion of evidential vs casual decision theory if you want, certainly, but I think that’s a bit off topic.
I have a couple of reactions to your point. My initial reaction is that evidential decision theory is superior in the case of Omega because nothing is known about em. Since Omega is a black box, the only thing that can really be done is gather evidence and respond to it.
But more generally, I think your example is somewhat strawman-ish. Just like in the smoking problem, there is other evidence suggesting that Asprin has the opposite effect. Saying that evidential decision theory has to ignore this is pretty unfair to it. Furthermore, you know that you can’t really rely on the evidence you have (that Asprin is correlated with heart attack, because you know you don’t have a random sample. Moreover, the evidence that Asprin is actually good for your heart was supposedly generated with some kind of statistical controls. It’s the same reason I mentioned in my analysis the assumption that Omega knows nothing more or less about you than ey did about any of the other people. The second you don’t have a representative sample, all of your statistics can be thrown out of the window.
If the goal is a simple analysis, why not this;
Let average_one_box_value = the average value received by people who chose one box.
Let average_two_box_value = the average value received by people who chose two boxes.
If average_one_box_value > average_two_box_value, then pick one box, else pick two.
As a bonus, this eliminates the need to assume Omega being right is constant over both one boxers and two boxers.
[Edit—just plain wrong, see Misha’s comment below] Minor quibble; It’s also not necessary to assume linear utility for dollars, just continuous. That is, more money is always better. However, I’m pretty sure that’s true in your example as well.
It is definitely necessary to assume linear utility for dollars. For example: suppose your (marginal) utility function for money is U($0) = 0, U($1000) = 1, U($1000000) = 2 (where $1000 and $1000000 are the amounts of money that could be in the two boxes, respectively). Furthermore, suppose Omega always correctly predicts two-boxers, so they always get $1000. However, Omega is very pessimistic about one-boxers, so only 0.2% of them get $1000000, and the average one-box value ends up being $2000.
It is then not correct to say that you should one-box. For you, the expected utility of two-boxing is exactly 1, but the expected utility of one-boxing is 0.2% x 2 = 0.004, and so one-boxing is a really stupid strategy even though the expected monetary gain is twice as high.
Edit: of course, there’s an obvious fix: compute the average utility received by people, according to your utility function, and optimize over that.
It’s been awhile, but isn’t that essentially what I did?
That was my goal, the same but less verbose. and without needing to factor out probabilities that are later factored back in.
My question was unclear, let me try again; (Why) is it necessary to go through all the work to arrive at a probability that Omega will predict you correctly?
[edit question: is there anyway to do strike-through text in markdown? Or embed html tags?]