I think answer is 4.596 assume a,b,c,d,e,f,g,h are at leave level in this order)
0 length =8(self people only) again because of two magical words (independent and uniform). since uniform irrespective of the fact any how probability of choosing two leaves is 1 so my concern is leaves only not internal node , I mean to say i will only choose from leaves not from any where
2 length =3 (a-a,a-b,b-b)(now if i choose c and b(no matter what 2 length is not there ,but 4 length is there(for this case it will be consider ,which i had covered latter)
4 length =10(a-a,a-b,a-c,a-d,b-b,b-c,b-d,------)(4+3+2+1) since unique path i ruled out those case in which a-b=b-a) some one might put up question what if i select b-a path rather then a-b (no matter what it will be unique path only, this is an assumption tell me if i am wrong here)
now expected length(average )=((36*6)+(10*4)+(3*2)+(0*8))/57=4.596~4.60 If I need correction please let me know. May be I have done blunder in this solution.