Six balls numbered from 1 to 6. Which is the probability that the 2nd is greater than 1st? 
One bag contains 6 balls numbered from one to six. They draw 2 without
  replacement. Calculate the probability that the second extracted ball
  has a greater number than the first one. Calculate the same
  probabilities in case of extraction with replacement.

Well, maybe is a stupid question, but i’m not sure if i’m thinking in the correct way and i’m a beginner.
I’ve supposed that the cases where the second extraction is greater than the first is like sum the probabilities of order this balls in all positive cases.
So, if I draw a 1 at first extraction, I have 5 different possibilities to extract a ball that is greater than the first. With a draw of 2, I have 4 
..and so on..
But how can I calculate it? What’s the result?

EDIT:
Maybe I got the point. The result is:
1/6 ( 5/5 + 4/5 + 3/5 + 2/5 + 1/5 ) = 1/2
Can anyone confirm this, please?
 A: 
So, if I draw a 1 at first extraction, I have 5 different possibilities to extract a ball that is greater than the first. With a draw of 2, I have 4 ..and so on..

So that's $\tfrac 16\tfrac 55+\tfrac 16\tfrac 45+\cdots+\tfrac 16\tfrac 05=$ an unsurprising result.

What is the probability that the process results in the second ball showing a lesser number than the first?
A: To begin with, we assume that every ball is equally likely to be drawn from the bag first; and then that every ball remaining in the bag is equally likely to be drawn second. You have already made these assumptions, and this is standard for problems of this kind. 
So (without replacement) you have a set of outcomes, each of which has a first number and a second number different from the first. Each outcome is equally likely according to the assumptions. 
Partition these outcomes into two subsets: subset $A$ in which the first number is greater than the second, and subset $B$ in which the second number is greater. Is it possible to map the members of $A$ to members of $B$ by swapping the first and second numbers? Is it a one-to-one mapping? What does that tell us about the relative sizes of the subsets and the relative probability of the events?
With replacement, you again have several outcomes, each a pair of numbers, each equally likely, but this time we partition the outcomes into three subsets: $A$ and $B$ as before, and a third subset $C$ containing outcomes in which the same number occurred twice. The same mapping between $A$ and $B$ applies, but their probabilities no longer add up to $1.$ Find the probability of $C,$ subtract that from $1,$ and now you can find the probabilities of each of the other two events. 
