Charlie Steiner comments on No Good Logical Conditional Probability

Charlie Steiner 4 Jun 2015 19:28 UTC
0 points
Is the idea that the proof necessary to use 1 is of infinite length, and you want your logic to be finitary? Hm. This seems odd, because $P (s | T_{\infty})$ is in some sense already a function with an infinitely long argument. How do you feel about using 2 in the form of $P (T_{\infty} | T_{0}) = P (T_{n} | T_{0}) \cdot [s o m e p r o b a b i l i t y]$ , therefore $P (T_{\infty} | T_{0}) \leq P (T_{n} | T_{0})$ , which has the same amount of argument as $P (s | T_{\infty})$ ? I’m confused about at least one thing.

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