Omega also knew the Lottery number before you saw it, and while making its prediction, and Omega likewise predicts correctly in 99.9% of the cases where the Lottery number happens to match Omega’s number.
(Omega’s number is chosen independently of the lottery number, however.)
I’m not sure that both these statements can be true at the same time.
What does Omega pick if my algorithm is, “if the number ends in 3, two-box, else one-box”? Seems it will be right if it either picks a composite ending in 3 (where I don’t get the million) or a prime not ending in 3 (where I do), so how does it decide?
Maybe I’m confused, but can’t it just… pick one however it wants? As long as the final tally comes to “correct in 99.9% of cases” the problem is satisfied and it doesn’t matter how it decided, does it? Why does it matter if there are muliple correct choices?
Edit: Nevermind, I see what you mean. If your algorithm depends on the Lotto, and Omega depends on your algorithm, then Omega depends on the Lotto. You can actually influence how often this “same number” scenario happens via your algorithm.
There are two equal-sized groups of large numbers:
Prime numbers
Composite numbers without any easy factorization
… such that you can’t be expected to distinguish them in the allowed time.
The lottery works by picking at one of the numbers at random.
Omega’s algorithm: for each number in those groups, predict whether you would two-box (given the already-determined lottery number). In the set of prime numbers for which you one-box, and composite numbers for which you two-box, pick a number at random and show it to you (so if it predicts you always two-box, it’s sure to pick a composite number, etc.).
In that setup, if your algorithm is “if the number ends in 3, two-box, else one-box”, then Omega will just pick a random number among the composites ending in 3 and the primes ending in 1, 3, 7, or 9.
And if your algorithm is “If the number is the same as the lottery, two-box, else, one-box”, then yeah the chosen number will not be completely independent of the lottery number (Omega can only pick the lottery number if it’s composite, or if it makes a prediction error), but that looks independent enough for the spirit of the post.
I’m not sure that both these statements can be true at the same time.
If you take the second statement to mean, “There exists an algorithm for Omega satisfying the probabilities for correctness in all cases, and which sometimes outputs the same number as NL, which does not take NL’s number as an input, for any algorithm Player taking NL’s and Omega’s numbers as input,” then this …seems… true.
I haven’t yet seen a comment that proves it, however. In your example, let’s assume that we have some algorithm for NL with some specified probability of outputting a prime number, and some specified probability it will end in 3, and maybe some distribution over magnitude. Then Omega need only have an algorithm that outputs combinations of primeness and 3-endedness such that the probabilities of outcomes are satisfied, and which sometimes produces coincidences.
For some algorithms of NL, this is clearly impossible (e.g. NL always outputs prime, c.f. a Player who always two-boxes). What seems less certain is whether there exists an NL for which Omega can always generate an algorithm (satisfying both 99.9% probabilities) for any algorithm of the Player.
This is to say, what we might have in the statement of the problem is evidence for what sort of algorithm the Natural Lottery runs.
Perhaps what Eliezer means is that the primeness of Omega’s number may be influenced by the primeness NL’s number, but not by which number specifically? Maybe the second statement is meant to suggest something about the likelihood of there being a coincidence?
I’m not sure that both these statements can be true at the same time.
What does Omega pick if my algorithm is, “if the number ends in 3, two-box, else one-box”? Seems it will be right if it either picks a composite ending in 3 (where I don’t get the million) or a prime not ending in 3 (where I do), so how does it decide?
Maybe I’m confused, but can’t it just… pick one however it wants? As long as the final tally comes to “correct in 99.9% of cases” the problem is satisfied and it doesn’t matter how it decided, does it? Why does it matter if there are muliple correct choices?
Edit: Nevermind, I see what you mean. If your algorithm depends on the Lotto, and Omega depends on your algorithm, then Omega depends on the Lotto. You can actually influence how often this “same number” scenario happens via your algorithm.
Excellent point.
A plausible setup:
There are two equal-sized groups of large numbers:
Prime numbers
Composite numbers without any easy factorization
… such that you can’t be expected to distinguish them in the allowed time.
The lottery works by picking at one of the numbers at random.
Omega’s algorithm: for each number in those groups, predict whether you would two-box (given the already-determined lottery number). In the set of prime numbers for which you one-box, and composite numbers for which you two-box, pick a number at random and show it to you (so if it predicts you always two-box, it’s sure to pick a composite number, etc.).
In that setup, if your algorithm is “if the number ends in 3, two-box, else one-box”, then Omega will just pick a random number among the composites ending in 3 and the primes ending in 1, 3, 7, or 9.
And if your algorithm is “If the number is the same as the lottery, two-box, else, one-box”, then yeah the chosen number will not be completely independent of the lottery number (Omega can only pick the lottery number if it’s composite, or if it makes a prediction error), but that looks independent enough for the spirit of the post.
If you take the second statement to mean, “There exists an algorithm for Omega satisfying the probabilities for correctness in all cases, and which sometimes outputs the same number as NL, which does not take NL’s number as an input, for any algorithm Player taking NL’s and Omega’s numbers as input,” then this …seems… true.
I haven’t yet seen a comment that proves it, however. In your example, let’s assume that we have some algorithm for NL with some specified probability of outputting a prime number, and some specified probability it will end in 3, and maybe some distribution over magnitude. Then Omega need only have an algorithm that outputs combinations of primeness and 3-endedness such that the probabilities of outcomes are satisfied, and which sometimes produces coincidences.
For some algorithms of NL, this is clearly impossible (e.g. NL always outputs prime, c.f. a Player who always two-boxes). What seems less certain is whether there exists an NL for which Omega can always generate an algorithm (satisfying both 99.9% probabilities) for any algorithm of the Player.
This is to say, what we might have in the statement of the problem is evidence for what sort of algorithm the Natural Lottery runs.
Perhaps what Eliezer means is that the primeness of Omega’s number may be influenced by the primeness NL’s number, but not by which number specifically? Maybe the second statement is meant to suggest something about the likelihood of there being a coincidence?