Game Theory: Finding the Value of a Relatively Prime Integer Game

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 $1 \cdot \frac{16}{45} + (-1) \cdot \frac{29}{45} = \boxed{-\frac{13}{45}}.$