"Perfect ten" dice game I have been modelling a dice game, trying to tweak the parameters to make it reasonably close to fair.
The rules are as follows:
The player pays a \$1 game fee. Then she throws one normal die repeatedly until the running total hits or exceeds $10$. If the total passes $10$ without equalling it, that's a loss - no prize. If the total reaches $10$ the prize is dependent on the highest single throw value in the sequence, as follows:


*

*Maximum single throw was $6$  - prize \$2

*Maximum single throw was $5$  - prize \$3

*Maximum single throw was $4$  - prize \$5

*Maximum single throw was $3$  - prize \$10

*Maximum single throw was $2$  - prize \$20

*Maximum single throw was $1$  - prize \$50


Of course the last two prizes are mostly there for decoration, or "sucker value" if I weren't trying to make the game fair
Anyhow, the tricky part is trying to get the actual probabilities of each prize, since the number of rolls is not fixed. So can anyone see a short or pretty way of doing that?

For guidance, the probabilities from simulation (not exact values) are
$\begin{array}{c|c}
\text{Outcome} & p \\ \hline
\text{win, max } 1 & 0 \\
\text{win, max }2 & 0.00058 \\
\text{win, max }3 & 0.01632 \\
\text{win, max }4 & 0.06095 \\ 
\text{win, max }5 & 0.10498 \\
\text{win, max }6 & 0.10768 \\
\text{loss} & 0.70949 \\
\end{array}$
 A: There are $35$ partitions of $10$ with maximal part $\leq6$. 
Example: The partition $(4,2,2,1,1)$ has  maximal part $4$, requires $5$ throws, and can be realized in $5\cdot{4\choose 2}=30$ ways. The probability that this partition is realized therefore comes to $30/6^5$.
Going through all $35$ partitions in this way leads to the following probabilities $p_k$ $(1\leq k\leq 6)$ of ending the game successfully with a maximal throw of $k$:

This means that $p_6={833\over7776}$, etc. The sum of the $p_k$ is about $0.289288$.
A: max of 6 happens in the following ways where the order doesn't matter [6,4]; [6,3,1]; [6,2,2]; [6,2,1,1]; [6,1,1,1,1];
Let's use the [6,2,1,1] case as an example. The exact ordering 6211 happens with probability $\frac{1}{6^{4}}$. Then it's a matter of figuring out how many distinct orderings there are. But this is $\frac{4!}{2!1!1!} = 12$ and thus, the probability of [6,2,1,1] occurring in any order is $\frac{12}{6^{4}}$
You can do this similarly for all the cases.
Here's an (expensive) python simulator that records the number of occurrences of each distinct case. You can find empirical probabilities by dividing by your sim size (n).

def sim(n):
    c = {}
    for i in range(n):
        t = []
        while sum(t) < 10:
            t.append(random.randint(1,6))
        if sum(t) == 10:
            if tuple(sorted(t)) not in c:
                c[tuple(sorted(t))] = 0
            c[tuple(sorted(t))] += 1
    return c

note that the case of ten $1s$ is a $1$ in $6^{10}$, so you might see one in $10^{8}$ sims.
A: There are six ways to end the game: scores of 10 through 15. We only care about the cases of a 10, and the rest are bust.
Drawing this as a digraph, we see the number of ways reaching:


*492

*484

*468

*436

*373, and

*248


for a total of 2501 possible roll combinations giving an outcome of the game.
Therefore, the odds of winning are $$\frac{492}{2501}=0.19672311\dots$$
The total of prizes should also be balanced by how likely that prize is to be obtained given that we did win, and we can find those likelihoods using partitions.
Now, how many of our winning scores have partitions, using parts no larger than 6, with a maximum number of each of 1 through 6? There are 35, in fact (42 full partitions of 10, minus 7 partitions that require a part greater than 6).
Because these rolls are also ordered while partitions are not, we then have to count the total ways to achieve each partition.
For example, the partition $6+4$ can be achieved in two ways of rolling, but the partition $2+2+2+2+1+1$ can be done in 15 ways of rolling. Generally the number of ways of rolling is based on a product of the choice functions $$\binom{l_i}{p_i}$$ where $l_i$ is the number of parts total less the parts already placed, $p_i$ is the number of parts of that size.
By hand calculation, there are 492 ways to partition 10 using parts of sizes 1 through 6 (hey, we saw that number already! we must be on the right track). For a maximum part size, there are:


*

*1 way;

*88 ways;

*185 ways;

*127 ways;

*63 ways; and

*28 ways


to obtain a score of 10.
