Reordering 26 letters, probability of certain outcomes My problem is based on the random ordering the letters of the alphabet {A, B, C, ... X, Y, Z}
It asks me to find the probability that A is placed to the left of C as well as the probability that both A and B are to the left of C. 
I know the there are 26! possible orderings of the letters. I know the 2nd probability is going to be the intersection of two symmetric events. The part I am having trouble with is defining the numerators for both of these events. Anybody have any strategies for approaching this kind of problem?
 A: Yes, it is correct that the probability of $A$ occurring to the left of $C$
is $\frac 12$, by symmetry. More specifically, the sample space,consisting of orderings of the alphabet, breaks into two mutually exclusive events : one in which $A$ comes before $C$, and one in which it comes after. These events can be put in bijective correspondence via switching $A$ and $C$, and now the ordinary formula for probability gives the result.
For the second situation, something similar applies, but with a lot more events. Consider just three letters, $A,B$ and $C$. Amongst each other, there are six ways to arrange these letters :
$$
ABC,ACB,BAC,BCA,CAB,CBA
$$
Now, we have some other letters $D , ..., Z$ which we can insert into the given $A-B-C$ strings, to give an ordering of the alphabet. But, this is the key point : the event that we want to study, involves only the positions of $A,B,C$ with respect to each other. This allows us to ignore the positions of the other letters, and therefore study in essence, a smaller problem. The smaller problem? Well, when do both $A$ and $B$ lie to the left of $C$? And the answer is given above : it happens precisely $\frac 16$ of the time.
Now, let us formally write this. Create the following subsets of the sample space : $S_{ABC} = \{o : \text{In $o$, the letters $A,B,C$ occur in the order ABC, maybe not consecutively}\}$.
Now, we can go on and define $S_{ACB}, S_{BAC}$ etc. and get six different sets. We will let $S_{xyz}$ denote any of the six sets above i.e. $x,y,z$ is some permutation of $A,B,C$.


*

*It is clear that the $S_{xyz}$ are disjoint, since $ABC$ cannot occur in two different orders in the same permutation. However, it must occur in some order. This gives, that the sample space is a disjoint union of the $S_{xyz}$.

*Next, there is a bijection between $S_{xyz}$ and $S_{x'y'z'}$ : take any order, and switch only $x,y,z$ amongst themselves so that the order becomes a member of $S_{x'y'z'}$. This is a bijection.
So, the break up of our sample space into six disjoint equiprobable events is complete. We desire the probability of one of these events, $S_{ABC}$ and hence the solution is $\frac 16$.
