Convergence of winning probability in a one-player dice-throwing game In this (one-player) game, the player starts with a total of $n$ points.  On each turn, they choose to throw either a four-, six-, or eight-sided die, and then subtract the number thrown from their point total.  The game continues until the player's point total reaches zero exactly, and they win, or falls below zero, in which case they lose.
The strategy does not seem obvious.  (Even for the simple case of $n=5$, one must do a calculation.)  And dynamic programming techniques show that the strategy is not straightforward.  For $n≤20$, the four-sided die is optimal, except for $n=5, 6, 13, 20$, where the six-sided die is optimal, and $n=8, 14, 16$, where the eight-sided die is optimal.
For large $n$, the probability of winning approaches $0.456615178766744$, which is not a number that I recognize.  (The ISC does not recognize it either.)
It's not particularly surprising that the probability of winning converges as $n$ increases, because the probability of winning with $n$ points is the mean of probability of winning with $n-i$ points, taken over a small range of $i$.  Considering the probabilities as a sequence, each element lies inside the range of the previous few elements, and tends to be in the middle of that range. As one goes farther out in the sequence, variations away from the mean tend to be damped out.
My questions are:


*

*What's this $0.456615178766744$?  (For a single $d$-sided die the probability of winning approaches $\frac 2{d+1}$, but for more dice the problem seems harder.)

*Is there any regularity to the optimal move, as $n$ increases?

*Is there any good way estimate a good strategy, short of exhaustive computer calculations?  For example, in the $n=7$ situation, the four-sided die is significantly better than the six-sided die.  Is there some way to see this, or at least to guess that it is so?

*Someone must have studied this before.  Does the problem have a name?  Can someone give me a reference to the literature?


 A: Yet more experimental results (to be honest I hardly understand how it works, I just know enough math to spot the mistakes in the previous experiments), but this time I believe I've got the correct answer.
I hereby thank @ShreevatsaR (and indirectly @RobPratt) for providing the notebook (and code) that allowed me to figure this out. Would never have done it by myself.
The (experimental, but probably correct) answer is $$\lim_{n\to\infty} V(n)=\frac{53593778027393979383062834089689687375851958941331473223707343944117988244200}{117371871369111168311975072842802096714042245341293610168598354761946856284327}.$$(That's 77 digits over 78 digits. No wonder nobody found it before!)
What I did: I added a line to the notebook that tried to approximate the fraction with a denominator only up to 1/3 as long as the precision, to avoid overfitting.
(Overfitting would probably start happening at around $\frac{N}{2}$ digits, where $N$ is the precision - this being the point where there's enough possible fractions of this length that most $N$-digit decimals are covered at least once. I could probably have stopped at $\frac{N}{2+\varepsilon}$ digits, but I decided to use $\frac{N}{3}$ just in case.)
At a precision of 1005 digits this got me a fraction far shorter than the expected 300 or so digits, so I guessed this was the correct answer.
But just in case I tried again with 2005 digits (this took about 15 minutes), and consequently 600+ digits allowed for the denominator, and got the same answer.
Incidentally, I agree with @Michael Lugo's conjecture that those values are probably rational (if sometimes with ridiculous denominators) for any finite set of integer dice.
A: Let $D$ be the set of dice, and let value function $V(n)$ be the maximum win probability in state $n$.  Then $V$ satisfies Bellman's equations:
$$V(n)=\begin{cases}
0 & \text{if $n<0$} \\
1 & \text{if $n=0$} \\
\max\limits_{d\in D} \frac{1}{d} \sum_{r=1}^d V(n-r) & \text{if $n>0$} \\
\end{cases}$$
For $D=\{4,6,8\}$, solving Bellman's equations appears to yield limiting value $$\lim_{n\to\infty} V(n)=\frac{1311501709}{2872225388}.$$
EDIT: This conjecture was based on post-processing a floating-point calculation, apparently with insufficient precision.

Here is Python code to compute $V(n)$ with exact arithmetic:
import functools

from fractions import Fraction

D = [4,6,8]

@functools.lru_cache()
def V(n):
    if n < 0:
        return 0
    if n == 0:
        return 1
    best = 0
    for d in D:
        expectation = 0
        for r in range(1,d+1):
            expectation += Fraction(V(n-r),d)
        best = max(best, expectation)
    return best

