The fair meat bazaar: Who has right? In a local town the girl scouts have been selling tickets, and now it is the time to pick winners. The number of winners is yet to be decided. The names are collected in $3$ books. With respectively


*

*$1000$

*$ 600$

*$ 400$


names in them. 
The question is without computer aid (eg the names are in physical books, too much hazzle to digitalize the results) how can one create a fair draw? 
My solution is as follows (please try before looking)

 Note that all books are divisible by 200. Hence pick $5$ names from book 1, $3$ names from book $2$ and $2$ names from book $3$. Amongst these $10$ names do another draw. Note one apparent problem with this solution is if you want more than $2$ winners. Then the smallest book is at an disadvantage. If you need $n$ winners then draw $\lceil n/2 \rceil \cdot (5,3,2)$ names$...\phantom{..............}\\$ Another solution is to have the names 1-2000 ready, and then draw $n$ names from this list. But in a practical sense, is not my solution easier than writing up $2000$ notes? (Remember no fancy technology allowed, the villages does not believe in this). 

The locals does not agree this is a fair solution. Instead they mean that it should not matter what book you are in. If a winner is picked from book 1 and you are in book 3, this should not affect you. They mean it is fair to draw between the books, then pick a a number in the range 1-1000. I think they are wrong in saying that is a fair draw. 
Who is right? And how can one explain the correct solution in laymans terms?
 A: Certainly a fair solution would be to pick $k$ numbers from $\{1,2,\ldots,2000\}$ and then 
\begin{array}{c|c|c}
\text{number} & \text{book} & \text{person in the book} \\\hline
1-1000 & \mathrm{I} & n \\
1001-1600 & \mathrm{II} & n - 1000 \\
1601-2000 & \mathrm{III} & n - 1600 
\end{array}
That is completely doable using pen and paper, if you don't have a way to pick a random number, throw a coin $11$ times: $2^{11} = 2048$ (if you happen to pick over $2000$ or a person picked previously, just try again; if you don't have a fair coin, throw the coin twice and interpret TH and HT as heads and tails respectively, while disregard any TT or HH as invalid throws).
As for whether you solution is fair: in general it is not. Assume there are to be two winners, and we would like to calculate the probability that some particular pair wins, which is $$\binom{2000}{2}^{-1} = \dfrac{2}{2000\cdot 1999}.$$ 
Observe that $1999$ is a prime number, so whatever calculation method we use, the denominator must has it as a factor. However, in your case the denominator can't have factors bigger than 1000 (unless you put it there artificially), so the prime number $1999$ does not happen, and so these numbers can't be equal.
I hope this helps $\ddot\smile$
