Would an infinite random sequence of real numbers contain repetitions? If random real numbers are selected from the set of all real numbers, for an infinite number of iterations, what is the likelihood of repetitions occurring?
 A: It's very hard to say just how to select a "random real number". Even selecting a random positive integer is tricky - any individual number will be chosen with zero probability.
So this is not an official mathematical answer to your question, since you haven't specified the selection method:
I'm quite sure that any reasonable way you do it you'll find the likelihood of repetitions is zero. @fleablood 's comment explains why. The probability for any particular number must be $0$ so the probability of a repeat in any finite sequence is $0$. That will also be true for any  countable sequence. 
The question does kind of specify countability when it says "an infinite number of iterations" but if that infinity is the cardinality of $\mathbb{R}$ the question is much murkier. 
A: There is no such thing as "random real numbers selected from the set of all real numbers".  Assuming what you mean is that you are taking independent samples $X_n$ from some continuous probability distribution, then yes, for any fixed $n \ne m$, the probability $\mathbb P(X_n = X_m)$ that $X_n = X_m$ is $0$, and so by countable additivity the probability of a repeat is also $0$.
A: (I read the problem too quickly and my entire first attempt was way off. Feel free to skip past the part that I've indented with block-quotes)

HINT: I would look at what happens when you pick numbers from "1" to
  "n", find a function that shows what happens when you repeat "n"
  times, then find the limit as "n" goes to infinity.
Do you know how to set this up?
ORIGINAL EDIT "No I don't know how to" Very well then, let's start
  with $n=5$
a) If you are referring to repetition anywhere in the sequence, then
  sequences like 1-4-3-5-3 or 2-5-4-1-2 would all count, leaving only
  the sequences that only contain each number one time: 1-3-4-2-5,
  2-4-3-5-1, 5-3-4-1-2...
Out of the total $5^5 = 3125$ combinations, this leaves $5! =
> 5*4*3*2*1 = 120$ that do not repeat contain repetition against
  $3125-120 = 3005$ combinations that do (3.84% without repetition,
  96.16% with). Whereas if $n=10$, then we have $10^{10} = 10$ billion combinations, only $10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800$ of which
  use each number exactly once without repetition (0.036288% without
  repetition, 99.963712% with).
As $n$ gets larger and larger, the probability $\frac{n!}{n^n}$ that
  each number from 1 to $n$ is used once and only once approaches 0,
  approaching a 100% chance as $n$ approaches infinity that there will
  be at least one repetition at some point in the sequence.
b) Or, if you only mean repetition from one number to the next
  (1-3-4-1-5 doesn't count as repetition, but 2-3-3-1-4 does), then you
  have $n-1$ pairs of adjacent numbers and a $\frac{1}{n}$ chance that
  any given pair is a match (whatever the first number is, there are $n$
  possibilities for what the next one will be).
The odds that one given pair does not match are
  $\frac{n-1}{n}=1-\frac{1}{n}$, and the odds that none of the $n-1$
  pairs match equals $(\frac{n-1}{n})^{n-1}= (1-\frac{1}{n})^{n-1}$. Do
  you know how to find $\lim \limits_{n \to ∞}(1-\frac{1}{n})^{n-1}$?

2nd EDIT: And I just realized I was supposed to be going by "all real numbers" in my previous two examples.
In any case, the set of all real numbers (integer, rational, irrational algebraic, irrational transcendental) is uncountably infinite, whereas the length of the sequence (the first number, the second, the third ... the billionth ... the googol-th ...) would be countably infinite.
Both versions of repetition that I looked at would actually have a 0% chance of occurring.
