The prior odds were a hundred-to-one against Voldemort surviving. What is the probability of Voldemort being alive? For educational purpose, I came across a probability problem and I can't make sense of it. Can someone put me in right direction on how I can solve this.
Dumbledore is worried if Voldemort is still alive because he noticed a suspicious mark on a hand.
If Voldemort dies, the mark continuous to exist with a 20% probability.
On the other hand, if Voldemort lives on, the mark will stay with 100% chance.
The prior odds were a hundred-to-one against Voldemort surviving. What is the probability of Voldemort  being alive??
 A: Studies have shown that people deal with proportions better than probabilities. Rewrite the question as

There 3030 fanfics. There are 100 times as many fanfics where Voldemort died as fanfics where Voldermort lived. In 100% of fanfics where Voldemort lived, the mark remained. In 20% of fanfics where Voldemort died, the mark remained. Of all the fanfics where the mark remained, in what percentage did Voldemort survive?"

A: CORRECTION: The old solution i gave is assuming that the prior odds were referring to the conditional probability of knowing the mark exists and you are calculating the overall probability of being alive. However if that is not the case and $p(A) = \frac{1}{101}$ then you can use:
$$p(A|M) = \frac{p(M|A)\times p(A)}{p(M|A)\times p(A) + p(M|D)\times p(D)}
= \frac{\frac{1}{101}}{1\times\frac{1}{101} + \frac{1}{5}\times\frac{100}{101}} = \frac{1}{1+\frac{100}{5}} = \frac{1}{21}$$
Or 20:1 odds.
OLD SOLUTION:
$$A = Alive, D= Dead, M = Mark$$
Currently we know that Voldemort has the mark (because Dumbledore noticed it). Therefore the prior becomes what we know which is the 1:100 odds.
$$p(A|M) = \frac{1}{101}$$
$$p(D|M) = \frac{100}{101}$$
We also learned the probability of the mark given that he is alive or dead:
$$p(M|A) = 1$$
$$p(M|D) = \frac{1}{5}$$
Using Bayes theorem and the fact that $p(D) = 1-p(A)$ we can solve the equation.
$$p(M|A) = \frac{p(A|M)\times p(M)}{p(A)}$$
$$p(M|D) = \frac{p(D|M)\times p(M)}{1-p(A)}$$
Keeping the fractions actually makes the math easier to do by hand. Solving for p(A) using these equations gives (Hopefully i did my math right but im getting):
$$p(a) = \frac{1}{501}$$
Which is 500:1 odds.
