## LESSWRONGLW

[anonymous]5y0

That's a really good explanation of part of the problem I was getting at. But that requires considering the three hypotheses as a group rather than in isolation from all other hypotheses to calculate 0.33.

But that requires considering the three hypotheses as a group rather than in isolation from all other hypotheses to calculate 0.33.

Not really. A hypothesis's prior probability comes from the total of all of your knowledge; in order to determine that P(HA)=0.33 Lumifer needed the additional facts that there were three possibilities that were all equally likely.

It works just as well if I say that my prior is P(HA)=0.5, without any exhaustive enumeration of the other possibilities. Then evidence E confirms HA if P(HA|E)>P(HA).

(One should be suspicious t... (read more)

1Lumifer5yNo, it does not. Let's say you have a hypothesis HZ. You have a prior for it, say P(HZ) = 0.2 which means that you think that there is a 20% probability that HZ is true and 80% probability that something else is true. Then you see evidence E and it so happens that the posterior for HZ becomes 0.25, so P(HZ|E) = 0.25. This means that evidence E confirmed hypothesis HZ and that statement requires nothing from whatever other hypotheses HA,B,C,D,E,etc. might there be.

# Open thread, Dec. 21 - Dec. 27, 2015

by MrMind 1 min read21st Dec 2015233 comments

# 2

If it's worth saying, but not worth its own post (even in Discussion), then it goes here.

Notes for future OT posters: