## LESSWRONGLW

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

No, 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.

[anonymous]5y0

How would you calculate that prior of 0.2? In my original example, my prior was 1, and then you transformed it into 0.33 by dividing by the number of possible hypotheses. You wouldn't be able to do that without taking the other two possibilities into account. As I said, the issue can be corrected for if the number of hypotheses is known, but not if the number of possibilities is unknown. However, frequently philosophical theories of bayesian confirmation theory don't consider this problem. From this paper by Morey, Romeijn, and Rouder:

Overconfident

# 2

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

Notes for future OT posters: