Assume we have an urn with $n$ different balls and we draw three times with replacement. What is the probability to draw the same ball at least twice?
My thoughts to this are:
There are $n^3$ possibilities to draw $3$ balls independently from an urn.
There are $n$ possibilities to draw the same ball exactly $3$ times.
There are $n\cdot(n-1)\cdot2+n\cdot1\cdot(n-1)=3\cdot n\cdot (n-1)$ possibilities to draw the same ball exactly two times, because:
(First case: Draw a random ball, then a different one, then the first one or the second one again. Second case: Draw a random ball, then the same again, then a different one)
So we get $$\frac{3\cdot n\cdot (n-1)+n}{n^3}=\frac{3}{n}-\frac{2}{n^2}$$ for the sought probability.
Now I feel like this is too complicated and one can probably find the solution pretty easily with a much better abstraction. Thank you!
The probability that the three balls are distinct is $$1\times \frac {n-1}n\times \frac {n-2}n=\frac {n^2-3n+2}{n^2}=1-\frac 3n+\frac 2{n^2}$$
Your desired result follows immediately.