Context
In a game players are divided into two groups, A and B. Players in A know identities of all players whereas players in B only know the identity of himself. In the first round Group A eliminates one player from Group B, then the eliminated choose one of the remaining players at random to eliminate. If the new victim is in B, the new victim would do the same until a player in A is eliminated, and a new round starts. The group that first have all players eliminated loses the game.
Suppose there are $n$ players in total and $x$ players in Group B. Let $p(n,x)$ denote the probability that Group B wins.
Question
What is the closed-form solution to the recurrence $$p(n+1,x+1)=\frac{n-x}{n}p(n-1,x)+\frac{x}{n}p(n,x)$$ with conditions $x\le n$ and $x,n \in\Bbb N$, and boundary conditions $p(0,0)=0$, $p(n,0)=0$ and $p(n,n)=1$ for $n\ne 0$?
At least a partial answer so far. Let $a_{n,k}=p(n,n-k)$ for $0\leqslant k\leqslant n$.
Then $a_{n,n}=0$ for $n\geqslant 0$, $a_{n,0}=1$ for $n>0$, and $$na_{n+1,k}=ka_{n-1,k-1}+(n-k)a_{n,k}$$ for $n>1$ and $1\leqslant k\leqslant n$. Let $g_k(x)=\sum_{n=k}^\infty a_{n,k}x^n$; then $\sum_{n=k}^\infty[\ldots]x^{n-1}$ yields $$\big(g_k(x)/x\big)'=kg_{k-1}(x)+g_k'(x)-kg_k(x)/x$$ with $g_0(x)=x/(1-x)$; solved for $g_k(x)$, this gives the recurrence $$g_k(x)=kx(1-x)^{k-1}\int_0^x\frac{g_{k-1}(t)}{(1-t)^k}\,dt.$$
Fortunately, this recurrence has a closed-form (!) solution: $$g_k(x)=\frac{x}{1-x}\big(x+(1-x)\log(1-x)\big)^k.$$ As $g_k(x)=\sum_{n=k}^\infty p(n,n-k)x^n$ (actually the terms are nonzero only for $n>2k$), this gives a way to get a closed form of $p(n,n-k)$ for each fixed $k$, in terms of Stirling numbers of the first kind.
But it remains to find out whether there is a closed form (with no summations) for arbitrary $k$.