Blinking Monty Hall solution

Question

http://allendowney.blogspot.com/2011/10/blinky-monty-problem.html

If we let the prior state be $P(A)=1/3,P(C)=2/3$ and the evidence be the blinking, isn't $$P(E)=P(E|A)P(A)+P(E|C)P(C)=3/5*1/3+3/13*2/3=23/65$$ so $$P(A|E)=\frac{P(E|A)P(A)}{P(E)}=\frac{3/5*1/3}{23/65}=13/23=0.565$$?

But apparently this is wrong because the website says $P(A|E)=0.51$.

BTW I did change the evidence from the webpage to exclude the choosing of door B, but conditioned everything to reflect that so I don't see why it shouldn't give the same answer.

Answer 1

Here, the evidence $E$ is that he opens door B and blinks. Assuming they are independent, $$P(E|A)=P(\text{opens door B}|A)P(\text{blinks}|A)=\frac12\times\frac35=\frac3{10}$$ This is where your calculation was incorrect.

EDIT:

Actually, I should have continued with the calcuation.

$$P(E|A)=P(\text{opens door B and blinks}|\text{car behind A})=\frac35\frac12=\frac3{10}\\P(E|C)=P(\text{opens door B and blinks}|\text{car behind C})=\frac3{13}\frac11=\frac3{13}\\P(A|E)=\frac{\frac3{10}\frac13}{\frac3{10}\frac13+\frac3{13}\frac23}=\frac{13}{23}=0.565...$$ which is exactly as you got.

So it appears your link indeed has it wrong.