When to consider the ordering for probability. A rather fundamental concept which I somewhat failed to grasp and now is jeopardising my further understanding/solving of probability problems..
In the case of this question, where we are to find the probability, that the minimum of two throws of a fair die equals $k$, $k \leq 6, k \in \mathbb{N}$, do we have to account for the ordering of the dice? 
I.e., assuming $k = 3$, is the probability $P(\{3\}) = \frac{1}{6}\times\frac{4}{6}\times 2$ in order to account for the fact that the first throw could be $3$ and the second throw anything from $3$ onwards OR vice versa (the first throw anything from $3$ onwards and the second throw $= 3$)? Or should it just be $P(\{3\}) = \frac{1}{6}\times\frac{4}{6}$ since the dice are similar and there is no mention that the two dice are unique (e.g. different in colour, size etc.).
The general question, hence, is, for cases where coins/dice are involved and are not uniquely labelled, should the order be regarded, if there is no additional mention of a first/second throw? 
Hope you all get my drift..
 A: Neither of your suggested answers is correct.
Imagine that you do roll the dice separately, so that you can talk about a first and a second die. Then it's clear that there are $6^2=36$ equally likely possible outcomes. Seven of them have $3$ as their minimum, $\langle 3,3\rangle,\langle 3,4\rangle,\langle 4,3\rangle,\langle 3,5\rangle,\langle 5,3\rangle,\langle 3,6\rangle,\langle 6,3\rangle$, so the probability of getting a minimum of $3$ is $\frac7{36}$. If there were two different ways to get a pair of $3$'s, your first answer would be right, but there aren't, and your calculation doesn't allow for this.
If you consider only the pairs of numbers that can result, you have to adjust for the fact that getting a $3$ and a $5$, say, is twice as likely as getting two $3$'s; it's simpler to distinguish the two dice. You can always do this. If the problem specifies that they're rolled simultaneously, imagine marking one of them. Or you can say that the 'first' die is the one that lands further to your left, or if that fails to distinguish them, the one that lands closer to you.
A: Brian M. Scott's answer points out the correct calculation for this problem.
You can't simply calculate the probability as $P({3}) = \frac{1}{6}\times\frac{4}{6}\times2$. You could divide it into three cases: (1) both outcomes are three, (2) the first outcome is three and the second one is greater than 3, and (3) the second outcome is three and the first one is greater than 3. For (1), you have $P_1 = \frac{1}{6}\times\frac{1}{6}$. For (2), you have $P_2 = \frac{1}{6}\times \frac{3}{6} = \frac{3}{36}$. For (3), you have the same result as (2), but in a different order. So, the probability is $P = (\frac{1}{6}\times\frac{1}{6}) + (\frac{1}{6}\times\frac{3}{6})\times 2 = \frac{7}{36}$.
Regarding the question of whether we have to take into account the ordering of the dice: in this case, the possible orders of the favorable outcomes need to be taken into account, because all the possible orders are included in the possible outcomes ($6^2 = 36$) (that is, the possible outcomes include both $(1,6)$ and $(6,1)$, for example). So, the "OR vice versa", as you wrote it, is important.
Let's suppose you didn't consider the order in this case, and you listed the favorable outcomes as $(3,3), (3,4), (3,5), (3,6)$. Since the possible outcomes are $6^2 = 36$, which include both $(3,6)$ and $(6,3)$, that would necessarily mean that the pairs $(3,6)$ and $(6,3)$ are somehow different, and, while $(3,6)$ is a possible outcome, $(6,3)$ is not, for some reason. But this isn't the case here, because both of them respect the condition "such that the minimum of two throws of a fair die equals 3". In other words, this restriction only says that the minimum of two throws should equal 3; it doesn't tell you that the minimum can only be the first dice, or the second dice. So, you have to take both possibilities into account.
A: While solving the problem at hand, as the classical definition is used to calculate probability, we need to ensure all the elements in the sample space to be "equally likely". Suppose the outcome (2,1) and (1,2) are regarded as same. Then the probability of getting the outcome (1,2) is 2/36 (see Note 1) while the probability of getting the outcome (1,1) is 1/36. Hence the classical method fails here as the elements are not "equally likely". However, if (1,2) and (2,1) are not considered the same, each element in sample space will have a probability of 1/36 and hence becomes "equally likely". 
Can the problem be solved without the classical definition so that we get the result without considering the ordering? It's difficult.
Note 1: At the first throw either 1 or 2 has to be obtained out of {1,2,3,4,5,6}. Hence the probability of favorable outcome during first throw is 2/6. At the second throw we need 1 if 2 was obtained in the first throw and vice versa. Hence the probability of the favorable outcome during second throw is 1/6. As the two throws are independent, the probability is 2/36) 
