You and an enemy play the following game each of you writes down an integer between 2 and 10 inclusive If the numbers are relatively prime1 you win $1 Find the value of the game Two positive integers

We can list the possible pairs of integers and their greatest common divisors (gcd) as follows: \begin{align*} (2,2) &: \gcd(2,2) = 2 \ (2,3) &: \gcd(2,3) = 1 \ (2,4) &: \gcd(2,4) = 2 \ (2,5) &: \gcd(2,5) = 1 \ (2,6) &: \gcd(2,6) = 2 \ (2,7) &: \gcd(2,7) = 1 \ (2,8) &: \gcd(2,8) = 2 \ (2,9) &: \gcd(2,9) = 1 \ (2,10) &: \gcd(2,10) = 2 \ (3,3) &: \gcd(3,3) = 3 \ (3,4) &: \gcd(3,4) = 1 \ (3,5) &: \gcd(3,5) = 1 \ (3,6) &: \gcd(3,6) = 3 \ (3,7) &: \gcd(3,7) = 1 \ (3,8) &: \gcd(3,8) = 1 \ (3,9) &: \gcd(3,9) = 3 \ (3,10) &: \gcd(3,10) = 1 \ (4,4) &: \gcd(4,4) = 4 \ (4,5) &: \gcd(4,5) = 1 \ (4,6) &: \gcd(4,6) = 2 \ (4,7) &: \gcd(4,7) = 1 \ (4,8) &: \gcd(4,8) = 4 \ (4,9) &: \gcd(4,9) = 1 \ (4,10) &: \gcd(4,10) = 2 \ (5,5) &: \gcd(5,5) = 5 \ (5,6) &: \gcd(5,6) = 1 \ (5,7) &: \gcd(5,7) = 1 \ (5,8) &: \gcd(5,8) = 1 \ (5,9) &: \gcd(5,9) = 1 \ (5,10) &: \gcd(5,10) = 5 \ (6,6) &: \gcd(6,6) = 6 \ (6,7) &: \gcd(6,7) = 1 \ (6,8) &: \gcd(6,8) = 2 \ (6,9) &: \gcd(6,9) = 3 \ (6,10) &: \gcd(6,10) = 2 \ (7,7) &: \gcd(7,7) = 7 \ (7,8) &: \gcd(7,8) = 1 \ (7,9) &: \gcd(7,9) = 1 \ (7,10) &: \gcd(7,10) = 1 \ (8,8) &: \gcd(8,8) = 8 \ (8,9) &: \gcd(8,9) = 1 \ (8,10) &: \gcd(8,10) = 2 \ (9,9) &: \gcd(9,9) = 9 \ (9,10) &: \gcd(9,10) = 1 \ (10,10) &: \gcd(10,10) = 10 \end{align*} Out of the 45 possible pairs, there are 16 pairs where the gcd is 1, so the probability of winning is $\frac{16}{45}$. Therefore, the value of the game is $$1 \cdot \frac{16}{45} + (-1) \cdot \frac{29}{45} = \boxed{-\frac{13}{45}}.$