You see two boxes and you can either take both boxes, or take only box B. Box A is transparent and contains $1000. Box B contains a visible number, say 1033. The Bank of Omega, which operates by very clear and transparent mechanisms, will pay you $1M if this number is prime, and $0 if it is composite. Omega is known to select prime numbers for Box B whenever Omega predicts that you will take only Box B; and conversely select composite numbers if Omega predicts that you will take both boxes. Omega has previously predicted correctly in 99.9% of cases.

Separately, the Numerical Lottery has randomly selected 1033 and is displaying this number on a screen nearby. The Lottery Bank, likewise operating by a clear known mechanism, will pay you $2 million if it has selected a composite number, and otherwise pay you $0. (This event will take place regardless of whether you take only B or both boxes, and both the Bank of Omega and the Lottery Bank *will* carry out their payment processes - you don't have to choose one game or the other.)

You previously played the game with Omega and the Numerical Lottery a few thousand times before you ran across this case where Omega's number and the Lottery number were the same, so this event is not suspicious.

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.)

You have two minutes to make a decision, you don't have a calculator, and if you try to factor the number you will be run over by the trolley from the Ultimate Trolley Problem.

Do you take only box B, or both boxes?

As wedifrid said, this is approximately transparent Newcomb plus distractions. Given Eliezer’s clarifications (Omega knows the lottery numbers and is accurate even when the Omega and lottery numbers match), I’ll ask how two algorithms would perform against Omega: OneBoxBot, which always one-boxes, and ConditionalBot, which one-boxes unless the lottery and Omega numbers match, in which case it two-boxes. I’ll ignore the tiny error rates in computing payoffs.

Case 1: the lottery is going to output a prime number.

Against OneBoxBot, Omega delivers a prime number, which may or may not match the lottery number. OneBoxBot gets a payoff of $1MM.

Against ConditionalBot, Omega must deliver a prime number different from the lottery number (if it matched the lottery Omega would be handing over a prime number with a predicted response of two-boxing from Conditional). So ConditionalBot one-boxes and gets a payoff of $1MM.

Case 2: the lottery will output a composite number.

OneBoxBot will one-box, so Omega must provide it with a prime number (which will not match the lottery). So OneBoxBot gets a payoff of $3MM every time the lottery randomly outputs a composite number.

Against ConditionalBot, Omega... (read more)

There is a big and implicit step that is worth explicating here, because most people who first approach Newcomb-like problems miss it completely:

TREAT HUMANS AS BOTSBy a bot I mean an algorithm, of course. An algorithm Omega can analyze for all possible combination of inputs.

That this step is valid follows from the problem's stipulation that Omega can predict your actions. In other words, it knows your output for any combination of inputs it cares to give you.

Whether Omega does this by running your algorithm in a sandbox or by analyzing your code does not affect the answer to the puzzle, since the end result is the same. But the sandboxing version can often make it easier to find the solution, because it lets one rely on the Reflective Equilibrium of sorts: you cannot tell when deciding what to do whether you are in an Omega's simulation of you or not, so you may as well assume that you are.

TL;DR: to Omega, you are a bot, so write down all relevant algorithms and analyze/run them before picking a winning one.I'm afraid that, as written, I cannot answer the problem's final question, as by the time it was asked...

... I'd been hit by the trolley.

(Initial remark: damn, I see that what I wrote got screwed up by asterisks getting treated as markup characters; perhaps everything would have been clear without that. Will fix once I've finished writing this.)

I'm not sure it really warrants dignifying with the term "method", but:

The relations I'm looking for are of the form n=a+b or n=a-b where a,b (1) are easy to factor because they're products of small primes, and (2) have no common factor. In that case, you know that n isn't a multiple of any of those prime numbers -- so, e.g., if a, but not b, is a multiple of p, then a+b and a-b are not multiples of p.

