How can I argue without people online and not come out feeling bad?

3 wantstojoinin 01 June 2012 02:24PM

I wont be the only one here who "wastes time" arguing about things they care about online (note: I am referring to web forums and things like subreddits, I am not including Less Wrong whose dynamic is completely different). It seems like something that is worth optimising in some direction.

The theory behind it is that one should expose themselves to counter-arguments allowing their claims to be attacked so they that have a chance to substantiate them or reject them upon realising they are mistaken.

In practice they generally follow a pattern that starts with people pointing out what they believe are mistakes then ignoring or intentionally misunderstanding the other party when he refutes or backs up claims.. and ends up with insults, patronising sarcastic remarks and nobody changing their mind about anything.

I don't particularly care about changing other peoples minds to make them agree with me (well, it would be great but I think it's practically impossible) so one thing I would like is for both people to at least end up feeling good.

So I'm interested in three things: Do other LWers recognize this pattern now that I have mentioned it? What decision did those that were already aware of it make, in order to optimise this activity?

Comment author: wantstojoinin 09 May 2012 09:53:32AM *  3 points [-]

Why isn't building a decision theory equivalent to building a whole AI from scratch?

Comment author: gRR 30 April 2012 10:34:09AM *  1 point [-]

n%3=0 is distinguishable from n%3=1∨n%3=2. If A="n%3=0", B="n%3=1", and C="n%3=2", then an isomorphism f that maps B∨C to A must satisfy f(B∨C) = f(B)∨f(C) = A, which is impossible.

Comment author: wantstojoinin 01 May 2012 02:56:44AM 0 points [-]

I understand, what I wrote was wrong. What if we use n%3=0 and ~(n%3=0) though?

Comment author: wantstojoinin 30 April 2012 09:06:33AM *  0 points [-]

A natural number n can be even or odd: i.e. n%2=0 or n%2=1.

If X = {n is natural number} then you showed that we can use P(n%2=0|X) + P(n%2=1|X) = 1 and P(n%2=0|X) = P(n%2=1|X) together to get P(n%2=0|X) = 1/2.

The same logic works for the three statements n%3=0,n%3=1,n%3=2 to give us P(n%3=0|X) = P(n%3=1|X) = P(n%3=2|X) = 1/3.

But then the same logic also works for the two indistinguishable statements n%3=0,n%3=1 \/ n%3=2 to give us P(n%3=0|X) = P(n%3=1 \/ n%3=2) = 1/2.

But 1/2 = 1/3 is a contradiction, so we find that axiom 3 leads to inconsistencies.

Comment author: wantstojoinin 17 March 2012 10:07:37PM *  0 points [-]

Isn't it just strategy stealing? Calling it tit-for-tat maybe focuses away from the fundamental reason why it wins.

Comment author: wantstojoinin 11 March 2012 12:19:50PM *  3 points [-]

I'd like to ask him for an explanation of what the hard problem is and why it's an actual problem, in a way that I can understand it (without reference to undefinable things like "qualia" or "subjective experience"). Would probably have to discuss it in person with him and even then doubt either of us would get anywhere though.