Q: Considering 2 slot machines. One machine has a 10% chance of paying out, the other has a 20% chance. One machine is red, the other green. You don't know which pays more. Initially, you're 50% sure that the red machine pays more. You try it once, and it doesn't pay out.
After this first attempt, how should you update the estimate that the red machine has a higher chance of paying out?
So, I did this, but I'm not sure if it's right. Any clarification appreciated!
$P(Red\ Pays\ More|No\ Pay\ Out) = P(No\ Pay\ Out|Red\ Pays\ More)P(Red\ Pays\ More)/P(No\ Pay\ Out)$
I assumed the $P(No\ Pay\ Out|Red\ Pays\ More) = 0.8$ (because there is a 0.8 chance of not being paid if the red machine is the one that pays more).
$P(Red\ Pays\ More) = 0.5$
$P(No\ Pay\ Out) = 0.85$
Yes. That is a-okay.
Let $R$ be the event that red pays more. Let $N$ be the event that the red machine doesn't pay out.
That is: $\mathsf P(N\mid R)= 1-0.20, \mathsf P(N\mid R^\complement)=1-0.10, \mathsf P(R)=0.50$
So: $\mathsf P(R\mid N) ~{= \dfrac{\mathsf P(R)\mathsf P(N\mid R)}{\mathsf P(R)\mathsf P(N\mid R)+\mathsf P(R^\complement)\mathsf P(N\mid R^\complement)} \\ = \dfrac{0.50(1-0.20)}{0.50(1-0.20+1-0.10)}\\=\dfrac {0.40}{0.85} \\= 8/17\\\approx 0.47{\small 0}}$
And likewise : $\mathsf P(R\mid N^\complement) ~{= \dfrac{\mathsf P(R)\mathsf P(N^\complement\mid R)}{\mathsf P(R)\mathsf P(N^\complement\mid R)+\mathsf P(R^\complement)\mathsf P(N^\complement\mid R^\complement)} \\ = \dfrac{0.50(0.20)}{0.50(0.20+0.10)}\\=\dfrac {0.10}{0.15} \\= 1/3\\\approx 0.33{\small 3}}$