The easiest case is where b is itself a small prime number like 17 or 13. Why 17 and 13? Because the number ends in 3, so subtracting something ending in 3 or adding something ending in 7 will give us (an easy-to-deal-with factor of 10, and) something nicely smaller. In some cases it's easier to remove digits from the start of the number rather than the end; for a trivial example, 1033 isn't a multiple of 103 because it's 1030 + 3.

Let's do 1033 a bit more thoroughly. It's between 32^2=1024 and 33^2=1089, so we need to check primes up to 31. 1033=1050-17 rules out 2,3,5... (read more)

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?

CDT:

Two box, obviously.

EDT:

Assuming this is your last game, two box. Two boxing is evidence that 1033 is composite, so you'll get more money.

If you will continue playing for a long time, one box. This is evidence that you will go with the "always one box" strategy, which will result in more money. More generally, it is evidence that you will go with a TDT-style strategy more often in the future, and get higher payouts as a result.

TDT:

One box. The always one box strategy has the highest payout.

I'm not sure if I have the right terminology with TDT, but these are the three obvious moves and the reasoning for them.

Although the title was originally selected in jest, I think this may actually

bethe Ultimate Newcomb's Problem because it tempts the largest number of people - EDTers, CDTers, and apparently a substantial portion of LWers - to two-box.Facebook discussion.

This seems to be just transparent Newcomb's problem combined with some redundant words. Unless I've missed something in the fine print this is a simple "one box" situation.

Reading this was like hearing Vincent Price saying "your payoff matrices are useless here! Ahahahaha!" That was a legitimate source of epistemic dread.

It took me about a minute of turning it over to fully grok the structure of the problem, at which point I settled on two-boxing, which

va guvf pnfr, jvgu guvf ahzore, yrnirf zr ubyqvat n zrer gubhfnaq qbyynef, orpnhfr guvf vf bar bs gur 0.01% bs havirefrf jurer Bzrtn jnf jebat.I am pretty solidly sold on two-boxing at the moment, under the absurd-sounding premise that I can influence whether or not the number is prime with my decision. I really hope time travel isn't possible.

The optimal choice if you're just told you're gonna play this game a bunch of times is one-boxing. (Edit: this is because because trying to control the lottery number will just result in less games where the numbers are the same). However, if you're guaranteed a run where the numbers come up the same, then there's some hidden control,

via the thought experiment itself.If I two box, Omega picked a composite, so in order for the scenario to have occurred at all (which by the thought experiment it's guaranteed to have) the lottery must have come up with a composite. Calling something random doesn't make it random. I probably wouldn't have noticed if I hadn't recently read GAZP vs GLUT.If Omega maintains a 99.9% accuracy rate against a strategy that changes its decision based on the lottery numbers, it means that Omega can predict the lottery numbers. Therefore, if the lottery number is composite, Omega has multiple choices against an agent that one-boxes when the numbers are different and two-boxes when the numbers are the same: it can pick the same composite number as the lottery, in which case the agent will two-box and earn 2,001,000, or it can pick a different prime number, and have the agent one-box and earn 3,001,000. It seems like the agent that one-boxes all the time does better by eliminating the cases where Omega selects the same number as the lottery, so I would one box.

I initially thought two-box, but on thinking about it more, I'm going for one-box.

For simple numbers, let's suppose that the lottery has a 50% chance of choosing a prime number, and that if Omega could select the same number as the lottery, he'll do so with 10% probability.

Three simple strategies:

1) Always one-box: Gets Omega's payout every time, wins the lottery 50% of the time. Average total payout $2M. (numbers are the same 10% of the time when the lottery is 'prime')

2) Always two-box: Omega never pays out, wins the lottery 50% of the time. Average tota... (read more)

Posting before reading comments:

If I one-box then (ignoring throughout the tiny probabilities of Omega being wrong) the number is prime. I receive $1M from Omega and $0 from the Lottery.

If I two-box then the number is composite, Omega pays me $1K, and the Lottery pays me $2M.

Therefore I two-box.

