TITLE: P(tomorrow is the end of the world) =?
QUESTION [12 upvotes]: Here we have $3$ prophets alpha, beta, gamma, they all predict that tomorrow is the end of the world. It's known that the accuracy of alpha's and beta's prediction is $90\%$, while that of gamma's is $4\%$. What is the probability that tomorrow is the end of the world?

Here's my calculations.
Let $A, B, C$ be alpha's, beta's and gamma's predictions.
P(tomorrow is the end of the world)
$= P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$
$= 0.9 + 0.9 + 0.04 - 0.81 - 0.036 - 0.036 + 0.0324$
$= 0.9904$
But this seems a bit too high, is my calculation correct? Thanks.

REPLY [6 votes]: To me, the most sensible way to look at the probability that a prophet's prediction is true, is to look at the ratio of times when their predictions come true over the times when their predictions don't come true:
$$
\Lambda(A)=\frac{P(A\mid X)}{P(A\mid\neg X)}
$$
Extending the ideas used in this answer, and assuming that the prophets are independent, we get that
$$
O(W\mid A\cap B\cap C)=\Lambda(A)\,\Lambda(B)\,\Lambda(C)\,O(W)
$$
Where $O(X)$ is the odds of $X$ defined as
$$
O(X)=\frac{P(X)}{1-P(X)}
$$
We are given that
$$
\begin{align}
\Lambda(A)=\frac{.9}{.1}&=9\\
\Lambda(B)=\frac{.9}{.1}&=9\\
\Lambda(C)=\frac{.04}{.96}&=\frac1{24}\\
\end{align}
$$
Thus, adding the information from the prophets would increase the odds by a factor of $\frac{81}{24}$:
$$
O(W\mid A\cap B\cap C)=\frac{27}{8}O(W)
$$

This matches Robert Israel's answer if we convert from odds to probabilities:
$$
\begin{align}
P(W\mid A\cap B\cap C)
&=\frac{\frac{27}{8}\frac{P(W)}{1-P(W)}}{\frac{27}{8}\frac{P(W)}{1-P(W)}+1}\\
&=\frac{27P(W)}{27P(W)+8(1-P(W))}\\
&=\frac{.0324P(W)}{.0324P(W)+.0096(1-P(W))}
\end{align}
$$