The PMF of the larger of two numbers selected at random from $1,\dots,12$ 
Two balls are chosen at random from a box containing 12 balls, numbered 1;2; : : : ;12. Let X be the
  larger of the two numbers obtained. Compute the PMF of X, if the sampling is done
(a) without replacement;
(b) with replacement

I understand the numerator for both cases. In case 1, we have 1 ball x such that X=x and x-1 balls smaller than x, so we have 1(x-1) = x-1. In case two, we have x choices for a ball, then x-1 choices, so x+x-1 = 2x-1. The denominator is troubling me and it's a core probability concept I never understood
In case 1, we have 12 options for the first choice and 11 for the second. Therefore, the total number of possibilities should be 12*11 = 132. But it's not, it's half of that, which is 66
In case 2, we have 12 options for the first choice and 12 for the second. Total possibilities is 12*12 = 144. This is correct
Why am I right in the 2nd case but wrong in the first?
We had a similar question on our midterm: We have 7 unique children and 20 identical cookies. How many ways can we distribute the cookies such that each child gets one?
I thought if each child gets one, there are 13 left. Each of those 13 cookies can go to one of 7 children, so the answer should be 7^13. After learning the stars and stripes method, I realize the correct calculation leads to 19C7. I'm wondering WHY it is that my calculation is wrong, why there are two possible calculations, and what my calculation represents. It is this concept that I never understood that is still giving me trouble. I can do PMF and density functions, but I have trouble with this simplest concept
 A: We look at the no replacement problem. There are two reasonable sample spaces.
Sample Space 1: Imagine the picking is done one at a time. The possible outcomes are all ordered pairs $(a,b)$, where $a$ is the first number chosen, and $b$ is the second. The sample space has $(12)(11)$ equally likely outcomes. 
Now let us count the favourables, the outcomes in which the larger of the two numbers chosen is $x$. The possible values of $x$ are $2$ to $12$. There are $x-1$ outcomes of the shape $(x,b)$, where $b\lt x$, and $x-1$ outcomes of the shape $(a,x)$, where $a\lt x$, for a total of $2(x-1)$. Divide.
Sample Space 2: Here we look at the unordered pair ("hand") that we got. There are $\binom{12}{2}$ equally likely ways to choose a hand of $2$ numbers from $12$. And precisely $x-1$ of these hands have $x$ as the larger of the two numbers.
The problem with your calculation is that your denominator used Sample Space 1, and your numerator used Sample Space 2. 
Remarks: $1$. For the replacement problem, the "hands" approach is more awkward, for the probability of getting a $3$ and a $7$, in some order, is not the same as the probability of getting a $3$ and a $3$. So an analogue of Sample Space 1 is the best choice, and it is the one that you made. 
$2$. On the cookies question, your $7^{13}$ treats the $13$ remaining cookies as distinguishable. 
$3$. The name Stars and Stripes may be preferred by Northerners, but the usual name is Stars and Bars.  
A: Without replacement, distinguishing between the two draws, there are $2(n-1)$ cases where $n$ is highest, out of $12\cdot11$ different draws. For example, for $n=3$, we have $$\{(3,1),(3,2),(2,3),(1,3)\}$$
This gives a PMF of
$$
\frac{2(n-1)}{12\cdot11}
$$
With replacement, there are $2n-1$ cases where $n$ is the highest, out of $12\cdot12$ different draws. For example, for $n=3$ we have $$\{(3,1),(3,2),(3,3),(2,3),(1,3)\}$$
This gives a PMF of
$$
\frac{2n-1}{12\cdot12}
$$
