Is a fair lottery possible? I'm trying to come up with a scheme for a lottery where each individual has roughly the same chance of becoming the winner, regardless of the number of tickets one holds. So no individual should have an incentive to buy a second ticket, but even if they decide to do so, it should not gain them any significant advantage over any of the other players.
Ofcourse, the simplest solution would be to just refuse to sell them multiple tickets, but since all players are anonymous, I cannot differentiate between ticketholders and tickets, so that's impossible.
A working solution would be to increase the house-edge exponentially (or decrease the jackpot) as soon as a new ticket is sold, so that buying another one lowers the expected profit more than is gained by having an extra ticket. No rational player would want do that, unless they have no previous tickets. But the problem with that is that only a very small amount of tickets can be sold untill the house has 100% chance of winning, so in practice nobody will ever win.
Is there a better formula possible? By dynamicly changing parameters like the ticket-price, jackpot, or win-chance with every new ticket? The lottery does not have to be economicly feasible, its no problem if the organizer always makes a loss from it for example.
I have a feeling I'm asking for the impossible, but since the proposal above would work fine for very small groups, there should be a formula which works in larger groups too.
 A: With the various restrictions in place the answer to the question as asked seems to me to be 'no, there is no such system'. 
It was said there is one prize to be distributed (one person to be selected) and it was also said the tickets could even be free. And, the lottery should be fair, so that different individuals have essentially the same chance to win they holding one or more ticktets regardless. And, of course the key point that one cannot distinguish tickets from individuals.
However, the prize can be adjusted in the process.  
So there are tickets $\{1, \dots, n\}$ distributed, and there is a partition of the set $\{1,\dots, n\}$ into disjoint nonempty sets $I_1, \dots, I_k$ corresponding to the tickets one individual holds. 
As described the only thing the organizer of the lottery knows is $n$. But, they have no control over the $I_j$. It is impossible to assign winning propabilties $p(i)$ to the different tickets so that $\sum_{i \in I_j}p(i)$, the probability individual $j$ will win, is about equal for all $j$ and each possible choice of $I_j$, except $p(i)= 0$ or $n\le 2$.
Now, the above assumes that tickets are distributed and then some 'drawing' is done possibly following some complicated system. 
If one does not insist on this there are of course ways to assign in a in some sense fair way one prize in a way that it is pointless or even bad to get more than one ticket (even if it is free). 
The "unique is the winner" was already mentioned, one could also do "first come, first served" (the first ticket sold always wins).
Finally, in OP and in some answers something else is discussed, which in my opinion is not really "fair" in the sense of the question. Merely, these would be systems to make it unreasonable to get more than one ticket but still if somebody would they would be in some sense better of than the others in the same game. