Rolling a die 15 times 
If I roll a die 15 times, what is the probability that each side (i.e. 1-6) will appear at least once?

The total amount of possible outcomes would obviously be $6^{15}$.  However, after this, I am a bit confused. 
 A: The chance that you get no $1$'s is $(\frac 56)^{15}$ and similarly for each other number.  Naively, you would then say that the chance you don't get some face is the sum of these, $6\cdot(\frac 56 )^{15}$, but you have counted twice the sets of rolls that have neither $1$'s nor $2$'s.  You need the inclusion-exclusion principle to cover that.  Basically you add back in the chance you missed two faces, but now have to think about how many times you have counted the ones missing three faces.  Since $(\frac 56)^{15}\approx 6.5\%$ is a small number and $(\frac 46)^{15}\approx 0.22\%$ is much smaller, the corrections don't amount to much.  The approximate answer will then be $1-6\cdot (\frac 56)^{15}\approx 61\%$
A: What is the probability that at least one of the numbers doesn't appear?
Number of combinations without a 1: $\ ^{15}C_5$
Number of combinations without a 2: $\ ^{15}C_5$
Number of combinations without a 1 or a 2: $\ ^{15}C_4$
Therefore, number of combinations that don't have both 1 and 2: $2\times\ ^{15}C_5 -\ ^{15}C_4$
Continue until you have the total number of combinations that don't have all numbers, and subtract from the total number of combinations. Then divide to get probability.
A: For $1 \le i \le 6$, let $g(i)$ be the number of ways to roll 15 dice such that a given set of $i$ different numbers all appear (and no other numbers appear).  For example, $g(1) = 1$.
You should be able to prove that, for all $k$,
$${k \choose 1} g(1) + \cdots + {k \choose k} g(k) = k^{15}$$
Your problem amounts to computing $g(6)$, which you should be able to do using the above formula.
