two integers are chosen at random between $0$ and $10$ what is the probability that they differ by no more than $5$? I've started studying geometric probability and I am having some difficulty with this version of problem :

Two integers are chosen at random between $0$ and $10$ inclusive. What is the
  probability that they differ by no more than $5$ ?

The integers restriction really makes it harder for me,without this restriction I would  tackle the problem like this(I am not sure it is correct):
Given two numbers $x,y$ we want $0\le y-x \le 5$ or $x \le y \le 5+x $
From the last restriction I have to satisfy the following inequalities $y \le 5+x$ and $y \ge x $ where $ 0 \le y,x \le 10$ (look image below)
Therefore the area I want is $75$
Thus the probability would be $\cfrac{75}{100} $
Now with the integers restriction I would have 

where the red filled circles indicate the integers which satisfy the restriction.
How do I count them now ?I can't simply count the dots as that would lead me to a probability higher than the previous one,which is impossible (I would get $\cfrac{42}{100}$)..
 A: There are $\overbrace{6+7+\dots+10+11}^{\frac{17}2\cdot6}+\overbrace{10+\dots+7+6}^{\frac{16}2\cdot5}=91$ ordered pairs that differ by at most $5$ out of $11^2$ ordered pairs. That gives a probability of
$$
\frac{91}{121}
$$

Clarification: I read "differ by no more than $5$" to mean $\left|x-y\right|\le5$. Then for $0\le x\le5$, there are $x+6$ choices for $y$ and for $6\le x\le10$, there are $16-x$ choices for $y$.

Why The Discrete Probability Might be Greater than the Continuous

In the image above, the red squares hang down and to the right of the associated points. Thus, the red area, $91$, divided by $121$ represents the discrete probability. The area inside the black hexagon, $75$, divided by $100$ represents the continuous probability.
The red area is definitely greater than the area of the hexagon, but it is being divided by a larger number, so it is hard to tell which will be greater.
A: There is no a priori reason why the probability should be less than the "continuous" value $37.5.$ When counting points you can use the general formula for the sum of the first $n$ nonzero natural numbers:
$$\sum_{i=1}^ni=\frac{n(n+1)}2$$
Also remember that the total number of grid points is 121 (not 100).
A: https://english.stackexchange.com/questions/118402/when-is-between-inclusive-and-when-exclusive
'Between 0 and 10' to me means the integers 1 through 9
So in that case 
p = 69/81
