Tickets, dominoes, dice: probability of winning with drawing $7$ Here is another problem from my probability textbook:

A has a bag of $6$ tickets numbered from $2$ to $7$, B has a box of dominoes numbered from double $1$ to double $6$, and C has a pair of common dice. A draws a ticket from [their] bag, then B draws a domino from [their] box, then C throws [their] dice, and they continue to do this in order, without replacing ticket or domino, until someone has drawn or thrown $7$, and [they are] declared the winner. Show that their respective chances of winning are $773347: 399303: 378622$.

Alright, here's what I did. For A it's got to be$${1\over6} + \left({5\over6}\right)\left({{25}\over{28}}\right)\left({5\over6}\right)\left({1\over5}\right) + \left({5\over6}\right)\left({{25}\over{28}}\right)\left({5\over6}\right)\left({4\over5}\right)\left({{24}\over{27}}\right)\left({{5\over6}}\right)\left({1\over4}\right) + \ldots$$And at this point I gave up, I don't see an easy way to evaluate this without writing it all out - same goes for B and C. Clearly the sums for each A, B, C are not infinite since we don't replace A's tickets or B's dominoes, but these sums are ugly, big, and huge, so given that the proportions of their respective chances of winning are relatively small, I am wondering if there is a clever way to to get those ratios without having to brute force the calculations naively.
 A: It would appear that the dominoes $B$ has to choose from are the $21$ without any blank faces. By my calculation this would produce the relative chances given as the answer.
Following the procedure suggested by user2661923 in the comments, let $\ f_0=1\ $,
$\ f_1\ $ be the probability that no-one wins in round $1$, and, for $\ i=2,3,\dots,6\ $, $\ f_i\ $ the conditional probability that no-one wins in round $\ i\ $ given that no-one has won in any of the previous $\ i-1\ $ rounds. Then
\begin{align}
f_i&=\frac{5(6-i)(19-i)}{6(7-i)(22-i)}\\
P\big(\text{$A$ wins in round $i$}\big)&=\frac{\prod_\limits{k=0}^{i-1}f_k}{7-i}\ ,\\
P\big(\text{$A$ wins}\big)&=\sum_{i=1}^6\frac{\prod_\limits{k=0}^{i-1}f_k}{7-i}\ ,\\
P\big(\text{$B$ wins in round $i$}\big)&=\frac{3(6-i)\prod_\limits{k=0}^{i-1}f_k}{(22-i)(7-i)}\ ,\\
P\big(\text{$B$ wins}\big)&=\sum_{i=1}^6\frac{3(6-i)\prod_\limits{k=0}^{i-1}f_k}{(22-i)(7-i)}\ ,\\
P\big(\text{$C$ wins in round $i$}\big)&=\frac{(19-i)(6-i)\prod_\limits{k=0}^if_k}{6(22-i)(7-i)}\ ,\\
P\big(\text{$C$ wins}\big)&=\sum_{i=1}^6\frac{(19-i)(6-i)\prod_\limits{k=0}^if_k}{6(22-i)(7-i)}\ .\\
\end{align}
Here is some Magma code to carry out the calculation:
pa:=0;
pb:=0;
pc:=0;
f:=1;
for i in [1..6] do
    pa:=pa+f/(7-i);
    pb:=pb+3*(6-i)*f/((22-i)*(7-i));
    pc:=pc+(19-i)*(6-i)*f/(6*(22-i)*(7-i));
    f:=5*(19-i)*(6-i)*f/(6*(22-i)*(7-i));
 end for;
 print pa,pb,pc,pa+pb+pc;

If you copy and paste this into the online Magma calculator it returns the values
\begin{align}
P\big(\text{$A$ wins}\big)&=\frac{773347}{1551312}\\
P\big(\text{$B$ wins}\big)&=\frac{14789}{57456}=
\frac{399303}{1551312}\\
P\big(\text{$C$ wins}\big)&=\frac{189331}{775656}=\frac{378622}{1551312}\ .
\end{align}
