Two diagonals of a regular nonagon (a $9$-sided polygon) are chosen. What is the probability that their intersection lies inside the nonagon? Two diagonals of a regular nonagon (a $9$-sided polygon) are chosen. What is the probability that their intersection lies inside the nonagon? 
This is similar to a previous problem that I posted about a dodecahedron. However, this provides more difficult, as a nonagon is  (in my opinion) more difficult to work with than a dodecahedron. A dodecahedron, in every sense of the object, is even (even edges, even vertices...). This cannot be said for the $2D$ Nonagon. 
Here is what I'm thinking. Draw a nonagon. Find all of the intersection points of the nonagon. A quick glance at online diagram of a nonagon (with diagonals drawn) shows that this method would fall flat quickly. Finding all of the intersection points can be found with $\binom{9}{4}=126$. This means that $126$ is the numerator. If only I could find the denominator....
Help is greatly appreciated!
 A: There are $27$ diagonals in a convex nonagon $P$, and you can choose $2$ of them in ${27\choose 2}=351$ ways. In order to intersect in the interior of $P$ the two diagonals have to have different endpoints, a total of four. Conversely any four vertices of $P$ determine exactly one pair of diagonals intersecting in the interior of $P$, hence there are ${9\choose4}=126$ such pairs. The probability in question therefore is ${126\over351}={14\over39}$.
A: There are $\frac{9\cdot(9-3)}{2}=27$ diagonals in a nonagon. There are 27C2 ways to choose two diagonals, which is $351$. This is our denominator.
The only way the diagonals can intersect inside the nonagon is if they share an endpoint. For each diagonal, there are $5$ other diagonals that share one endpoint, and 5 that share the other for a total of $10$ ways for a certain diagonal to share an endpoint with another. $27$ diagonals means $\frac{10\cdot27}{2}=135$ ways to have adjacent diagonals. The probability for the diagonals to intersect is $\frac{135}{351}=\frac{5}{13}$.
Taking the complement of this gives us an answer of $\frac{8}{13}$.
The question appears to be somewhat vague in the phrase, "their [two diagonals'] intersection"; two diagonals of a nonagon do not necessarily have to intersect.
If the question means to ask for the probability of two uniformly randomly chosen diagonals intersecting inside (but not on) the nonagon, see Christian Blatter's answer.
This answer answers a different question than what is presumably being asked.
A: Call the points $1$ through $9$ clockwise.  The first point can be $1$ by symmetry and the second can be $3,4,5$.  If it is $3$ the diagonals cross if $2$ is a point of the other diagonal, which is $6$ of the $26$ choices.  If the second point is $4$ the diagonals cross if one point is $2,3$ and the other is $5-9$, which is $10$ of the $26$.  If the second point is $5$ the diagonals cross if one is $2-4$ and the other is $6-9$, which is $12$ cases of $26$.  The overall probability is $\frac 13\cdot \frac {6+10+12}{26}=\frac {28}{78}=\frac {14}{39}$ 
This agrees with your $\frac {126}{351}$, which is correct.
