Question 394879
Sorry, I had misread the minus as a plus and nearly thought that the probability was {{{(n^2 - 3n + 2)/(3(n^2  -3n  +2))}}} which simplifies to 1/3 (and doesn't make sense). I have revised my solution here, but didn't have enough time to finish, as there is probably going to be a ton of algebra. However I have got you started, you can finish the algebra from then on.




I'm quite sure that the best way is to fix one of the n points {{{a[1]}}}, {{{a[2]}}}, ..., {{{a[n]}}} at {{{a[1]}}}. We go case by case on what the second point of diagonal {{{D[1]}}} is.


Case 1: {{{D[1]}}} includes points {{{a[1]}}} and {{{a[3]}}}.


Then, there is 1 point between {{{a[1]}}} and {{{a[3]}}} and n-3 points from {{{a[3]}}} to {{{a[1]}}} (counting all the numbers 4, 5, ..., n inclusive).


We can generalize this to say:


If {{{D[1]}}} comprises of points {{{a[1]}}}, {{{a[i]}}}, then there are {{{(n-i)(i-2)}}} possible diagonals. Given this, we can sum them up from i = 3 to i = n-1. If P is the total number of diagonals, then


{{{P = sum((n-i)(i-2), i = 3, n-1) = sum(-i^2 + (n+2)i - 2n, i = 3, n-1) = sum (-i^2 + (n+2)i - 2n, i = 1, n-1) - (1-n)}}} (subtracting the case where n = 1, 2)


After this, use the sum identities {{{sum(i^2, i = 1, k) = k(k+1)(2k+1)/6}}} and {{{sum(i, i = 1, k) = k(k+1)/2}}}. After that, it's pretty much algebra bashing time to find an expression for P.


Suppose Q is the number of sets of two diagonals. Be careful that one of the diagonals of Q must have an endpoint at {{{a[1]}}} (due to our previous assumption).


The probability that two chosen diagonals intersect is then P/Q, which should turn out to the expected value (however it's a whole lot of algebra and simplifying from here on).
 




The other solution I considered involved mathematical induction. Basically, you show the base case (n=4, trivial), and show that n = k implies n = k+1. However, I had to split this problem into two cases, one case where the diagonals in the (n+1)-gon were contained within the n-gon, and the other case where one diagonal contains the (n+1)th point. However this solution is probably a bit longer than this one.