I was originally trying to solve the following problem:
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?
Just like many others, I too tried to solve this using the stars and bars method which is wrong. However, while reading through the answers/comments for the problems (here, and here), I see that @true-blue-anil and @satish-ramanathan have said that "cases yielded by stars and bars not equi-probable".
I understood that the issue with my approach (stars and bars) is it does not consider the robberies as distinguishable. I also understand why is it necessary to consider each robbery as an unique event.
Moreover, I also get why the statement "cases yielded by stars and bars not equi-probable" stands true (thanks to the example in this solution).
But my concern is, I don't quite understand the statement's relevance to this particular problem. I mean, using stars and bars, the number of possible ways to rob 6 districts 6 times is $\binom{11}{6}$. Whereas if we consider all the robberies as distinct, we get $6^6$ possible ways to rob the 6 districts 6 times. So where does "cases yielded by stars and bars not equi-probable" come into picture?
Is it that since (6−0−0−0−0−0) is not equally likely as (1−2−1−1−1−0), we don't consider it? No, we still consider it. $6^6$ encompasses all the possible ways of robbing 6 districts. The only difference I see in both these approaches is that stars and bars does not honor ordering whereas the latter does.
I am pretty confused with this, any help is appreciated.
Best Answer
Yes, the events $(6,0,0,0,0,0)$ and $(1,2,1,1,1,0)$ are both considered by sticks and stones and by the "all are distinct" method. Nobody is contradicting that statement when they say "cases yielded by [sticks and stones] not equi-probable".
What they mean is precisely and simply that the probability of $(6,0,0,0,0,0)$ is not equal to the probability of $(1,2,1,1,1,0)$. That's all. And that's a fatal flaw with the application of sticks and stones to your problem.
It isn't good enough to merely consider both outcomes; you also have to consider them with the correct probabilities assigned to them.
The technique where you count number of "successful" outcomes divided by the total number of outcomes is valid when all outcomes have the exact same probability, probability $1/n$ where $n$ is the total number of outcomes. If there are any outcomes that are more likely than others, you have to do something more sophisticated than just counting.
Since some outcomes counted by sticks and stones are more likely than others (that is, "cases yielded by [sticks and stones] not equi-probable"), you can't use sticks and stones as a counting method for probability in this example.
That's why sticks and stones leads to a wrong result. Here's why we get a correct result by considering $6^6$ equally likely outcomes instead.
The problem statement says,
In order to even talk about "which" robbery occurred where, we have to be able to distinguish one robbery from another. So let's number the robberies $1, 2, 3, 4, 5, 6$ in the sequence in which they occurred during the week.
Further, number the districts $1, 2, 3, 4, 5, 6$ and let $d(n)$ be the number of the district where robbery number $n$ occurred.
Then if you write out all the values of $d(n)$, you will know exactly which robbery occurred where. For example, if it turns out that $$ d(1) = 1, \ d(2) = 1, \ d(3) = 1, \ d(4) = 1, \ d(5) = 1, \ d(6) = 1 $$ then all robberies occurred in district $1$ and you have the event that you denoted by $(6,0,0,0,0,0)$ when using sticks and stones. And this is the only way to define $d(n)$ that produces the event $(6,0,0,0,0,0)$; if we write anything but exactly these values of $d(n)$ for each $n$, we get at least one robbery in a district other than district $1$, which means $(6,0,0,0,0,0)$ did not happen.
But some other possibilities for the definition of $d(n)$ are $$ d(1) = 1, \ d(2) = 2, \ d(3) = 2, \ d(4) = 3, \ d(5) = 4, \ d(6) = 5 $$ and $$ d(1) = 5, \ d(2) = 4, \ d(3) = 3, \ d(4) = 2, \ d(5) = 2, \ d(6) = 1, $$ each of which matches the outcome $(1,2,1,1,1,0)$ in the sticks and stones method. And we've still barely begun to list all the possible different ways we could define the function $d(n)$ and still end up with $(1,2,1,1,1,0)$.
But if you know where each robbery occurred, you know exactly how the function $d(n)$ is defined -- you know every value of $d(n)$ in a list like the ones written above. And if you have a full definition of $d(n)$, showing all its values, you know exactly where each robbery occurred.
This gives us a particular one-to-one correspondence between the set of possible definitions of $d(n)$ and the set of possible cases of which robbery occurred where. And remember, each of those cases has the same probability as every other case because the problem explicitly statement says it does.
Now we just need to count the number of cases to determine what the probability of each case is. The number of cases is the same as the number of possible ways to define $d(n)$. Every possible $d(n)$ is a function from the set $\{1,2,3,4,5,6\}$ to the set $\{1,2,3,4,5,6\}$, and every function from $\{1,2,3,4,5,6\}$ to $\{1,2,3,4,5,6\}$ could be $d(n)$, so the set of cases, which has the same number of elements as the set of possible definitions of $d(n)$, also has the same number of elements as the set of all functions from $\{1,2,3,4,5,6\}$ to $\{1,2,3,4,5,6\}$.
The number of functions from a set $A$ to a set $B$ is $\lvert A\rvert^{\lvert B\rvert}$. (The notation $\lvert S\rvert$ is defined as the number of elements in $S$.) In our problem, $A = B = \{1,2,3,4,5,6\}$ and $\lvert A\rvert = \lvert B\rvert = 6$, so the number of possible definitions of $d(n)$ is $6^6$.
That's how we get $6^6$ possible outcomes, and how we know they are equally likely (because in this question the problem statement says they are). The likelihood of each of these outcomes is $$\dfrac{1}{6^6}.$$
With sticks and stones, on the other hand, you have only $\binom{11}{6}$ possible outcomes (each of which says only how many robberies occurred in each district, not which robberies occurred in each district). One of these outcomes is denoted $(6,0,0,0,0,0)$, which as we have found can only occur when the function $d(n)$ is defined in one particular way, which is something that happens with probability ${1}/{6^6}.$ So unless you come up with some way to get the sticks and stones method to produce the result that the probability of $(6,0,0,0,0,0)$ is ${1}/{6^6},$ you can't get correct probabilities by this method.