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

First, we note that the only way two numbers can be relatively prime is if neither is even (i.e. both are odd) and they are not divisible by the same odd number (e.g. 9 and 15 are not relatively prime because they are both divisible by 3). /n/nThere are a total of 5 odd numbers between 2 and 10 inclusive, so there are $5^2 = 25$ possible pairs of numbers that can be chosen. Of these, 4 pairs will have both numbers even and therefore not relatively prime. /n/nFor each odd number $n$ between 3 and 9 inclusive, there are two odd numbers that are not relatively prime with $n$ (i.e. the odd multiples of $n$ between 3 and 9 inclusive, excluding $n$ itself). So, for each odd number, there are 2 out of the 4 remaining odd numbers that are not relatively prime with it. /n/nTherefore, there are $4 - 2 = 2$ pairs of odd numbers that are not relatively prime, out of a total of 21 pairs of odd numbers. So, the probability of choosing a pair of numbers that are not relatively prime is $/frac{2}{21}$, and the probability of choosing a pair of numbers that are relatively prime is $1 - /frac{2}{21} = /frac{19}{21}$. /n/nThe value of the game is the expected value of the winnings, which is $1 /cdot /frac{19}{21} + 0 /cdot /frac{2}{21} = /boxed{/frac{19}{21}}$.