Constructing a Multivariate Probability Distribution Formula
That's the correct formula for the probability that a specified play scores exactly $a$ points in $n$ rounds, but I see that it helps you much.
If a specified player wins, then he scores $10$ points and each of the others scores at most $9$ (so $10\leq n\leq 28)$. So the probability that he wins is $$\frac13\sum_{k=0}^9\sum_{j=0}^9\binom{k+j+9}{k,j,9}\left(\frac13\right)^{k+j+9}$$ because the winning toss, when the specified player gets his tenth point, must be the last one.
Now you just have to modify this formula to account for the fact that the game lasts $n$ rounds, that is $k+j+10=n$.
To calculate the probability of the game ending on a particular score $k-j-10$, you need the score to reach $k-j-9$ in any order of scoring, and then the particular player to have the final score.
As saulsplatz has said in effect, the probability that the final score is $k - j - 10$ is $$\frac{(k+j+9)!}{k!\,j!\, 9!}\left(\frac13\right)^{k+j+10}$$ with $j$ and $k < 10$, though note that $j - k -10$ is also possible $($if $k \not=j)$ with the same probability, as are $k - 10 -j$ and $j-10-k$ and $10 -k-j$ and $10-j-k$, so the overall probability of that pattern is $6$ times that expression if $k \not=j$ and $3$ times that expression if $k=j$.
So the probabilities for individual scores are (with a total of $\frac13$ since we have assumed a particular player wins)
j 0 1 2 3 4 5 6 7 8 9
k
0 0.00002 0.00006 0.00010 0.00014 0.00015 0.00014 0.00012 0.00009 0.00006 0.00004
1 0.00006 0.00021 0.00041 0.00060 0.00070 0.00070 0.00062 0.00050 0.00038 0.00026
2 0.00010 0.00041 0.00090 0.00140 0.00174 0.00186 0.00176 0.00151 0.00119 0.00088
3 0.00014 0.00060 0.00140 0.00233 0.00310 0.00351 0.00351 0.00318 0.00265 0.00206
4 0.00015 0.00070 0.00174 0.00310 0.00439 0.00527 0.00556 0.00530 0.00464 0.00378
5 0.00014 0.00070 0.00186 0.00351 0.00527 0.00668 0.00742 0.00742 0.00680 0.00579
6 0.00012 0.00062 0.00176 0.00351 0.00556 0.00742 0.00865 0.00907 0.00869 0.00772
7 0.00009 0.00050 0.00151 0.00318 0.00530 0.00742 0.00907 0.00993 0.00993 0.00919
8 0.00006 0.00038 0.00119 0.00265 0.00464 0.00680 0.00869 0.00993 0.01034 0.00996
9 0.00004 0.00026 0.00088 0.00206 0.00378 0.00579 0.00772 0.00919 0.00996 0.00996
and the probability of a pattern is (with a total of $1$, since we do not care who wins or comes second)
j 0 1 2 3 4 5 6 7 8 9
k
0 0.00005
1 0.00034 0.00062
2 0.00062 0.00248 0.00269
3 0.00083 0.00359 0.00837 0.00698
4 0.00090 0.00419 0.01046 0.01860 0.01318
5 0.00084 0.00419 0.01116 0.02108 0.03163 0.02003
6 0.00070 0.00372 0.01054 0.02108 0.03338 0.04451 0.02596
7 0.00053 0.00301 0.00904 0.01908 0.03179 0.04451 0.05440 0.02979
8 0.00038 0.00226 0.00715 0.01590 0.02782 0.04080 0.05213 0.05958 0.03103
9 0.00025 0.00159 0.00530 0.01236 0.02267 0.03476 0.04634 0.05517 0.05977 0.02988
and the probability that a game lasts a particular number of rounds is $($adding up to $1)$
10 0.00005
11 0.00034
12 0.00124
13 0.00331
14 0.00718
15 0.01339
16 0.02232
17 0.03402
18 0.04819
19 0.06425
20 0.08123
21 0.09750
22 0.11066
23 0.11787
24 0.11668
25 0.10592
26 0.08620
27 0.05977
28 0.02988
This means that
- the equally most likely individual final scores are $10-8-8$ or $8-10-8$ or $8-10-8$, each with a probability of about $0.01034$,
- the most likely final pattern is $10-9-8$ with a combined probability of about $0.05977$, and
- the most likely number of rounds in the game is $23$ with a combined probability of $0.11787$. The expected number of rounds is about $22.34469$ and the median also $23$, all rather less than the $27$ or $28$ involved in the most likely scores.
The expected final score per player is about $7.44823$, also rather less than the most likely scores might suggest.