Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral ?

Let the six points 1, 2, 3, 4, 5, 6

All convex quadrilateral  are all selection(4 Numbers)  in ascending order from the set 1, 2, 3, 4, 5, 6

$$\rm{convex~quadrilateral } = \bordermatrix{Points& 1 & 2 & 3 & 4 & 5 & 6 \cr
1. &1 &2 &3 &4 & & \cr
2. &1 &2 &3 & &5 & \cr
3. &1 &2 &3 & & &6 \cr
4. &1 &2 & &4 &5 & \cr
5. &1 &2 & &4 & &6 \cr
6. &1 &2 & & &5 &6 \cr
7. &1 & &3 &4 &5 & \cr
8. &1 & &3 &4 & &6 \cr
9. &1 & &3 & &5 &6 \cr
10.&1 & & &4 &5 &6 \cr
11.& &2 &3 &4 &5 & \cr
12.& &2 &3 &4 & &6 \cr
13.& &2 &3 & &5 &6 \cr
14.& &2 & &4 &5 &6 \cr
15.& & &3 &4 &5 &6 \cr} = \binom64$$

Point-connections:

$$\small{\text{$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
&1&2&3&4&5&6 \\
\hline
1&x&1\Rightarrow2&1\Rightarrow3&1\Rightarrow4&1\Rightarrow5&1\Rightarrow6 \\
2& &x&2\Rightarrow3&2\Rightarrow4&2\Rightarrow5&2\Rightarrow6 \\
3& & &x&3\Rightarrow4&3\Rightarrow5&3\Rightarrow6 \\
4& & & &x&4\Rightarrow5&4\Rightarrow6 \\
5& & & & &x&5\Rightarrow6 \\
6& & & & & &x \\
\hline
\end{array}
$}} = \binom62=15$$

Four  chords from 15 pont-connections

  $$=\binom{\binom62}{4}=\binom{15}{4}$$

the probability that the four chords form the sides of a convex quadrilateral

 $$\small{\text{$
\begin{array}{rcl}
&=&\dfrac{\binom64}{\binom{15}{4}}
=\dfrac{ \dfrac{6!}{4!\cdot(6-4)!} }{ \dfrac{15!}{4!\cdot(15-4)!} }
=\dfrac{6!}{2!} \cdot \dfrac{11!}{15!}
=\dfrac{3\cdot 4\cdot 5\cdot 6}{ 12\cdot 13 \cdot 14 \cdot 15}
=\dfrac{36}{3276}
=0.01098901099
\end{array}
$}}$$

I don't know Mellie, your questions are hard.  :))

Fristly any quadrilateral that is formed will be a convec quadrilateral because opposite angels in a cyclic quads are supplementary - so none can be greater than 180 degrees.

First lets see how many diagonals that there are 

From S there are 5, then from A there are 4, then 3 then 2 then 1 = 5+4+3+2+1=15 diagonal

How many ways can we choose 4 of these 15 diagonals.    15C4

 $${\left({\frac{{\mathtt{15}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)} = {\mathtt{1\,365}}$$         ways

I'm going to attempt to list all the cyclic quad possibilities

SABCS,SABDS,SABES, SACDS, SACES, SADES, SBCDS, SBCES, SBDE, SCDES      that is 10

Now  I think I have exhasted all the ones with the point S in them  

ABCDA, ABCEA, ABDEA, ACDEA     that is 4

Now  I think I have exhasted all the ones with the point S or A in them.

BCDEB  that is the only one left    

Total = 10+4+1 = 15

So I get     $${\frac{{\mathtt{15}}}{{\mathtt{1\,365}}}} = {\frac{{\mathtt{1}}}{{\mathtt{91}}}} = {\mathtt{0.010\: \!989\: \!010\: \!989\: \!011}}$$

That is my best guess at the moment.   It might even be correct     

Very nice, Melody and heureka......!!!

This WAS a tough one   !!!!

One observation......  if we laid out ABCDES in a "line"......we only need to choose 4 of any of the 6 letters. Then, putting each resulting set in rotational order will  give us all the possible cyclic quads.

That is how I apprached it and Heureka too I think.

I tried to use the table fuction on the forum for layout.  But it does not work anywhere good enough and it looks like a mess.

I expect Heureka's and mine are very similar but mine is probably riddled with errors. :)