Can you play the acchordion?

A total of $6$ points are chosen on the circle, and are joined in pairs to form three chords. The probability that any two of these points are the same is $0$ , and so the points may be assumed to be distinct. The exact placement of these points does not matter - all that matters is which pairs of points are connected, as below:

If the vertices are labelled $ABCABC$ in anticlockwise order around the circle (as in the middle picture), and then joined up alphabetically, there will be $3$ intersections. If the vertices are labelled $AABCCB$ (as in the right-hand picture), there will be $0$ intersections. Note that configurations where three or more chords are concurrent at the same point (for example, in the $ABCABC$ picture with the six points at the vertices of a regular hexagon) occur with probability $0$ , and so can be ignored. We can make these labellings unique by requiring us to list in the vertices in order, starting from a particular point $X$ on the circle, and labelling the first vertex we meet with $A$ and the first vertex we meet that does not belong to the $A$ chord with $B$ . In the above diagrams, the point $X$ could be situated at about $10$ o'clock, so the appropriate labelling for the left-hand picture is $ABCBCA$ . Again, the probability that any one of the vertices is actually at the point $X$ is $0$ , so we can assume that this labelling scheme is uniquely and unambiguously defined.

Since the six points are chosen independently and uniformly around the circle, the different vertex labellings are equally likely. There are $\frac{6!}{2! \times 2!\times 2!} = 90$ ways of ordering the symbols $A,A,B,B,C,C$ , and these reduce to $15$ sets of $6$ , where the $15$ sets can be classified by assuming that $A$ and $B$ are the first and second distinct letters seen. Given the conventions adopted above for making chord labellings unique, these $15$ labelling types are the only possiblities, and each of the $15$ types is equally likely to occur. Going through these $15$ labelling types, we easily see how many chord intersections each type yields:

$\begin{array}{cccccccc} \mathrm{Type} & \mathrm{Intersections} & \hspace{0.5cm} & \mathrm{Type} & \mathrm{Intersections} & \hspace{0.5cm} & \mathrm{Type} & \mathrm{Intersections} \\ AABBCC & 0 & & AABCBC & 1 & & AABCCB & 0 \\ ABABCC & 1 & & ABACBC & 2 & & ABACCB & 1 \\ ABBACC & 0 & & ABCABC & 3 & & ABCACB & 2 \\ ABBCAC & 1 & & ABCBAC & 2 & & ABCCAB & 1 \\ ABBCCA & 0 & & ABCBCA & 1 & & ABCCBA & 0 \end{array}$ Thus we deduce that $p(0) = \tfrac{5}{15} \hspace{1cm} p(1) = \tfrac{6}{15} \hspace{1cm} p(2) = \tfrac{3}{15} \hspace{1cm} p(3) = \tfrac{1}{15}$ and hence $p(0) + p(1) - p(2) - p(3) = \tfrac{7}{15}$ , making the answer $7+15 = \boxed{22}$ .

The first point of the first chord is assumed to be 0 to simplify the expression some what.

The outermost $\text{Boole}$ function returns 1 if the desired count has been achieved. The five way multiple integral counts the chord intersections.

The inner $\text{Boole}$ functions return 1 if an intersection between the respective chords. There is one $\text{Boole}$ function for each chord pair, of which there are 3 pairs.

The integral is executed for each possible intersection count, from 0 to 3. The results are converted to an Association (hash table).

I added an assistant function:

$\text{intersect}= \\ \ \ \text{Block}[\{p=((\text{\$\#\$2}-\text{\$\#\$1}) \bmod 1),a=((\text{\$\#\$3}-\text{\$\#\$1}) \bmod 1),b=((\text{\$\#\$4}-\text{\$\#\$1}) \bmod 1)\}, \\ \ \ \ \ \text{Boole}[(0\leq a\leq p\land p\leq b\leq 1)\lor (0\leq b\leq p\land p\leq a\leq 1)]]\&;$

$p=\text{Association}\left[\text{Table}\left[n\to \text{Assuming}\left[(\text{\$\theta \$1b}|\text{\$\theta \$2a}|\text{\$\theta \$2b}|\text{\$\theta \$3a}|\text{\$\theta \$3b})\in \mathbb{R}, \\ \ \ \int _0^1\int _0^1\int _0^1\int _0^1\int _0^1 \\ \ \ \ \ \text{Boole}[ \\ \ \ \ \ \ \ \text{intersect}(0,\text{\$\theta \$1b},\text{\$\theta \$2a},\text{\$\theta \$2b})+ \\ \ \ \ \ \ \ \text{intersect}(0,\text{\$\theta \$1b},\text{\$\theta \$3a},\text{\$\theta \$3b})+ \\ \ \ \ \ \ \ \text{intersect}(\text{\$\theta \$2a},\text{\$\theta \$2b},\text{\$\theta \$3a},\text{\$\theta \$3b})=n] \\ \ \ d\text{\$\theta \$1b}\,d\text{\$\theta \$2b}\,d\text{\$\theta \$2a}\,d\text{\$\theta \$3b}\,d\text{\$\theta \$3a}\right], \\ \{n,0,3\}\right]\right] \longrightarrow \\ \text{Association}\left[0\to \frac{1}{3},1\to \frac{2}{5},2\to \frac{1}{5},3\to \frac{1}{15}\right]$

The requested probability formula is computed.

$r=\frac{p(0)+p(1)-p(2)-p(3)}{\text{Total}[p]} \Longrightarrow \frac{7}{15}$

The requested answer is computed.

$\text{Denominator}[r]+\text{Numerator}[r] \Longrightarrow 22$