Correct approach to evaluate probability? A city with $6$ districts has $6$ robberies in a particular week. Assume the robberies are
located randomly, with all possibilities for which robbery occurred where equally likely.
What is the probability that some district had more than $1$ robbery?
For the above problem, the solution provided is this:
There are $6^6$ possible configurations for which robbery occurred where. There
are $6!$ configurations where each district had exactly $1$ of the $6$, so the probability of
the complement of the desired event is $\frac{6!}{6^6}$
. So the probability of some district having
more than $1$ robbery is
$1 − \frac{6!}{6^6} ≈ 0.9846$.
Note that this also says that if a fair die is rolled $6$ times, there’s over a $98\%$ chance
that some value is repeated!
But this is my approach: 
I thought that this is equivalent of dividing 6 dots into 6 boxes, there are $11\choose6$ ways of distributing, out of which there is only $1$ way in which $1$ dot (equivalent to robbery) is there in each box (equivalent to district). Hence the probability of more than one robbery in any district is $1 - \frac{1}{11\choose6}$
I am unable to digest how the above situation is equivalent to the dice situation. Can anyone explain me, what is wrong with my approach, and how the dice thing is equivalent ?
 A: A model of tossing a die six times is appropriate here, since it is stated that "the robberies are located randomly, with all possibilities for which robbery occurred where equally likely."  
Cases yielded by stars and bars are not  equi-probable
You would expect a much lower frequency for all robberies  taking place in one district, $(6-0-0-0-0-0)$ compared to a more even distribution of robberies, e.g.  $(1-2-1-1-0-0)\;\;$ The first has $6!\over 5!$ $= 6$ permutations, the second has $6!\over{3!2!}$ $= 60$
Btw, note that if you write down all possible stars and bars solutions, and add up the permutations of each such solution, you will get $6^6$
A: Lets name robberies A,B,C,D,E,F and districts 1,2,3,4,5,6. As per assumption: the robberies are located randomly, with all possibilities for which robbery occurred where equally likely means that 
P(A occuring in 1) = P(A occuring in 2) = P(A occuring in 3) ....
So A  has 6 possibilities 
similarly B has 6 ....
by multiplication rule.
Number of ways in which robberies can occur is 6^6
A: As mentioned in the solution provided, this is isomorphic to the problem of rolling a fair die six times. We can build intuition around this by thinking of each robbery as analogous to each one of the six rolls of the die, while each district corresponds to each one of the six faces of the die. Further, we can imagine each of the six robbers rolling a die to pick a district to rob. This yields 6^6 possibilities by the Multiplication Rule.
Note that we cannot think of the robberies and the districts interchangeably here, as each of the six robberies will occur regardless of whether any of the districts do not get robbed. Similarly, each of the six die rolls are inherent to the die roll experiment, while the six faces of the die may or may not all be selected.
As also pointed out in the other answers, it is dangerous to apply the "dots and bars" result (Bose-Einstein) in calculating probabilities using the naive definition of probability because the outcomes from this result are not equally likely.
