Say I have a hand of X playing cards drawn from a standard 52-card deck, where X can be 1 through 10.
Then say I also have a second deck of the 52 standard playing cards, and draw a card from there. (alternately, say I rolled 1d13)
I need to calculate the odds (at least approximately) that at least 1 card in my X-card hand will A: MATCH the number drawn/rolled, or B: BE WITHIN ONE OF that value. (So if I drew/rolled a 6, B would be satisfied if the cards in my hand included a 5, 6, or 7)
On
Assuming that you mean ranks $\pmod {13}$, so that Aces are next to both $2's$ and Kings:
Then, for part $A$ we note that there are exactly $4$ cards that match the preferred rank. Thus the probability that none of your $X$ cards matches the preferred rank is $$\binom {48}X\Big /\binom{52}X$$ It follows that the answer you want is $$1\,-\,\binom {48}X\Big /\binom{52}X$$
Similarly for $B$ the answer is $$1\,-\,\binom {40}X\Big /\binom{52}X$$
If you don't want Aces to be next to both $2's$ and Kings then $A$ stays the same, but you have to modify $B$ to account for the possibility that the preferred rank only has one neighbor.
When $X=1$ it is easy (assuming the ranks are circular so ace is next to both two and king). You have $1/13$ chance of a match and $3/13$ chance of being within one.
When $X=2$ part A should be easy at $2/13$ except that the two cards you drew might match. The chance of a match is $\frac 3{51}$ because after you draw the first card there are $3$ of the remaining $51$ that match. The chance the die matches one of the two cards is then $\frac {48}{51}\cdot \frac 2{13}+\frac 3{51}\cdot \frac 1{13}=\frac {99}{663}=\frac {33}{221}\approx 0.1493$ Part B has more trouble with overlap. If the two cards match the chance is $3/13$, if they are one apart $4/13$, if they are two apart $5/13$ and if the are further $6/13$. Getting the exact value is a fair mess, and for $X$ larger it is messier yet.