TITLE: Iteratively replacing $3$ chocolates in a box of $10$
QUESTION [32 upvotes]: 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?

REPLY [3 votes]: 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
$$
T\left( 4 \right) = \frac{1}{4} \, \left( {\matrix{
   0 & 0 & 0 & 1 & 0  \cr 
   0 & 0 & 0 & 3 & 1  \cr 
   0 & 0 & 0 & 2 & 2  \cr 
   0 & 0 & 0 & 1 & 3  \cr 
   0 & 0 & 0 & 0 & 1  \cr 
 } } \right)
$$