Question about the expected number of sampling until one item out of k is found )
I have k items (let's say balls) that all have a different color, but I do  not know the color until I pick them. I would like to know how many balls I have to pick until I get one with a specific color.
My line of thought is as follows.
For one item, I will need 1 pick in expectation.
For two items, I will need 1/2*1+1/2*2=1.5 picks in expectation.
For three items, I will need 1/3*1+1/3*1/2*2+1/3*1/2*3 = 2 picks in expectation.
For four items, I will need 1/4*1+3/4*1/3*2+3/4*2/3*1/2*3+3/4*2/3*1/2*4 = 2.5 picks in expectation.
My feeling is that the general formula is (1/k)(k(k+1)/2) = (1/k)*(k^2+k)/2 = (k+1)/2.
Do you think that this is correct? Is there maybe a common english name for this formula where I could read more about it (english is not my first language, sorry :-( )
Thank you!

Hint:
Think of the balls as if they are randomly placed in a line. After that you start picking from left to right. The specific ball can take place $1,2,\dots,k$ and every spot has equal probability to be possessed by that ball. What is its expected position?
