Intuitive/Visual solution for: There are $k$ balls in a bowl. How many draws does it on average take, to get one specific ball?

Let $E$ denote the result.

Consider the first trial. You win with probability $\frac 1k$. With probability $\frac {k-1}k$ you start over. It follows that $$E=\frac 1k\times 1 +\frac {k-1}k\times \left(E+1\right)\implies E=k$$

You are looking for the expected number of draws, if assuming you only draw repeatedly until you get the right ball.

This is given by: $$ E[\#\text{draws}]=\sum_{i=0}^\infty i P(i) $$

With $P(i)$ the probability of taking $i$ draws. Computing for some cases:

$$ P(1) = 1/k $$ $$ P(2) = ((k-1)/k)(1/k) $$

$$ P(3) = ((k-1)/k)^2(1/k) $$ ...

$$ P(i) = ((k-1)/k)^{(i-1)}(1/k) $$

Thus:

$$ E[\#\text{draws}]=\sum_{i=1}^\infty i ((k-1)/k)^{(i-1)}(1/k)=(1/k)\sum_{i=1}^\infty i ((k-1)/k)^{(i-1)} $$

Which is an Arithmetic-Geometric sum. The partial sum of such sequence is given by:

$$ S_n =\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r)^2} $$

Then for a ratio $r$ such that $|r|<1$, the series converges to:

$$ S =\frac{a}{1-r}+\frac{dr}{(1-r)^2} $$

In the current case:

• $r= (k-1)/k$ so $(1-r)=1/k$

• $a=1$

• $d=1$

So the arithmetic geometric sum is given by: $$ S =\frac{1}{(1/k)}+\frac{(k-1)/(k)}{(1/k)^2}=k+k(k-1)=k^2 $$

Going back to our previous summation: $$ E[\#\text{draws}]=(1/k)\sum_{i=1}^\infty i ((k-1)/k)^{(i-1)}=(1/k)\left(k^2 \right)=k $$