Problem
In a gin variant, you win immediately if you are dealt a hand with no sets (3+ of a kind or suited run of 3+ cards) and no "potential sets", i.e. no additional card added to the hand could make a set. What's the probability of this happening for a hand of $n$ cards? I actually play this game ($n=9$) and my friend and I were curious about this probability, which is the motivation for this question. It's easy to compute numerically for any $n$ and some special $n$ values are also easy (e.g. $n$=1, $n>13$) but an analytic solution for general $n$ seems deceptively difficult.
(Bad) Approach Idea
I cannot seem to find a nice way to do it analytically at all. Obviously we want the total number of ways to make this special hand, $N_h$, divided by $N={52 \choose n}$ possible hands.
The no 3+-of-a-kind/potential 3+-of-a-kind requirement is simple enough because a potential 3-of-a-kind is just a pair. This means we must select $n$ distinct values from the $13$ values in the deck $\to {13 \choose n}$.
However the suited runs are much more difficult. Depending on the distribution of the $n$ values across the $4$ suits, the number of ways of making runs/potential runs vary. The only thing I can think of is to enumerate the probabilities of runs/potential runs for each number of cards in a suit. E.g. $1$ card in a suit never makes a run/potential run. $2$ cards has $n-m+1$, $m$ card runs and $n-2$ "gapped" potential runs. $n=3$ then is more complicated with the placement of the $3$rd relative to the other $2$. There are also complications with double counting since these things are not mutually exclusive. I have some vague notions that these enumerations have some relationship to the polyhedral numbers but haven't pursued that in depth because I'm convinced there's a better approach and I'm hoping someone here can help me find it!
Numerics
It's easy to write a program to do this numerically for some $n$. Just check that there are no pairs and that within a suit all cards are at least $2$ away from every other card. Here's a quick Mathematica code that does it
deal[n_] := QuotientRemainder[RandomSample[Range[0, 51], n], 13]
valid[hand_] := DuplicateFreeQ[Transpose[hand][[-1]]] &&
Min[GroupBy[hand,First,Min[Abs[Differences/@Subsets[Last/@#,{2}]]]&]]>2
prob[n_, num_] := N@Count[Table[valid[deal[n]], {i, num}], True]/num
For the game I play $n=9$ and dealing half a billion hands gives $P\approx 0.00185276$