Suppose the 8 seats are labeled
A,B,C,D,E,F,G,H
with
1. seat A opposite seat E,
2. seat B opposite seat F,
3. seat C opposite seat G,
4. seat D opposite seat H
The two people can choose a pair of seats in any of C(8,2) = 28 ways.
They can choose a pair of opposite seats in only 4 ways.
So the probability is 4 out of 28 ways or
which reduces to
.
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Explanation:
That's these four choices of ways to sit opposite each other:
{A,E} {B,F} (C,G}, {D,H}
out of these 28 ways to sit opposite or not:
{A,B), {A,C}, {A,D}, {A,E}, {A,F}, {A,G}, {A,H}
{B,C), {B,D}, {B,E}, {B,F}, {B,G}, {B,H}, {C,D}
{C,E), {C,F}, {C,G}, {C,H}, {D,E}, {D,F}, {D,G}
{D,H), {E,F}, {E,G}, {E,H}, {F,G}, {F,H}, {G,H}
Edwin