Iteratively replacing $3$ chocolates in a box of $10$ 
In the fridge there is a box containing 10 expensive high quality Belgian chocolates, which my mum keeps for visitors. Every day, when mum leaves home for work, I secretly pick 3 chocolates at random, I eat them and replace them with ordinary cheap ones, that have exactly the same wrapping. On the next day I do the same, obviously risking to eat also some of the cheap ones. How many days on average will it take for the full replacement of the expensive chocolates with cheap ones?

I would say $10/3$ but this is very simplistic. 
Also, the total number of ways to pick 3 chocolates out of 10 is 
$\binom {10} 3=\frac {10!}{3!7!} = 120$
which means that after 120 days I will have replaced all chocolates but I don't think it is correct.
Any help?
 A: Sketch (not a complete solution, but a road map towards one):
We proceed recursively. Let $E_i$ denote the expected number of day it takes given that you have exactly $i$ good ones left.  The answer you want is $E_{10}$.
Let's compute $E_1$, for example.  Each day the good one gets selected with probability $\frac {3}{10}$.  Thus $$E_1=\frac {10}3$$
Now let's consider $E_2$.  The first draw you gets $0,1$ or $2$ of the good ones.  We quickly deduce that $$\binom {10}3 \times E_2=\binom 83\times (E_2+1)+\binom 82\times \binom 21 \times (E_1+1)+ \binom 81\times \binom 22\times (1)$$
This resolves to $$E_2=\frac {115}{24}\approx 4.7917$$
Similarly, for $i>2$ $$\binom {10}3\times E_i=\binom {10-i}3\times (E_i+1)+\binom {10-i}2\times \binom i1\times (E_{i-1}+1)$$ $$+\binom {10-i}1\times \binom i2\times (E_{i-2}+1)+\binom {10-i}0\times \binom {i}3\times (E_{i-3}+1)$$
And we can solve the system step by step.  The computation requires a bit of attention since in the recursion we must define $\binom nm=0$ when $n<m$.
In the end I get about $\boxed {9.046}$ but as the comments will clearly indicate, a great many careless errors were made en route so I advise checking carefully.
A: Following on from @lulu I wrote the following program to compute the expected number of days required to take $n$ 'expensive chocolates' given that you can only take $k$ in one day:
from fractions import Fraction

#n - number of 'expensive chocolates' at start
#g - number of 'expensive chocolates' in current state of recursion
#k - number of chocolates taken each turn
def expected(n,g,k):
    if g<=0:
        return Fraction()
    if k>=n:
        return Fraction(1,1)
    ex=ncr(n-g,k)
    for i in range(1,k+1):
        ex+=ncr(n-g,k-i)*ncr(g,i)*(expected(n,g-i,k)+Fraction(1,1))
    return ex/(ncr(n,k)-ncr(n-g,k))

def ncr(n,r):
    if r>n:
        return Fraction()
    c=Fraction(1,1)
    while r>0:
        c*=Fraction(n,r)
        n-=1
        r-=1
    return c

For example, running expected(10,10,3) gives Fraction(8241679, 911064) which is equivalent to $$\frac{8241679}{911064}\approx9.046212999\text{ days}$$
for the case in question.
A: HINT for a Markov approach
If at a certain point you have $A$ ordinary and $B$ belgian chocolates,  when you select the three chocolates
you can select :
 - $[a,a,a]$ (three ordinary ch.) with prob.
$$
{A \over {A + B}}{{A - 1} \over {A - 1 + B}}{{A - 2} \over {A - 2 + B}} = {{A^{\,\underline {\,3\,} } } \over {\left( {A + B} \right)^{\,\underline {\,3\,} } }}
$$
 - $[a,a,b]$ with prob.
$$
{A \over {A + B}}{{A - 1} \over {A - 1 + B}}{B \over {A - 1 + B}} = {{A^{\,\underline {\,2\,} } B^{\,\underline {\,1\,} } } \over {\left( {A + B} \right)^{\,\underline {\,2\,} } \left( {A - 1 + B} \right)}}
$$
 - $[a,b,a]$ with prob.
$$
{A \over {A + B}}{B \over {A - 1 + B}}{{A - 1} \over {A - 1 + B - 1}} = {{A^{\,\underline {\,2\,} } B^{\,\underline {\,1\,} } } \over {\left( {A + B} \right)^{\,\underline {\,3\,} } }}
$$
and so on, obtaining the development of
$$
\left( {A + B} \right)^{\,\underline {\,3\,} }  = \sum\limits_j {\left( \matrix{
  3 \hfill \cr 
  j \hfill \cr}  \right)A^{\,\underline {\,3 - j\,} } B^{\,\underline {\,j\,} } } 
$$
The falling factorial at the numerator will ensure that we are not going to pick more pieces than available, and we shall 
ensure that the fraction be null, whenever the numerator is null.
At the end, we are going to restore three ordinary chocolates, i.e. bring $A$ to $A+3$ and restore the total to the original $N=A+B$
It is therefore possible to write the probabilities that after a cycle of picking and restoring the number of ordinary ch. passes from $A$ to $A+0,\,  A+1, \, A+2, \, A+3$
(with $B$ being the complement to $N$).
That means that we can set up a 4-diagonal Markov Matrix for $A$, or else for $B$, and study the characteristics of that.
Performing the calculations as above the $A$-th row of the transition matrix will be
$$
\begin{array}{*{20}c}
   {} & | &   \cdots  & A & {A + 1} & {A + 2} & {A + 3}  \\
\hline
   A & | &   \ddots  & {\frac{{A^{\,\underline {\,3\,} } }}{{N^{\,\underline {\,3\,} } }}}
 & {3\frac{{A^{\,\underline {\,2\,} } \left( {N - A} \right)^{\,\underline {\,1\,} } }}{{N^{\,\underline {\,3\,} } }}}
 & {3\frac{{A^{\,\underline {\,1\,} } \left( {N - A} \right)^{\,\underline {\,2\,} } }}{{N^{\,\underline {\,3\,} } }}}
 & {\frac{{\left( {N - A} \right)^{\,\underline {\,3\,} } }}{{N^{\,\underline {\,3\,} } }}}  \\
\end{array}
$$
Example
To contain the dimensions of the matrix, let's consider the case $N=4$, then the transition matrix is
$$
T\left( 4 \right) = \frac{1}{4} \, \left( {\matrix{
   0 & 0 & 0 & 1 & 0  \cr 
   0 & 0 & 0 & 3 & 1  \cr 
   0 & 0 & 0 & 2 & 2  \cr 
   0 & 0 & 0 & 1 & 3  \cr 
   0 & 0 & 0 & 0 & 1  \cr 
 } } \right)
$$
