Alan

It won't reach 180ft!

Since they are both on the unit circle, a and b differ only in angle.  Without loss of generality we can set the angle of b to be zero, so that b = 1.

Then \(|\frac{a-b}{1-\bar{a} b}|=|\frac{a-1}{1-\bar{a}}|\)  

If we let  \(a=e^{i\theta}\)  then \(|\frac{a-1}{1-\bar{a}}|=|\frac{e^{i\theta}-1}{1-e^{-i\theta}}|=|e^{i\theta}|=1\)

So, as long as \(\theta \ne 0\)  the ratio is 1

Here's another way of looking at it:

On the webcalc calculator it's as seen below: 

Set x/2 = -8, for the left limit, and x/2 = 8 for the right-hand limit.  Solve for x in both cases.

Here's a more detailed derivation:

You are right Melody (though your list chooses the higher score for the host rather than the lower score - but the principle is the same!). 

I assumed a large number of potential rounds, with the random allocation being different each time.

I agree that there are a possible 32! pairings in total, and that the maximum number of possible matches in a draw hosted by the seeded team is 31.  This is the same as saying the minimum number of matches hosted by the unseeded team in a draw is 1.  Since there are 32 unseeded teams the number of possible draws that result in just 1 unseeded team hosting is 32, so I make the probability p = 32/32! or p = 1/31!.

1.  a*ln(b) = ln(b^a)   so 9*ln(x-3) - 11*ln(x) can be written as  ln(x-3)^9 - ln(x^11)

2. ln(c) - ln(d) = ln(c/d)  so  ln(x-3)^9 - ln(x^11) can be written as  ln( (x-3)^9/x^11 )    i.e. ln( ( x-3)⁹x-¹¹ )

(Edited to correct mistaken sign)

Like so: