The probability to hire At least twice is (n-1)/n
.
Assuming you have a random permutation of the candidates, you hire only one candidate if and only if the first candidate is also the best. The probability for it to happen is 1/n
. Thus, the probability that it won't happen (and you will hire twice or more) is 1-1/n = (n-1)/n
To hire exactly twice:
- Choose a location (not first) for the 'best':
n-1
possibilities. Let the place be i - For each such place, choose i-1 candidates to be placed before the best:
Choose(n-1,i-1)
- The first from these i-1 candidates is the best out of them. Need to permute the rest:
(i-2)!
- In addition, still need to permute for (n-i-1) candidates after the 'best': (n-i-1)!.
This gives us the total number of 'valid' permutations that exactly two candidates are hired are:
f(n) = Sum[ Choose(n-1,i-1)*(i-2)!*(n-i-1)! | for i=2,...,n]
And the probability is simply P=f(n)/n!