Choose 3 strings and output the exact minimum Hamming distance between any pair Consider \$3\$ binary strings of length \$n\$ chosen independently and uniformly at random.   We are interested in computing the exact expected minimum Hamming distance between any pair. The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
Task
Given an input n between 2 and 10, inclusive, consider \$3\$ binary strings of length \$n\$ chosen independently and uniformly at random. Output the exact expected minimum Hamming distance between any pair.
Output should be given as a fraction.
Your code should run in TIO without timing out.
Pari/GP, 92 90 bytes

2 bytes saved thanks to @Nicolas!

n-&gt;sum(a=0,n,sum(b=0,n-a,sum(c=0,n-a-b,n!/a!/b!/c!/(n-a-b-c)!*min(a+b,min(a+c,b+c)))))/4^n


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$$ f(n) = \frac{1}{4^n} \sum_{\substack{a+b+c+d=n\\a,b,c,d\geq 0}} \binom{n}{a,b,c,d}\min(a+b,a+c,b+c)$$

A more direct brute force approach may be shorter in some languages, but I'm not sure if that's true for Pari. Anyway, it's a shame that Pari's min only accepts two arguments.

I can avoid the min and be about 6 times faster, but the compensating additional code makes it longer. Maybe in another language this is the shorter way. And maybe it is possible to simplify further and get rid of the innermost sum.

n-&gt;sum(a=0,n,sum(b=a,n-a,sum(c=b,n-a-b,n!/a!/b!/c!/(n-a-b-c)!*(a+b)*6/(1+(a==b)+(b==c))!)))/4^n


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