To properly decide I need to know if I am Jones or Leftie.

EDIT: Given that the Ultimate Trolley only directly kills Jones or Lefty, any agent that tries factoring either Omega's or the lottery's number must be Jones or Leftie. To decide whether it's better to factor the number and die by trolley it's necessary to know the answer to the Ultimate Trolley Problem and whether it's better for Jones or Leftie to die. The expected utility of the final outcome of the Ultimate Trolley Problem certainly dwarfs the expected utility of either the lottery or Omega's bank.

I would claim that the correct probability to hold is somewhere around 0.999 in favor of composite if you take both boxes [...]

EDIT: Looks like I was right about probabilities, but too hasty about thinking that meant you should two-box. Omega can be malicious:

Suppose we do this Primecomb + lottery experiment a jillion times. What algorithm maximizes payout over those jillion times?

One-boxing sure seems like a good plan - usually the lottery will pay out, sometimes not, but no biggie since you can't affect it. And since there aren't that many prime numb... (read more)

That's impossible (given your other specifications) against an agent who two-boxes iff the lottery number and Omega's number match. If the lottery number is prime, this

causesOmega to ensure that adifferentprime number is placed in its box.Edit: Realized; it should be "The lottery number is chosen independently of Omega's number."

Edit again: Just realized that this phrase sneaks in a causal postulate: Omega can't change the output of the lottery! Starting here in reason... (read more)

Att... (read more)

I want to pre-commit to 1-boxing as long as the numbers are different, just like in the standard Newcomb problem. Since Omega knows that I will notice if the numbers are the same, I can decide to make a special case for this situation without affecting the standard case. But I still want to pre-commit to 1-boxing in this case, for the same reason I want to pre-commit in standard Newcomb: I can predict that Omega will be much more likely to put $1,000,000 in box B if I do so, and the causality here doesn't allow me to influence the outcome of the lottery.

(Note: I haven't checked yet to see if 1033 is prime)

So... basically, it's the standard Newcomb's problem, one box or two, one boxing means it's a prime number and two boxing means it's a composite number being displayed for the lottery, in this singular case.

I'd still probably one box here. If 1033 is prime, and I two box... well, then, Omega probably wouldn't have picked it and we wouldn't be discussing this scenario.

Put another way, I don't see how the lottery number matching Omega's number gives me any useful information about Omega's accuracy, since the value of one number in no way depends on the other.

I'll flip a coin. Heads I 2-box, tails I 1-box. That's got to be pretty good in expectation. Gotta make omega work for that 99.9% accuracy!

Since this is too complicated for me to figure out in any deep sense in 2 minutes (and I'm not sure exactly what computation Omega is doing or if it's even well-defined), I'm falling back on EDT, and two-boxing to get the $2M lottery. EDT might be suboptimal in a lot of cases (smoker's lesion), but at least it's

unlikelyto choose wrong, which is better than I can say for what I'd pick if I tried to use TDT or CDT-combined-with-uncertainty-about-who-I-am (whether I'm a copy simulated by omega to determine payoffs for the real copy of me).Written before reading comments; The answer was decided within or close to the 2 minute window.

I take both boxes. I am uncertain of three things in this scenario: 1)whether the number is prime; 2) whether Omega predicted I would take one box or two; and 3) whether I am the type of agent that will take one box or two. If I take one box, it is highly likely that Omega predicted this correctly, and it is also highly likely that the number is prime. If I take two boxes, it is highly likely that Omega predicted this correctly and that the number is composite. I... (read more)

I wonder if we're all burying the lead. You've played these games thousands of times, so why aren't you already a billionaire? Are you being suckered into overpaying for the numerical lottery and that's eating your winnings from Omega? Are you paying to play against the Omega Bank? Have you been two-boxing this whole time? We're not getting the whole story here.

(None of that changes my answer, I'm one-boxing all the way, but we need answers!)

