"[@2] I think there should b..."


by Tim Egan May 13 2016

Eliezer Yudkowsky I think there should be a small change here? Variable f becomes x and back to f, and I believe it should just be f?

Or, writing out the whole operation from scratch:

$~$\mathbb P(f\mid e\!=\!\textbf {THT}) = \dfrac{\mathcal L(e\!=\!\textbf{THT}\mid f) \cdot \mathbb P(f)}{\mathbb P(e\!=\!\textbf {THT})} = **\dfrac{(1 - x) \cdot x \cdot (1 - x) \cdot 1}{\int_0^1 (1 - x) \cdot x \cdot (1 - x) \cdot 1 \** \operatorname{d}\!f} = 12 \cdot f(1 - f)^2$~$