Candies

The answer is 1/2, and it will always be 1/2 regardless of how many of each kind of candy is in the bag, assuming there is at least one of each type.

Here is a "quick version" of the proof:

Let P(c,m) be the probability that the last candy will be caramel given c caramels and m mints at the start of the day.
P(1,1) = 1/2
Assume P(1,n) = 1/2 for all n <= some integer m
If there are m mints and 1 caramel in the bag at the start of the day, there is a 1 / (m + 1) chance of the caramel being the first, second, ..., (m+1)th candy being drawn.
If it is first, then the next one is mint, which is put back; all of the candies are now mints, so the probability of the last one being a caramel = 0
If it is last, then the bag is empty, and it goes back into the bag, so it will be the last one drawn (the next day), and the probability of the last one being a caramel = 1.
In each of the other (m - 1) possibilities, there will be 1 caramel and some number k mints remaining; under our assumption, the probability that the last one will be a caramel.
The total probability = 1 / (m + 1) x (0 + 1/2 + 1/2 + ... + 1/2 + 1); there are (m + 1) terms, so the total is 1/2.
Therefore, P(1,n) = 1/2 for all positive integers n.
Note that, by symmetry, P(m,c) = 1 - P(c,m), so P(n,1) = 1 - P(1,n) = 1/2 for all positive integers n.

For 2 caramels and m mints in the bag, note that P(2,2) = 1/2; if we assume that P(2,n) = 1/2 for all n <= some m, then if we start with 2 caramels and (m + 1) mints, then we will either get to P(2,n) or P(n,m+1) for some n, depending on whether the first candy drawn was a mint or a caramel. All of the terms are 1/2 except for P(2,0) = 1 and P(0,n-1) = 0, so they add up to 1/2. The same goes for 3, 4, 5, ... caramels in the bag.