So, if I am correct in assuming "relatively prime" simply means that one number is not evenly divisible by the other, then here are my initial thoughts:

You have a 1/n chance of first drawing a "1"...in which case, you will win automatically (1/1 chance)

You have a 1/n chance of first drawing a "2"...in which case, you will win if the second number is even OR a factor of 2 (aka: 1)

You have a 1/n chance of first drawing a "3"...in which case, you will win if the other number is a multiple of 3 OR a factor of 3 (aka: 1)

You have a 1/n chance of first drawing a "4"...in which case, you will win if the other number is a multiple of 4 OR a factor of 4 (aka: 1 or 2)

...

You have a 1/n chance of drawing "x"...in which case, you will win if the other number is a multiple of x OR a factor of x

So let FACT(x) = number of factors of number x...we can then write our chance of winning to be:

SUM (1/n * (1/x + FACT(X)/n)), x=1..n

Which clearly just get worse and worse the more balls there are in the bin...in which case, I would only take the bet if there were visibly "few" (less than 20 or so) balls in the bin.

Now, I suppose my above reasoning only holds if we assume all the numbers 1-n are in the bin...if you have completely random/arbitrary numbers...then, my reasoning probably completely falls through. Again, just my first thoughts. I am also curious to see what others come up with.