Prior to seeing the fact that the Lottery numbers matched, I would have

likedto have pre-committed to one boxing in all cases. That's how I set my deterministic algorithm.Therefore, I

willsurely one box. This seems more or less identical to the classic Newcomb. Yes, Iknowthe number is primenow, so Iwouldlike to get away with taking the second box and I should tryas hard as I canto override my initial programming and two box...but unless my algorithm is unsuccessfully pre-committed, Iwillfail to do so.Some folks seem to think you aught to pre-co... (read more)

I think most of the commenters aren't getting that this is a parody. Edit: It turns out I was wrong.

"Unlike these other highly-contrived hypothetical scenarios we invent to test extreme corner-cases of our reasoning,

thishighly-contrived hypothetical scenario is a parody. If you ever find yourself in the others, you have to take it seriously, but if you find yourself in this one, you are under no such obligation."I'm at the current MIRI workshop, and the Ultimate Newcomb's Problem is not a parody.

I don't get paid on the basis of Omega's prediction given my action. I get paid on the basis of my action given Omega's prediction. I at least need to know the base-rate probability with which I actually one-box (or two-box), although with only two minutes, I would probably need to know the base rate at which Omega predicts that I will one-box. Actually, just getting the probability for each of P(Ix|Ox) and P(Ix|O~x) would be great.

I also don't have a mechanism to determine if 1033 is prime that is readily available to me without getting hit by a trolley (... (read more)

I can’t decide anything in 2 minutes, so I’d just one-box it because I remember it as the correct solution to the original Newcomb’s problem — and hope for the best.

This game seems to have a roughly 50% chance of a fatality.

+EV of one boxing swamps my uncertainty over the status of 1033.

Problem lacks specification: ought we assume that Omega also predicted TNL's number? Or was that random both to me, Omega-past, and TNL? Omega predicting correctly in 99.9% of previous cases doesn't determine this.

(eta: Ah, was answered on the Facebook thread; Omega predicted the lottery number. Hm.)

...so if I google 'is 2033 a prime number' and receive the answer that it isn't, all in under two minutes, and put the money from box A into box B and

thenchoose box B, do I get any money?:)EDIT:

Reading through comments it appears the idea is that you're attempting to control whether or not omega matches the number with your choice of strategy, rather than what you get from matched or unmatched lotteries. So the idea seems to be that always taking onebox yields lower payoffs from matching lotteries but causes lotteries to match less often, which is beneficial because unmatching lotteries have better payoffs than matching lotteries.

I haven't looked at the comments yet. Two minutes was enough thinking time to get me to one-box, put not enough time to verbalize my intuition. There was a post earlier on lw about making decisions with imperfect memory that is relevant, I think.

My intuition is something along the lines of 'my decisions only affect whether omega gives me a million dollars or not, so the lottery doesn't matter'.

Further thoughts: your strategy when numbers match change the information that numbers matching conveys. If you one box, it tells you that the particular round loses... (read more)

(Answering before reading any other responses)

Both boxes -- I want the number to be composite, so I want Omega to have selected a composite number, which he'd have more chances of doing if I two-boxed.

EDIT:wedrifid's explanation has now mostly convinced me that one-boxing is correct instead. (My expressed logic was too much EDT-influenced, I think)The number 1033 is prime. The rest of the hypothetical scenario is pretty confusing, and it'll take me longer to analyze it and be confident about my conclusion than it did to determine that 1033 is prime.

This isn't a fair problem (in the sense that you defined somewhere else): two people both choosing to take both boxes, but via different algorithms (only one of whom factoring the number), will get different rewards.

I don't know what "trying to factor" even would be for a number so small. It just

looks like a prime. I may have seen it on a prime number list, or as a prime factor of something, or who knows where. There's easy to construct rules for determining divisibility by it's potential factors.One could also use Miller-Rabin primarity test, which I in fact happen to have implemented before. Much of the public key cryptography depends on how testing a prime is easier than factoring a number. I'm pretty sure there is no general algorithm for determining w... (read more)

I'm not clear on what constitutes trying to factor the number. It would seem that noticing if it was odd wouldn't count as trying to factor it, but what about forms of inductive reasoning or non-exhaustive heuristics?

