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Comment author: Oscar_Cunningham 01 November 2017 02:45:56PM 4 points [-]

However, if T is an explanatory theory (e.g. ‘the sun is powered by nuclear fusion’), then its negation ~T (‘the sun is not powered by nuclear fusion’) is not an explanation at all.

The words "explanatory theory" seem to me to have a lot of fuzziness hiding behind them. But to the extent that "the sun is powered by nuclear fusion" is an explanatory theory I would say that the proposition ~T is just the union of many explanatory theories: "the sun is powered by oxidisation", "the sun is powered by gravitational collapse", and so on for all explanatory theories except "nuclear fusion".

Therefore, suppose (implausibly, for the sake of argument) that one could quantify ‘the property that science strives to maximise’. If T had an amount q of that, then ~T would have none at all, not 1-q as the probability calculus would require if q were a probability.

There are lots of negative facts that are worth knowing and that scientists did good work to discover. When Michelson and Morley discovered that light did not travel through luminiferous aether that was a fact worth knowing, and lead to the discovery of special relativity. So even if you don't call ~T an explanatory theory it seems like it still has a lot of "the property that science strives to maximise"

Also, the conjunction (T₁ & T₂) of two mutually inconsistent explanatory theories T₁ and T₂ (such as quantum theory and relativity) is provably false, and therefore has zero probability. Yet it embodies some understanding of the world and is definitely better than nothing.

A Bayesian might instead define theories T₁' = "quantum theory leads to approximately correct results in the following circumstances ..." and T₂' "relativity leads to approximately correct results in the following circumstances ...". Then T₁' and T₂' would both have a high probability and be worth knowing, and so would their conjunction. The original conjunction, T₁ & T₂, would mean "both quantum theory and relativity are exactly true". This of course is provably false, and so has probability 0.

Furthermore if we expect, with Popper, that all our best theories of fundamental physics are going to be superseded eventually, and we therefore believe their negations, it is still those false theories, not their true negations, that constitute all our deepest knowledge of physics.

Right, right. The statement T₁ is false; but the statement T₁' is true.

What science really seeks to ‘maximise’ (or rather, create) is explanatory power.

Does Deutsch write anywhere about what a precise definition of "explanation" would be?

In response to Just a photo
Comment author: Lumifer 20 October 2017 12:36:28AM 0 points [-]

It's one of those ambiguity things of which this is the canonical example.

In response to comment by Lumifer on Just a photo
Comment author: Oscar_Cunningham 20 October 2017 11:11:15AM *  0 points [-]

It's also similar to this image:

https://i.pinimg.com/736x/60/16/5b/60165bfd56829bb95563a36cd69a5825--art-optical-optical-illusions.jpg

It's difficult to see it as anything until it "snaps" and then it's impossible to not see it.

Comment author: Thomas 25 September 2017 07:41:18AM 1 point [-]
Comment author: Oscar_Cunningham 25 September 2017 10:03:03AM *  1 point [-]

If you fail to get your n flips in a row, your expected number of flips on that attempt is the sum from i = 1 to n of i*2^-i, divided by (1-2^-n). This gives (2-(n+2)/2^n)/(1-2^-n). Let E be the expected number of flips needed in total. Then:

E = (2^-n)n + (1-2^-n)[(2-(n+2)/2^n)/(1-2^-n) + E]

Hence (2^-n)E = (2^-n)n + 2 - (n+2)/2^n, so E = n + 2^(n+1) - (n+2) = 2^(n+1) - 2

Comment author: Thomas 01 September 2017 07:57:32AM *  0 points [-]

1.round____1_____8_0.125

2.round____15_____64_0.234

3.round___164____512_0.32

4.round___1,585____4,096_0.387

5.round__14,392___32,768_0.439

6.round__126,070___262,144_0.481

7.round_1,079,808__2,097,152_0.515

8.round_9,111,813__16,777,216__0.543

9.round_76,095,176__134,217,728__0.567

Comment author: Oscar_Cunningham 01 September 2017 09:57:53AM *  0 points [-]

I think you must just have an error in your code somewhere. Consider going round 3. Let the probability you say "3" be p_3. Then according to your numbers

164/512 = 15/64 + (1 - 15/64)*(1/2)*p_3

Since the probability of escaping by round 3 is the probability of escape by round 2, plus the probability you don't escape by round 2, multiplied by the probability the coin lands tails, multiplied by the probability you say "3".

But then p_3 = 11/49, and 49 is not a power of two!

Comment author: Thomas 31 August 2017 07:02:05PM *  0 points [-]

Interesting. Do you agree that every number is reached by the z function defined above, infinite number of times?

And yet, every single time z != sleeping_round? In the 60 percent of this Sleeping Beauty imprisonments?

Even if the condition x>y is replaced by something like x>y+sqrt(y) or whatever formula, you can't go above 50%?

Extraordinary. Might be possible, though.

You clearly have a function N->N where eventually every natural number is a value of this function f, but f(n)!=n for all n.

