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?
from Hot Weekly Questions - Mathematics Stack Exchange
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