2 distinct vertices of an octagon are chosen at random.

Here's my  best  attempt :

Call  the vertices  (A,B,C,D,E,F,G,H)

The  number of possible  combos of choosing any two vertices  (without regard to order)  =  8C2  = 28

Note  that  the  vertices chosen - again disregarding order - where there  is only one  vertex bt\etween them  are :

ABC , BCD, CDE, DEF, EFG, FGH, GHA, HAB   =  

AC, BD, CE, DF, EG, FH , GA, HB

So....the probability =  8/28 =   2/7