Occam's Razor is non-Bayesian? Correct me if I'm wrong, but I thought it falls naturally out of Bayesian model comparison, from the normalization factors, or "Occam factors." As I remember, the argument is something like: given two models with independent parameters {A} and {A,B}, the P(AB model) \propto P(AB are correct) and P(A model) \propto P(A is correct). Then P(AB model) <= P(A model).
Even if the argument is wrong, I think the result ends up being that more plausible models tend to have fewer independent parameters.
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How does deciding one model is true give you more information? Did you mean "If a model allows you to make more predictions about future observations, then it is a priori less likely?"
Let's assume a strong version of Bayesianism, which entails the maximum entropy principle. So our belief is the one that has the maximum entropy, among those consistent with our prior information. If we now add the information that some model is true, this generally invalidate our previous belief, making the new maximum-entropy belief one of lower entropy. The reduction in entropy is the amount of information you gain by learning the model. In a way, this is a cost we pay for "narrowing" our belief.
The upside of it is that it tells us something useful about the future. Of course, not all information regarding the world is relevant for future observations. The part that doesn't help control our anticipation is failing to pay rent, and should be evacuated. The part that does inform us about the future may be useful enough to be worth the cost we pay in taking in new information.
I'll expand on all of this in my sequence on reinforcement learning.