Omega can be replaced by amnesia

26th Jan 2011

11lucidfox

0Douglas_Knight

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7tenshiko

6Vladimir_Nesov

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4Vladimir_Nesov

4cousin_it

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3Emile

1cousin_it

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0cousin_it

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3jimrandomh

3Perplexed

2timtyler

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4Perplexed

3Snowyowl

1timtyler

1[anonymous]

2Bongo

4timtyler

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2lucidfox

7[anonymous]

2ata

1Bongo

0timtyler

0Snowyowl

4[anonymous]

1Snowyowl

0Broggly

0[anonymous]

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1DanielLC

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44 comments, sorted by Click to highlight new comments since: Today at 3:58 AM

This is unsurprising: CDT relies on explicit dependencies given by causal definitions, while what you want is to look for logical (ambient) dependencies for which the particular way the problem was specified (e.g. physical content defined by causality) is irrelevant. After you find the dependencies as a result of such analysis, all that's left is applying expected utility, at which point any CDT-specificity is gone (see Controlling Constant Programs).

If that's really all the information you want to preserve, then I don't understand why you bother with amnesia in Newcomb's Problem. Just offer the player two boxes, the first one contains $1K, the second contains $1M, taking both boxes triggers a bomb that destroys the second box. I'm not sure what insight into decision theory we're supposed to get from such translations.

offer the player two boxes, the first one contains $1K, the second contains $1M, taking both boxes triggers a bomb that destroys the second box.

Hmm. This form has the same expected winnings for all strategies, but the $0 and $1,001,000 outcomes are impossible, unlike in the transformed Newcomb and the original Newcomb (given an Omega that doesn't punish mixed strategies). Also, expected winnings doesn't equal expected utility. For some utility functions, your problem has different expected utility than the normal or amnesiac Newcomb even if you play the same strategy in each. So it's not really equivalent.

Another example: consider (the tranformation of) Parfit's Hitchiker. If you use a coinflipping strategy there, the expected utility is

```
0.5*U(die)+0.5*(0.5*U($0)+0.5*U(-$100) = 0.5*U(die)+0.25*U($0)+0.25*U(-$100)
```

While the expected utility in the version where you simply plop the player in front of an ATM and drive them to the desert and dump them there if they don't pay $100 is:

```
0.5*U(die) + 0.5-U(-$100)
```

Which is clearly different.

I don't think they're weird. I think Omegas that go out of their way to discriminate against mixed strategies are weird. A strategy that one-boxes with probability 0.999 never gets a million, while one that one-boxes with probability 1 always gets a million. You could call that a discontinuity.

And I thought 1 was not a probability anyway! Any real rational one-boxing agent will *expect* to one-box with probability ~1, not with "probability" 1. Does that mean that the agent is using a mixed strategy? On the other hand, any agent that isn't using quantum randomness will in fact either one-box or two-box, even if it flips coins and stuff. Does that mean the agent is using a pure strategy? I can't answer this off the top of my head.

I assume the following is the key thing about Eliezer's original Omega:

Omega has been correct on each of 100 observed occasions so far - everyone who took both boxes has found box B empty and received only a thousand dollars; everyone who took only box B has found B containing a million dollars

I didn't see Eleizer saying that Omega doesn't tolerate mixed strategies. If there were coinflippers among that 100, presumably Omega predicted the results of their coinflips and set up box B accordingly. To the extent that I can't duplicate the conditions perfectly to make sure any coin will land the same way both times, I can't do that. To the extent that I can, I can.

This is equivalent to Newcomb's Problem in the sense that any strategy does equally well on both, where by "strategy" I mean a mapping from info to (probability distributions over) actions.

I don't see this. For example, the mixed strategy of one-boxing half the time and two-boxing half the time generates very different results in the transformed problem than in the original Newcomb's Problem.

Though I suppose there may be some ambiguity in the question of what amnesia is supposed to do to the 'seed' in your pseudo-random number generator.

Does CDT return the winning answer in such transformed problems?

It does fine on your transformation of Newcomb. I won't venture a guess on more general problems, because I don't understand how the general transformation is imagined to work. What is the transformation of the Hitchhiker, for example?

I don't see this. For example, the mixed strategy of one-boxing half the time and two-boxing half the time generates very different results in the transformed problem than in the original Newcomb's Problem.

Conventionally, you are not allowed access to a random number generator in Newcomb's Problem - and so can't use a mixed strategy. Any such usage would tarnish Omega's reputation. Omega - being a mind-reading superintelligence - can fairly easily discourage such a tactic by punishing randomising agents economically - and letting the punishment strategy be known.

I don't see this. For example, the mixed strategy of one-boxing half the time and two-boxing half the time generates very different results in the transformed problem than in the original Newcomb's Problem.

Nope? Let's say you flip a coin. Then your expected winnings are

- 0.5x(0.5x1000000+0.5x(1000000+1000))+0.5x(0.5x0+0.5x(0+1000)) = 500500 dollars

in both versions if Omega follows the rule:

- if you one-box with probability p, Omega fills box B with probability p

What is the transformation of the Hitchhiker, for example?

Put the player in front of an ATM and give them the amnesia drug. If they don't pay you $100, take them to the desert and dump them there. If they paid, put the money from the first round back into their bank account and give them the amnesia drug again. If they pay you again, keep their money. And the player knows these rules.

I don't have the general transformation down yet.

if you one-box with probability p, Omega fills box B with probability p

Really? I thought Omega would correctly predict the results of the coin flip and whether I called heads or tails. I guess this shows that Omega is better at predicting what I do than I am at predicting what he does.

In any case, thank you for the thought experiment. I agree with Snowyowl that your version is philosophically different from the original, but if we want our philosophical concepts to pay rent, they are going to have to have different consequences than some cheap amnesia drug. Otherwise, why keep them around?

I don't think it's quite the same. The underlaying mathematics are the same, but this version side-steps the philosophical and game-theoretical issues with the other (namely, acausal behaviour).

Incidentally; If you take both boxes with probability p each time you enter the room, then your expected gain is p*1000 + (1-p) *1000000. For maximum gain, take p=0; i.e. always take only box B.

EDIT: Assuming money is proportional to utility.

I could change the rules and decide not to stand for such tricks (mixed strategies) either. **EDIT: No, I couldn't.**

And on the other hand, Omega *could* deal with mixed strategies perfectly well, and I don't really understand why people make it so that he explicitly doesn't tolerate mixed strategies in their problems. For example, in Newcomb's Problem, if you one-box with probability p, Omega can just fill box B with probability p - for example if p=0.5 your expected winnings in Newcomb's Problem are $500,500.

Hang on a minute though

*1 box then 2 box = $1,001,000
*1 box then 1 box = $1,000,000
*2 box then 2 box = $1,000
*2 box then 1 box = $0

$2,002,000 divided by 4 is $500,500. Effectively you're betting a million dollars on two coinflips, the first to get your money back (1-box on the first day) and the second to get $1000 (2-box on the second day). Omega could just use a randomizer if it thinks you will, in which case people would say "Omega always guesses right, unless you use a randomizer. But it's stupid to use one anyway."

Where p is the probability of 1 boxing, $E = p^2 * $1,000,000 + p(1-p * $1,001,000 + (1-p)^2 * $1,000 = $999,000 p + $1000.
So the smart thing to do is clearly always one-box, unless showing up Omega who thinks he's so big is worth $499,500 to you.

I don't see that the scenario is the same. If you one-box everytime in your thought experiment, you are guaranteed to get the million; if you two box everytime, you will certainly not get the million. With Omega, there is a high probability but not certainty.

Also, what you do in the first round *causes* what happens in the second round, but with Omega, it is debatable whether what you end up doing *causes* there to be a million dollars or not.

You point out perhaps the only potentially meaningful difference, and it is the main salient point in dispute between one-boxers and two-boxers in the Omega problem.

First subpoint: With Omega, you are told (by Omega) that there is certainty--that he is never wrong--and you have a large but finite number of previous experiments that do not refute him. Any uncertainty is merely hoped for/dreaded. (There are versions in which there is definite uncertainty, but those are clearly not similar to the OP.)

Second subpoint: If there is truly, really, actually no uncertainty, then correlation is perfect. It is hard to determine cause and effect in such conditions with no chance to design experiments to separate them. I'd argue that cause is a low-value concept in such a situation.

This is equivalent to Newcomb's Problem in the sense that any strategy does equally well on both, where by "strategy" I mean a mapping from info to (probability distributions over) actions.

I suspect that any problem with Omega can be transformed into an equivalent problem with amnesia instead of Omega.

Does CDT return the winning answer in such transformed problems?

Discuss.