Having been preparing for my upcoming exam, I have stumbled upon a seemingly simple problem:
A smallest enclosing circle for a set of points P, |P| > 2 is determined by a subset of exactly 3 points of P. What is the probability of chosing exactly those points when picking 3 points at random with equal probability.
My first train of thought was to see it as a modified "urn problem": For the first pick, I have all 3 points to chose from, i.e. the probability is $\frac{3}{n}$. For the second pick, you have 2 "good" points left and, i.e. probability $\frac{2}{n}$. Lastly, you are left with 1 "good" points, so $\frac{1}{n}$. This gives me $\frac{6}{n^3}$, which does not match the solution our professor had.
The way she did it, the problem was abstracted into "chosing a subset of size 3 of n points at random with equal probability", i.e. $\frac{1}{\binom{n}{3}}$.
My question: Why is my first idea incorrect? I have solved a seemingly similar problem with my first train of thought:
You have a bag full of colourful crayons, each coloured with one of n colours. If you draw m crayons out of the bag uniformly at random (without replacement), what is the probability of all of them being of the same colour?
All I did there was multiply: $\frac{m}{n} * \frac{m-1}{n-1} * ... \frac{1}{n- (m-1)}$.
As the problems seemed so similar, I expected that leaving the denominator constant would give the solution for the first problem...
What am I missing here?
Thank you!
Felix
For the first choice for the triangle you indeed have $n$ points to choose from. But then you have taken a point away, and the second choice is from $n-1$ points (you don't want to pick the same point twice).