Then there are a couple billion dollars in my bank account, and the marginal utility of one more million wouldn't be that large. :-)

I hadn't considered this angle, but I agree with this. If I'm going to get trollied for thinking the wrong thought, I would want to try hard not to think at all since I might accidentally think about how I already know whether the number is prime.

On the other hand, I don't think this is a consequence the post intended.

Omega knows I'm smart enough to two box when I see that the #'s match up. So the # will be composite, thereby fulfilling his goal of predicting my actions and responding appropriately. The Lottery Bank doesn't care about my decisions.

So I'll two box.. I'll do so, and would predictably do so. So Omega has selected a composite #, and I receive 1000$ from the boxes. Since its the same # the lottery will give me 2 million dollars.

Immediate thoughts, before reading comments: One-box. I had started to think more deeply until I read the part about being run over for factoring, and for some reason my brain applied it to reasoning about this topic as a whole and spit out a final answer.

Intuitively, it seemed one boxing would get me a million, as per standard Newcomb. The lottery two million seemed like gravy above that (diminishing marginal utility of money), with a potential for 3 million total. Since they're independent, the word "separately" and its description made it seem like the lottery was unable to be affected by my actions at all. Thus, take box B, and hope for a lottery win. Definitely don't over think it, or risk a trolley encounter.

Posting before checking the comments.

If I take only box B I will either make 1M$ or 2M$. Omega, with its 99,9% accuracy, will likely have selected a prime number. Expected utility is 0.999

1M + 0.0012M $ = 1M+1K $.If I take both, I will either get 1M+1K $ or 2M +1K $. Already I'm grabbing both boxes, because the expected utility is clearly higher. Omega would likely have selected a composite number. Expected utility is therefore 0.999

2001 K + 0.0011001 K = 2M $.In cases where the Lottery number doesn't match Omega's, I have a number of general strate... (read more)

2-box on one-off [Edit: on rereading, this comes off as more confident than I intended. This was what I thought I would do in the 2 minutes, which in retrospect were spent unproductively], but the one-off nature of the problem is modified by the third paragraph, which means that strategy might mean one-boxing previously got me no money. I would interpret that as less-than-perfect accuracy (which might be the source of the 99.9% probability) How does Omega deal with mixed strategies?

I expect to maintain control over Omega(today, my decision), even after learning Omega(today, my decision) = Lottery(today). I take both boxes for ~$2,001,000.

Since I don' t know the definition of the process generating the lottery number, my prior distribution over the primality of today's lottery number (before learning that Omega's number in Box B and the Lottery Number are equal) would be taken from the distribution of primes (1/log(1033) ~ 0.332). I don't know why my intuition says to throw that number out after I condition on the equality. I feel les... (read more)

I don't think I understand the problem. While reading the Numerical Lottery (NL) paragraph, I decided choosing only box B was the obvious answer, which led me to think I've misunderstood something.

In box B is $1,000, a composite number. If I pick a composite number, the NL gifts me $2 million it-doesn't-really-matter-what-currency

Oh, I misread the problem. In box

Ais $1,000. Well, now I think I've understood the problem, and chosen the right answer for mistaken reasons.If the NL pays "[me] $2 million

if it has selecteda composite number, and oth... (read more)Written before reading comments:

This is not a well formed problem. If my strategy is to 1-box iff the numbers match, then omega, choosing his number independently of the lottery number must choose a composite number, since I will 2-box 99.9% of the time. Therefore if omega is correct 99.9% of the time when the numbers match, then the lottery number must be composite at least 99.9% of the time.

However, if my strategy is to 2-box iff the numbers match, then omega, choosing his number independently of the lottery number must choose a prime number, since I wi... (read more)