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

We can solve this problem using probability. There are 9 choices for each player, so there are a total of 81 possible outcomes. We can use complementary probability to find the probability that the two numbers are not relatively prime.

First, we count the number of outcomes where the two numbers share a common factor greater than 1. We can do this by counting the number of outcomes where one of the numbers is even (and therefore has a factor of 2 in common with any other even number) or where one of the numbers is a multiple of 3, 4, 5, 6, 7, 8, 9, or 10. We can use the principle of inclusion-exclusion to count the total number of these outcomes:

• There are 4 choices that are even: 2, 4, 6, and 8.
• There are 3 choices that are multiples of 3: 3, 6, and 9. (We don't need to count 2 and 5 separately, since any multiple of 2 or 5 is also a multiple of 4 or 10, respectively.)
• There are 2 choices that are multiples of 4: 4 and 8.
• There are 2 choices that are multiples of 5: 5 and 10.
• There is 1 choice that is a multiple of 6: 6.
• There is 1 choice that is a multiple of 7: 7.
• There is 1 choice that is a multiple of 9: 9.

Using the principle of inclusion-exclusion, we can count the total number of outcomes where at least one of the numbers is even or a multiple of 3, 4, 5, 6, 7, 8, or 9:

$4+3+2+2+1+1+1=14$

However, we have overcounted the outcomes where one of the numbers is a multiple of both 2 and 3 (i.e. 6). There is only 1 such outcome, so we need to subtract it once from our count. We have also overcounted the outcomes where one of the numbers is a multiple of both 2 and 4 (i.e. 4 or 8), both 2 and 5 (i.e. 10), or both 3 and 9 (i.e. 9). There are no outcomes where one of the numbers is a multiple of both 2 and 4, or both 2 and 5, so we don't need to subtract anything for those cases. There is only 1 outcome where one of the numbers is a multiple of both 3 and 9 (i.e. 9), so we need to subtract it once from our count. Using the principle of inclusion-exclusion again, we get:

$14-1-1=12$

So there are 12 outcomes where the two numbers share a common factor greater than 1. Therefore, there are 81-12=69 outcomes where the two numbers are relatively prime. The probability of winning is the probability that the two numbers are relatively prime, which is:

'69/81 = 23/27'

The value of the game is the expected value of the winnings. If we win with probability '23/27' and win $1, then the expected value of the winnings is:

'23/27 * 1 + 4/27 * 0 = 23/27'

Therefore, the value of the game is '23/27'.