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I was wondering... If two people do multiple choice test, and the questions are taken from a pool of question (the size of the pool is unknown). Both people must be given 30 questions. They both end up with 15 of the same questions (and the other 15 different). What is the most likely size of the pool?

This question is not from a book or anything, it is a question that I'm just wondering about.

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+1 for playing around with math. –  Ross Millikan Jun 17 '13 at 3:09

2 Answers 2

up vote 4 down vote accepted

Let the number of questions be $M$. We make two draws of $30$ without replacement, but allowing replacement between the draws and ask the probability of overlap of $15$. The naive answer is that since half the questions overlap, we are seeing half the questions, so $M=60$.
We now prove this (under one large assumption). This probability is $\frac {{30 \choose 15}{M-30 \choose 15}}{M \choose 30}$ because we make the first draw, then choose $15$ questions of the first $30$ and choose $15$ of the rest. In a classic reversal (the probability of $M$ given $15$ overlap is the probability of $15$ overlap given $M$-this assumes that the probability distribution of $M$ is uniform, but there is no uniform distribution over all the naturals), we look for the maximum of this to determine $M$. It is not surprising that this is maximized near $M=60$ in this Alpha graph. If we imagine a change of $M$ from $60$, going upward by $1$ we multiply the probability by $\frac {\frac {61}{46}}{\frac {61}{31}} \lt 1$. Similarly if we go down by $1$ we multiply by $\frac {\frac {30}{15}}{\frac {60}{30}}=1$ and exact computation shows tha $M=59$ is equally probable, but then it falls off.

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There are $K$ questions chosen from a pool of size $N$ for one questionnaire. This defines a set of "successes" for a selection without replacement for the next questionnaire.

Sampling without replacement is a Hypergeometric distribution, so the chance of getting $k$ matches from a sample of size $s$ is:

$$P_N(X=k)=\frac{{{K}\choose {k}}{{N-K}\choose{K-k}}}{{N}\choose{K}}$$

I will go no further because I am on the same track as @RossMilikan

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