Wordy Bayesian Question Five million boys below the age of five live in Erewon. The priests of Erewon are sure that one of them (chosen by fate at random) embodies the spirit of Captain Coin Tosser, who can predict heads or tails of a flipped fair coin with perfect accuracy. In order to find this boy, they travel the country giving a test: each boy being tested is asked to predict which way a flipped coin will land, eighteen times.
After months of grueling search, they find a boy who passes their test, correctly predicting all eighteen tosses. Taking the priests' belief as a given, what is the probability that this boy is in fact the embodiment of the spirit of Captain Coin Tosser?  
So:
$$\mathbb{P}(C) = \frac{1}{5,000,00},\;    \mathbb{P}(T)= \frac{1}{(0.5)^{18}}$$
And we want $\mathbb{P}(C\mid T) = \frac{\mathbb{P}(T\mid C)\mathbb{P}(C)}{\mathbb{P}(T)}$ but we don't have $\mathbb{P}(T\mid C)$ from what I can see.
 A: The calculation we want is as follows:
$$
\mathbb{P}(C \mid T) = 
\frac{\mathbb{P}(T\mid C)\mathbb{P}(C)}
{\mathbb{P}(T\mid C)\mathbb{P}(C) + \mathbb{P}(T\mid \text{not }C)\mathbb{P}(\text{not }C)}
$$
We have 


*

*$\mathbb{P}(T\mid C) = 1$

*$\mathbb{P}(T\mid \text{not }C) = \frac{1}{2^{18}}$

*$\mathbb{P}(C) = \frac{1}{5,000,000}$

*$\mathbb{P}(\text{not }C) = 1 - \frac{1}{5,000,000}$


For the sake of efficient computation, we can rewrite that as 
$$
\mathbb{P}(C \mid T) = \left(1 + \frac{\mathbb{P}(T\mid \text{not }C)\mathbb{P}(\text{not }C)}{\mathbb{P}(T\mid C)\mathbb{P}(C)} \right)^{-1}
$$
You should get a probability of approximately $5\%$.
A: (By the way, the symbol "$¬$" is used to mean "not".)
You said that $\mathbb{P}(T)$ is $\frac{1}{0.5^{18}}$. First off we should correct this to just $0.5^{18}$ (which is the same as $\frac{1}{2^{18}}$), after all the probability is supposed to be less than one. But also, this isn't really $\mathbb{P}(T)$, it's actually $\mathbb{P}(T|¬C)$ the probability that the boy would get it right if he wasn't Captain Coin Tosser. Is he was Captain Coin Tosser then the probability he would get it right is 100%, so $\mathbb{P}(T|C)=1$.
From this we can calculate the value of $\mathbb{P}(T)$ by taking a combination of $\mathbb{P}(T|C)$ and $\mathbb{P}(T|¬C)$.
$$\mathbb{P}(T)=\mathbb{P}(C)\mathbb{P}(T|C)+\mathbb{P}(¬C)\mathbb{P}(T|¬C)$$
This formula is called the "Law of Total Probability". And now I think you have all the factors that you need to use Bayes' theorem.
(To check your answer you can do a quick approximation as follows: $1/2^{18}$ is $1/262144$, or roughly $1/250000$. This is the probability that a boy would get it right by chance. They test $5000000$ boys, so about $5000000/250000$ will get it right by chance. This gives 20, or 21 when we include Captain Coin Tosser. So $21$ boys will get it right and only one is Captain Coin Tosser, so the answer is about $1/21$.)
