Suppose I have a die with six arithmetic operations--- $${-}2, {-}1, \times 0, +1, +2, +3$$ ---and that each roll of the die is uniformly distributed.
To any finite sequence of rolls of the die, assign the value given by successively applying the operations to a starting value of $0$. So, for example, for the $5$-term sequence $(+1,+3,{-}2,\times 0,-1)$, the value would be: $$((((0 + 1) + 3) - 2) \times 0) - 1 = -1 .$$ For any number $n$ of rolls, we can ask for the fraction $P_n(k)$ of $n$-roll sequences that have value $k$.
How can I find an explicit expression for the function $$f(k) := \lim_{n \to \infty} P_n(k)?$$
This answer expands on A. Kriegman's and folds in some of my comments thereunder.
Let $P_n(k)$ denote the fraction of values of $n$-term sequences with value $k$, which we can interpret as the probability that the value of a uniformly randomly selected $n$-term sequence has value $k$.
The limiting probabilities $p_k := \lim_{n \to \infty} P_n(k)$ are stable under the application of a uniformly selected die roll, giving an infinite set of equalities: $$\begin{array}{rcll} p_k &=& \frac{1}{6}(p_{k - 3} + p_{k - 2} + p_{k - 1} + p_{k + 1} + p_{k + 2}), & k \neq 0 \\ p_0 &=& \frac{1}{6}(p_{- 3} + p_{- 2} + p_{- 1} + p_{1} + p_{2} + 1) . \\ \end{array}\qquad (\ast)$$
The first equation defines a linear recurrence with characteristic polynomial $$p(r) = r^5 + r^4 - 6 r^3 + r^2 + r + 1,$$ and so the half-infinite sequences $\{p_k\}_{k \leq 0}$ and $\{p_k\}_{k \geq 0}$ can be given as linear combinations of powers $\alpha_i^k$ of the roots $\alpha_i$ of $p$ (possibly with different coefficients for $k > 0$ and $k < 0$).
The roots of $p$ are : $$ \alpha = 0.82140\ldots, \quad \beta = -0.27496\ldots+i 0.38561 \ldots, \quad \bar\beta, \quad \gamma = 1.77912\ldots, \quad \delta = -3.05060\ldots . $$ Since $0 \leq p_k \leq 1$ for all $k$, the coefficients of $\gamma, \delta$ (whose real parts have absolute value $> 1$) must be zero for the sequence $\{p_k\}_{k \geq 0}$, and the coefficients of $\alpha, \beta, \bar\beta$ (whose real parts have absolute value $< 1$) must be zero for $\{p_k\}_{k \leq 0}$, and so $$\boxed{\begin{array}{rcll} p_k &=& A \alpha^k + B (\beta^k + \bar\beta^k), &k \geq 0 \\ p_k &=& C \gamma^k + D \delta^k , &k \leq 0 \end{array}\qquad (\ast\ast)}$$ for some constants $A, B, C, D$. (NB we can rewrite $\beta^k + \bar\beta^k$ as a manifestly real expression, namely, as $2 e^{\operatorname{Re}(\beta) k} \cos (\operatorname{Im}(\beta) k)$.) We can find those constants by producing an independent linear system in those variables and solving; one option is to substitute the expressions $(\ast\ast)$, $k = -1,0,1$ in $(\ast)$. We get one equation each from substituting the first and second equations in $(\ast\ast)$ in $(\ast)$, or we can replace one of those two equations with the condition $A + 2 B = C + D$ given by substituting $k = 0$ in both of the equations in $(\ast\ast)$.
Appealing to a C.A.S. produces explicit formulae for $A, B, C, D$ as rational polynomials in $\alpha, \beta, \gamma, \delta$, but the expressions are unwieldy (hundreds of thousands of characters among them), and it is not evident that they can be simplified further. Their numerical values are: $$\boxed{\begin{align*} A &= 0.13210\ldots\\ B &= 0.04359\ldots\\ C &= 0.15602\ldots\\ D &= 0.06328\ldots . \end{align*}}$$ In particular, $p_0 = 0.21930\ldots$.
Since $A, C \neq 0$, the limiting behaviors of $p_k$ are \begin{align*} p_k \sim A \alpha^k ,&\quad k \to \phantom{-}\infty \\ p_k \sim C \gamma^k ,&\quad k \to -\infty . \end{align*}
Remark One might ask whether we can produce exact expressions for the roots $\alpha, \beta, \ldots$ of the (quintic) polynomial $p$. If we restrict ourselves to algebraic expressions, we cannot: By reducing modulo $2$ we can efficiently deduce that $p$ is irreducible over $\Bbb Q$, so its Galois group contains a $5$-cycle. On the other hand, we've seen that $p$ has exactly $2$ nonreal roots, and hence the complex conjugation map is a transposition in the Galois group of $p$. But a transposition and a $5$-cycle generate all of $S_5$, which is hence the Galois group. In particular, it is not solvable, so the roots $\alpha, \beta, \ldots$ are not expressible in terms of radicals.