Bayes' Theorem in Stephen Baxter's Manifold: Time I am currently reading the sci-fi novel Manifold: Time by Stephen Baxter, which contains the following problem.
You are given a box which has either 10 marbles or 1000 marbles. By pressing a lever on the box, one marble is randomly taken out and given to you. You know that there is exactly one red marble.
After pressing the lever three times, you obtain a red marble. The book claims that this information implies, using Bayes' theorem, that the probability that there are 10 marbles in the box is 2/3.
Can anyone explain how this computation actually works out, or at least how one is supposed to set up Bayes' equation to get this answer? Thanks.
 A: Let $N$ be the unknown number of marbles in the box.
The question is ambiguous on whether (i) you press the lever three times and obtain the red marble on one of the three tries, or (ii) you press the lever three times and obtain the red marble only on the third try.
Assuming the latter case, the probability that you get the red marble on the third try is
$$P(k=3|N=n)=\frac{n-1}n\cdot\frac{n-2}{n-1}\cdot\frac1{n-2}=\frac1n.$$
So $P(k=3|N=10) = 1/10$ and $P(k=3|N=1000) = 1/1000$. By Bayes' theorem,
$$\begin{align}
P(N=10|k=3)&=\frac{P(k=3|N=10)P(N=10)}{\sum_n P(k=3|N=n)P(N=n)}\\
&= \frac{\frac1{10}P(N=10)}{\frac1{10}P(N=10) + \frac1{1000}P(N=1000)}.
\end{align}$$
This only turns out to be $\frac23$ if your prior on the box having ten marbles is $P(N=10)=\frac1{51}$.
A: Something doesn't sound complete about this problem the way it is posited. No prior is known. The probability of N=10 if the author says the probability is $\frac{2}{3}$ works out only if the prior is $\frac{1}{51}$ as mentioned above. 