That would be easier if it would be f(n)>>n almost always. But sometimes is bigger, sometimes is smaller.

Comment author: Oscar_Cunningham 31 August 2017 08:08:18PM 0 points [-]

Do you agree that every number is reached by the z function defined above, infinite number of times?

Yes, definitely.

Even if the condition x>y is replaced by something like x>y+sqrt(y) or whatever formula, you can't go above 50%?

Yes. I proved it.

You clearly have a function N->N where eventually every natural number is a value of this function f, but f(n)!=n for all n.

Well, on average we have f(n)=n for one n, but there's a 50% chance the guy won't ask us on that round.

Comment author: Thomas 31 August 2017 12:53:38PM 0 points [-]

Define flip values as H=0 and T=1. You have to flip this fair coin twice. You increase x=x+value(1) and y=y+value(2) and z=z+1. If x>y you stop flipping and declare - It's the z-th round of the game.

For example, after TH, x=1 and y=0 and z=1. You stop tossing and declare 1st round. If it is HH, you continue tossing it twice again.

No matter how late in the game you are, you have a nonzero probability to win. Chebyshev (and Chernoff) can help you improve the x>y condition a bit. I don't know how much yet. Neither I have a proof that then the probability of exiting is > 1/2. But at least that much it is. Some Monte-Carloing seems to agree.

Comment author: Oscar_Cunningham 31 August 2017 06:00:01PM 0 points [-]

Based on some heuristic calculations I did, it seems that the probability of escape with this plan is exactly 4/10.

Comment author: cousin_it 31 August 2017 10:08:19AM *  2 points [-]

It seems to me that there's no difference in kind between moral intuitions and religious beliefs, except that the former are more deeply held. (I guess that makes me a kind of error theorist.)

If that's true, that means FAI designers shouldn't work on approaches like "extrapolation" that can convert a religious person to an atheist, because the same procedure might convert you into a moral nihilist. The task of FAI designers is more subtle: devise an algorithm that, when applied to religious belief, would encode it "faithfully" as a utility function, despite the absence of God.

Does that sound right? I've never seen it spelled out as strongly, but logically it seems inevitable.

Comment author: Oscar_Cunningham 31 August 2017 12:06:49PM 0 points [-]

It seems to me that there's no difference in kind between moral intuitions and religious beliefs,

That just doesn't seem true to me. I agree that there's often difference between religious beliefs and ordinary factual beliefs, but I don't think that religious beliefs are the same sort of thing as moral intuitions. They just feel different to me.

For one thing religious beliefs are often a "belief in belief" whereas I don't think moral beliefs are like that.

Also moral beliefs seem more instinctual, whereas religious beliefs are taught.

Comment author: halcyon 29 August 2017 08:23:17PM 1 point [-]

Integrals sum over infinitely small values. Is it possible to multiply infinitely small factors? For example, Integration of some random dx is a constant, since infinitely many infinitely small values can sum up to any constant. But can you do something along the lines of taking an infinitely large root of a constant, and get an infinitesimal differential in that way? Multiplying those differentials will yield some constant again.

My off the cuff impression is that this probably won't lead to genuinely new math. In the most basic case, all it does is move the integrations into the powers that other stuff is raised by. But if we somehow end up with complicated patterns of logarithms and exponentiations, like if that other stuff itself involves calculus and so on, then who knows? Is there a standard name for this operation?

Comment author: Oscar_Cunningham 29 August 2017 10:57:17PM 2 points [-]

Good question!

The answer is called a Product integral. You basically just use the property

log(ab) = log(a) + log(b)

to turn your product integral into a normal integral

product integral of f(x) = e ^ [normal integral of log(f(x))]

Comment author: Thomas 28 August 2017 11:47:00AM 0 points [-]

If you always say "It's my first time" you will be freed with the probability 1/2, yes.

I'll give the best strategy I know before the end of this week. Now, it would be a spoiler.

Comment author: Oscar_Cunningham 28 August 2017 05:20:01PM *  4 points [-]

Let p_n be the probability that I say n. Then the probability I escape on exactly the nth round is at most p_n/2 since the coin has to come up on the correct side, and then I have to say n. In fact the probability is normally less than that since there is a possibility that I have already escaped. So the probability I escape is at most the sum over n of p_n/2. Since p_n is a probability distribution it sums to 1, so this if at most 1/2. I'll escape with probability less than this is I have any two p_n nonzero. So the optimal strategies are precisely to always say the same number, and this can be any number.

Comment author: halcyon 24 August 2017 05:46:54PM *  1 point [-]

Thanks. You're right, that part should be expanded. How about:

At this point, you have two choices: Either 1. one randomly selected door, or 2. one door among two doors, chosen by the host on the basis of the other not having the prize.

You would have better luck with option 2 because choosing that door is as good as opening two randomly selected doors. That is twice as good as opening one randomly selected door as in option 1.

Comment author: Oscar_Cunningham 25 August 2017 08:58:36AM 0 points [-]

Yeah, I like that.

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