I was asking myself how lower than P(S|R) P(S|~R) must be in order for the expected value of option R to be (strictly) higher than the expected value of ~R, given the following value assignments to the different outcomes:
~R & S = 10; ~R & ~S = 1; R & S = 9; R & ~S = 0
I'm using the following (standard) formula for calculating the expected value (EV) of an option: EV(R) = P(S|R) × V(S & R) + P(~S|R) x V(~S & R).
I was also wondering if there is a general method for solving this kind of problem.
Many thanks in advance for your help!
As $P(\neg S | R) = 1 - P(S | R)$ and $P(\neg S | \neg R) = 1 - P(S | \neg R)$, we have $$\operatorname{EV}(R) = 9 \cdot P(S | R) + 0 \cdot (1 - P(S | R)) = 9 \cdot P(S|R)$$ and $$\operatorname{EV}(\neg R) = 10 \cdot P(S | \neg R) + 1 \cdot (1 - P(S | \neg R)) = 9 \cdot P(S | \neg R) + 1$$ Now, substituting this expressions into $\operatorname{EV}(R) > \operatorname{EV}(\neg R)$ and solving inequalities, we get $P(S | R) > P(S | \neg R) + \frac{1}{9}$.