Is it possible to estimate the odds of winning a multi-entry contest, when I don't know the breakdown of entries? Suppose I am entered into a contest, with the following rules:


*

*Every person may get up to 6 entries

*All the entries will be pooled, and 25% of the entries will be selected to be winners, with a maximum of 25.

*Each person can only win once, regardless of the number of their entries. If someone's name gets drawn again, it is discarded and a new name drawn.

*I know how many entries I have (the maximum, 6)

*I know how many total entries there are, broken down by type of entry

*I do not know how many of the entries are repeat entries by the same person.


The count of entries by type is as follows:

Type 1: 42
  Type 2: 72
  Type 3: 119
  Type 4: 217
  Type 5: 156
  Type 6: 178

Is it possible to estimate my odds of winning in this situation?  I'm a bit confused by the fact that I can't predict how the early winners will effect my chances, since I don't know how many entries each winner will remove from the pool.
I'm interested in the solution given the data set, but I'm also interested in the proper procedure/algorithm for calculating it.
 A: 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.
A: If I did the math right, you have between 19.43% and 21.15% chance of winning a prize
The 19.43% is the best-case scenario, where every entrant has 6 tickets
The 21.15% is the worst-case scenario, where every entrant has 1 ticket except you
Both scenarios are extremely unlikely, so your actual odds of winning probably fall somewhere in between, however a roughly 1/5 chance at winning seems like a fairly solid number to go by
The details on how those numbers were obtained can be found in this Google spreadsheet, however to summarize how they were obtained:


*

*Start with Total # of Entries (784) and Your Entries (6)

*Get chance at winning (6 / 784 = 0.77%)

*Subtract 6 for best-case, or 1 for worst-case from TotalEntries

*Get chance of winning (6/778 for best case 6/783 for worst case)

*Repeat steps 3-4 until you have 25 percentages

*Add the 25 percentages together to find out your overall chance at winnning something


Here's an alternative way to get the approximate percentage that is simpler, but is not as accurate since you are not removing duplicate entries every time you draw a winner.
6 (your tickets) / 784 total tickets = 0.00765
0.00765 chance to win * 25 prizes = 19.14 % chance to win

EDIT: I'm fairly sure I'm missing something in my math and that you cannot simply add percentages like this (or multiply percent chance to win by # of prizes), although I think I'm close
Whobar's comment gives a 17.4% chance of winning, although I still need to figure out the formula he gave and make sure it's accurate for the contest. Perhaps a weekend project :)
