Question
Ten apples are to be distributed among three people without restriction.
(a)$\quad$ What is the probability that one person receives all ten apples if each way of distributing the apples is equally likely?
(b)$\quad$ What is the probability that one person receives all ten apples if each apple is distributed uniformly at random?
(c)$\quad$ Why are the two answers so different?
My working
(a)$\quad$ If each way of distributing the apples is equally likely, then the required probability is $$\frac {\binom {3} 1} {\binom {12} 2},$$ where the numerator comes from the fact that there are only three ways to distribute all ten apples to a single person (i.e. just choose one of the three people) and the denominator comes from the total number of ways to distribute all ten apples without restriction.
Is my reasoning for part (a) correct? Moreover, I have no clue as to how I should approach (b) and (c). Any intuitive explanations will be greatly appreciated!
In the second problem, the first apple has a one-third chance of going to person $A$ as it is distributed uniformly at random. The same is true for all the apples. Therefore, the chance that $A$ receives every apple is just $\frac 1 {3^{10}}$. $B$ or $C$ have the same chance to get all of them, giving $3\times \frac{1}{3^{10}}=\frac{1}{3^9}$.
To understand the difference, look at the following game: $A$ tosses coins. Heads gives $+1$ and tails gives $-1$. In two tosses, he can be $+2,-2$ or $0$. Problem 1 asks us to consider all these states as equally likely. Therefore, the probability is $1/3$ for each. However, in Problem 2, the probability is $1/4$ for the first two and $1/2$ for the last one because the states are not, in reality, equally likely.