The possible chances lie between 17.7% and 18.7%.
The worst case occurs when everybody but you has exactly one entry in the lottery: this is a configuration consistent with the data (although unlikely!).
Let's count the number of possibilities in which you do not win. This is the number of ways of drawing $25$ tickets out of the $784-6$ remaining tickets, given by the Binomial coefficient $\binom{784-6}{25}$. (It's a huge number). The total number of possibilities--all of them equally likely in a fair drawing--is $\binom{784}{25}$. The ratio simplifies to $(784-25)\cdots(784-30) / [(784)\cdots(784-5)]$, which is about 82.22772%: your chances of not winning. Your chances of winning in this situation therefore equal 1 - 82.22772% = 17.7228%.
The best case occurs when there are as few individuals involved in the lottery as possible and as many as possible have $6$, and then $5$, etc, tickets. Given that the "gem" counts are $(42, 72, 119, 156, 178, 217)$ (in ascending order), this implies
At most $42 = a_6$ people can have $6$ entries each.
At most $72-42=30 = a_5$ people can have $5$ entries each.
...
At most $178-156=22 = a_2$ people can have $2$ entries each.
$217-178=39 = a_1$ people have $1$ entry each.
Let $p(\mathbf{a}, l, j)$ designate the chance of winning when you hold $j$ (between $1$ and $6$) tickets in a lottery with data $\mathbf{a}=(a_1,a_2,\ldots,a_6)$ and $l=25$ draws. The total number of tickets therefore equals $1 a_1 + 2 a_2 + \cdots + 6 a_6 = n$. Consider the next draw. There are seven possibilities:
One of your tickets is drawn; you win. The chance of this equals $j/n$.
Somebody else's tickets are drawn. The chance of this equals $(n-j)/n$. If they hold $i$ of them, then all $i$ tickets are removed from the lottery. If $l \ge 1$, drawing continues with the new data: $l$ has been decreased by $1$ and $a_i$ has been decreased by $1$ as well. The chance that some person with $i$ tickets in the lottery is chosen, given that yours are not, equals $ia_i/(n-j)$. This gives six disjoint possibilities for $i=1,2,\ldots,6$.
We add these chances because they partition all outcomes with no overlap.
The calculation continues recursively down this probability tree until all the leaves at $l=0$ are reached. It's a lot of computation (about $25^6$ = 244 million calculations), but it only takes a few minutes (or less, depending on the platform). I obtain 18.6475% chances of winning in this case.
Here's the Mathematica code I used. (It is written to parallel the preceding analysis; it could be made a little more efficient through some algebraic reductions and tests for when $a_i$ is reduced to $0$.) Here, the argument a does not count the $j$ tickets you hold: it gives the distribution of counts of tickets everyone else holds.
p[a_, l_Integer, j_Integer] /; l >= 1 := p[a, l, j] = Module[{k = Length[a], n},
n = Range[k] . a + j;
j/n + (n - j)/n ParallelSum[
i a[[i]] / (n - j) p[a - UnitVector[k, i], l - 1, j], {i, 1, k}]
];
p[a_, 0, j_Integer] := 0;
(* The data *)
a = Reverse[Differences[Prepend[Sort[{42, 72, 119, 217, 156, 178}], 0]]];
j = 6; l = 25;
(* The solution *)
p[a - UnitVector[Length[a],j], l, j] // N
As a reality check, let us compare these answers to two naive approximations (neither of which is quite correct):
25 draws with 6 tickets in play should give you around 6*25 out of 784 chances of winning. This is 19.1%.
Each time your chance of not winning is about (784-6)/784. Raise this to the 25th power to find your chance of not winning in the lottery. Subtracting it from 1 gives 17.5%.
It looks like we're in the right ballpark.
