(The answer was changed, since the initial OP text was not clear for me. I will restate, to show which two problems are solved.)
Let us count exactly among all sets $S=\{x,y,z\}$ of three different cards among $52$, showing "numbers"
$|x|$,
$|y|$,
$|z|$ in $\{1,2,\dots,13\}$,
how many are favorable for the following properties $(A)$ (as in the OP now) and $(B)$ (as it may have been digested in a former version):
Property $(A)$:
There exist at least two consecutive numbers among $|x|$,
$|y|$,
$|z|$, but all three numbers are not consecutive (after reordering them).
Property $(B)$:
There exists exactly one pair among the pairs $(x,y)$, $(x,z)$, $(y,z)$, so that restricting to numbers we obtain two consecutive numbers (after reordering them).
Note: The first solution here was addressing $(B)$. Since there are already solutions to $(A)$, i will go an other way, this way is using $(B)$ to get the final result. For this reason, $(B)$ will be considered first.
The total number of cases is $$\binom {52}3=\frac 16\cdot 52\cdot 51\cdot 50=22100\ .$$
Solution for $(B)$:
While counting the favorable cases, the "good cases", we may and will assume that $x$ and $y$ are consecutive in this order, $|x|+1=|y|$ and that $|z|$ avoids $|x|-1$, and $|x|=|y-1|$, and $|x|+1=|y|$, and $|x|+2=|y|+1$.
So we need to consider a special counting, depending on the choice of $x$.
- If $|x|=1$, $|y|=2$, then $|z|$ avoids $1,2,3$, so there are $4\cdot 4\cdot 4(13-3)$ (good) cases.
- If $|x|=2$, $|y|=3$, then $|z|$ avoids $1,2,3,4$, so there are $4\cdot 4\cdot 4(13-4)$ cases.
- If $|x|=3$, $|y|=4$, then $|z|$ avoids $2,3,4,5$, so there are $4\cdot 4\cdot 4(13-4)$ cases.
- This is so for all values of $|x|$ till $|x|=11$, $|y|=12$, in each case we have $4\cdot 4\cdot 4(13-4)$ cases.
- The final case is similar to the first one,
$|x|=12$, $|y|=13$, then $|z|$ avoids $11,12,13$, so there are $4\cdot 4\cdot 4(13-3)$ (good) cases.
Putting all together, there are
$$
2\cdot 4\cdot4\cdot4(13-3) +
(12-2)\cdot 4\cdot4\cdot4(13-4)
=4\cdot4\cdot 4(2\cdot 10 + 10\cdot 9)
=7040
$$
good cases.
The probability is thus
$$
\frac{7040}{22100}
=\frac {352}{1105}
\approx
0.318552036199095\dots\ .
$$
Solution for $(A)$:
We have already counted $7040$ $(B)$-favorable cases.
To get the $(A)$-favorable cases, we will count and add the cases with (reordered) $|x|,|y|,|z|$ of the shape $n,n,n\pm 1$.
(We do not need to count and eliminate the consecutive shape $n,n+1,n+2$, since no such constellation was considered.)
We add thus twice the number of all $S=\{x,y,z\}$ matching the shape $n,n,n+1$, so we add totally
$$
2\cdot 12\cdot\binom 42\cdot \binom 41=576\ .
$$
The result is $7040+576= 7616$, so the probability is
$$
\frac{7616}{22100}
\approx
0.344615384615385\dots \ .
$$
Alternative solution for $(A)$ following the counting idea from the OP.
Consider an $(A)$-favorable situation $\{x,y,z\}$, where $x,y$ are making $|x|, |y|$ fit the pattern $n,n+1$. We count the corresponding possibilities to make a choice of $(\ \{x,y\}\ , z)$ first,
separating the cases $n=1$, $n+1=13$, and $1<n<12$ (so $n$ has $10$ possible values)
$$
\begin{aligned}
& 10\cdot 4\cdot 4\cdot(52-1-1-8)
\\
+ & 1\cdot 4\cdot 4\cdot(52-1-1-4)
\\
+ & 1\cdot 4\cdot 4\cdot(52-1-1-4)
\\
=& 8192
\end{aligned}
$$
In this generation and count of all $(\{x,y\}, z)$, during the passage to
$\{x,y,z\}$
some resulted events are hit and finally counted twice. Namely those
of the shape $\{x,y,x'\}$ and/or
$\{x,y,y'\}$ with the same number in $x,x'$, respectively in $y,y'$.
There are $2\cdot 12\cdot 4\cdot 5=576$ such events, already counted above. So the good count delivers $8192-576=7616$.
Computer check in sage:
X = cartesian_product( [[1..13], [1..4]] )
C = Combinations(X, 3).list()
A = [ (x,y,z) for (x,y,z) in C
if 1 in [abs(x[0]-y[0]), abs(y[0]-z[0]), abs(z[0]-x[0])]
and {1,2} != {abs(x[0]-y[0]), abs(y[0]-z[0]), abs(z[0]-x[0])} ]
B = [ (x,y,z) for (x,y,z) in C
if 1 == [abs(x[0]-y[0]), abs(y[0]-z[0]), abs(z[0]-x[0])].count(1)]
N = ZZ( len(C) )
pa = len(A) / N
pb = len(B) / N
print "P(A) = %s ~ %s" % (pa, pa.n())
print "P(B) = %s ~ %s" % (pb, pb.n())
This gives:
P(A) = 112/325 ~ 0.344615384615385
P(B) = 352/1105 ~ 0.318552036199095
