$4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are at least two successive numbers among those $4$ is $4$ integers are randomly selected from the numbers from $1$ to $10$.  The chance that there are atleast two successive numbers among those $4$ selected is 
$(A)\frac{5}{6}\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{2}{3}\hspace{1cm}(D)\frac{1}{2}\hspace{1cm}$
I calculated answer as $\frac{24}{\binom{10}{4}}$ but this wrong.  Please help me find the right answer.
 A: Take out $4$ numbers, 6 numbers (N) are left with $7$ possible gaps including ends
$_ N _ N _ N _ N _ N _ N _$
We can replace the $4$ numbers  in forbidden way in any of 7 gaps (including ends) in $\dbinom{7}{4}$ ways
thus indicated $Pr = 1 - \dfrac{\dbinom{7}{4}}{\dbinom{10}{4}}$ 
A: I came up with the following reasoning :

Pattern 1: At least two successive numbers,
$$
\eqalign {
& (**)(*)(*) \cr
& (*)(**)(*) \cr
& (*)(*)(**) \cr
& choices \ = 3 \times (2!\cdot 9) \times 8 \times 7 = 3024
}
$$
Pattern 2: At least three successive numbers,
$$
\eqalign {
& (***)(*) \cr
& (*)(***) \cr
& choices \ = 2 \times (3!\cdot 8) \times 7 = 672
}
$$
Pattern 3: At least four successive numbers,
$$
\eqalign {
& (****) \cr
& choices \ = 1 \times (4!\cdot 7) \times 6 = 1008
}
$$
The total number of legal possible choices, using the inclusion exclusion principle is :
$$
3024-672+1008 = 3360
$$
and the required probability is
$$
{3360 \over 10 \cdot 9 \cdot 8 \cdot 7} = {2 \over 3}
$$
A: Here is a simulation in R with 100,000 performances of the
experiment (choices without replacement), which verifies 
the Answer $5/6$ of @trueblueanil within simulation error.
 m = 10^5;  d = numeric(m)
 for(i in 1:m) {
   x = sample(1:10, 4);  sort(x)
   d[i] = min(diff(sort(x)))  }
 mean(d == 1)  # 'd==1' is logical vector of T's and F's
 ## 0.83222    #  its 'mean' is proportion of T's
 1 - choose(7,4)/choose(10,4)  # exact analytic answer
 ## 0.8333333
 2*sqrt((.83)*(.17)/m)  # 95% margin of simulation error
 ## 0.002375710

To get greater accuracy, use m=10^6, but run shown is
sufficient to distinguish among the four answers proposed.    
