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While doing the exercises in Grinstead & Snell's Introduction to Probability I came across this question:

11) There are n applicants for the director of computing. The applicants are interviewed independently by each member of the three-person search committee and ranked from 1 to n. A candidate will be hired if he or she is ranked first by at least two of the three interviewers. Find the probability that a candidate will be accepted if the members of the committee really have no ability at all to judge the candidates and just rank the candidates randomly. In particular, compare this probability for the case of three candidates and the case of ten candidates.

I know of another way of doing this exercise, but I was wondering what is wrong with the following answer:

Let's assume that a candidate $k$ will be chosen for hire. We know that he has to be in the first place of at least two rankings so we can randomly place all of the other candidates in the first two rankings in $(n-1)!$ ways. We don't care if he is the first in the third ranking so we can just set it randomly in $n!$ ways. Afterwards we can assign the rankings to committee members in 3 ways.

Simplifying the probability should be:

$$3\frac{(n-1)!(n-1)!n!}{n!n!n!} = \frac{3}{n^2}$$

However the correct answer to the question is: $\frac{3n-2}{n^3}$

Why the above doesn't work?

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Your probability overcounts the case where all three raters rank the candidate first.

If the raters are called $A, B, C$, then you are regarding $(A,B)$ ranking Candidate $k$ as first, then $C$ ranking the same candidate first, as distinct from the outcome where $(A,C)$ ranks Candidate $k$ first, then $B$ ranks $k$ first. That is why your probability exceeds the true probability by exactly $2/n^3$.

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  • $\begingroup$ This is what I initially thought of, but I can't intuitively get that it introduces exactly $2/n^3$ additional probability. Could You please explain? $\endgroup$ – wesolyromek Jul 22 '15 at 18:27

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