What is symmetry? In probability, I have often seen the argument "and this probability equals $\frac 1n$ by symmetry" and I have never really understood the formality behind that statement. One of the biggest problems I have had with this line of reasoning is when solving the following problem: 

There are $n$ small and $m$ large pills in a bottle, every day, one of them is taken at random. If a large pill is taken, it is broken into two small pills, one of them is eaten and the other returned to the flask. If a small pill is taken out, it is eaten. What is the expected value of small pills remaining in the bottle after the last large pill has been taken?

The solution used the linearity of expectation and $n+m$ indicator variables, and to calculate the expected value of the first $n$ indicator variables (corresponding to the initial $n$ small pills) it said that it sufficed to consider only the $1+m$ pills consisting of that small pill and the $m$ big ones, and then, by symmetry, the probability that it survived the $m$ large ones was $\frac {1}{m+1}$.

The problem is I do not really understand why it suffices to consider only those $m+1$ pills and why they they have probability $\frac {1}{m+1}$ of being chosen last.
What are the formalities behind this argument?
 A: "By symmetry" usually refers to a situation in which multiple viewpoints will behave exactly the same. For example, say we want to know the probability of getting at least $5$ heads out of $9$ coin flips. The options are either there are at least $5$ heads or there aren't. If there aren't, there are at least $5$ tails. So the two options are either there are $5$ heads or there are $5$ tails. But there's no "real" difference between "heads" and "tails" - mathematically, they behave exactly the same. So the probability of getting at least $5$ heads should be equal to the probability of getting at least $5$ tails. Since they're disjoint, and together cover all the possibilities, these probabilities must add to $1$; so each must be $\frac{1}{2}$. "By symmetry" is a shorter way of saying all of that.
For the pills: Consider which pill gets to be chosen last. Either it's the small pill, or it's one of the large ones. But for the purposes of "picking", there's no difference between them - each should have the same chance of being chosen last. Since there are $m + 1$ pills, each has a $\frac{1}{m+1}$ chance of being chosen last, "by symmetry", because they all have the same chance and those probabilities must add up to $1$.
Concentrating on those $m+1$ pills isn't a symmetry issue; it's just observing that the expected number of small pills remaining is the probability of any one small pill outlasting the large ones, multiplied by the number of small pills. The latter is known, so we just need to know the probability of one small pill surviving.
Once you've calculated that, you have to deal with the issue of new small pills appearing because large pills were broken; I assume that's dealt with later in the argument.
A: I am unable to wrap my head around the argument, given elsewhere, for the assertion that if we have, say, $(k+1)$ small pills and $m$ large pills and if we choose one specific small pill and find the probability that it survives, it comes out to be $1/(m+1)$. Let's call this probability $p_{k,m}$ If that is true then obviously "by symmetry" for each of the other $k$ small pills $p_{k,m} = 1/(m+1)$. This is where the symmetry argument helps us. But for the first statement, I don't know, how do we justify that. Here's my approach:
$$  p_{k,l} = \frac{k}{(k+l+1)}p_{k+1,l-1} + \frac{l}{k+l+1}p_{k-1,l}$$
The base cases are $p_{i,0} = 1, i = 1, 2, ..., m+n; p_{0,j} = \frac{j}{j+1}p_{1,j-1}$
Now, start filling up the matrix in bottom-up manner and you will find $ p_{k,l} = 1/(l+1)$
