# A riddle about estimating the number of errors in a book

There is a riddle which I heard a while ago and haven't came up with a satisfying approach yet.

Say a book is written by person $$A$$, two other people, $$B$$ and $$C$$ review the text and each of them find 40 and 30 errors respectively and among their founded errors, there are 15 common words. Estimate the true number of errors in the book with this given information.

I know there might be various approaches, so any ideas would be appreciated.

• please add [self-study] tag and tell us what have you tried and where are you stuck.
– Tim
Commented Apr 26, 2021 at 13:40
• The task is impossible: the number of errors that two other people can find does not necessarily indicate the "true number" of errors. Moreover, at best this can be framed as a question of mathematics (answer: between them, $B$ and $C$ found $40+30-15$ errors). It is not a statistical question because we have no information at all about how the number of errors discovered by any reviewer might be related to the total number of errors in the book.
– whuber
Commented Apr 26, 2021 at 15:04
• sounds like this could be solved as a capture-recapture problem en.wikipedia.org/wiki/Mark_and_recapture but instead of capturing animals, you are capturing errors. Commented Apr 27, 2021 at 14:35

I think the idea is that as B has found half the errors that C found, you could estimate that B detects half of all errors. This makes the dubious assumption that whether B finds an error is independent of whether C finds that error. Realistically, B finding an error is evidence that the error is easy to find, which increases the probability that C finds it.

Reference: Probabilities in proofreading, G. Polya https://www.cs.princeton.edu/courses/archive/fall07/cos323/papers/Polya76.pdf

Let the total number of errors be $$N$$.

Then let's assume that reviewers B and C find errors independently, that is to say if C found the error, B is no more likely to find it. (So no given error is any easier/harder to find than another, an unrealistic assumption as fblundun points out)

Then $$P\left(\text{B finds error i}\right) = \frac{30}{N}$$, $$P\left(\text{C finds error i}\right) = \frac{40}{N}$$, and so \begin{align} \frac{15}{N} & = P\left(\text{B and C find error i}\right) \\ & = P\left(\text{B finds error i}\right) \times P\left(\text{C finds error i}\right) \\ & = \frac{30}{N} \times \frac{40}{N} \end{align} And then we can rearrange to find $$N = 80$$ as an estimate.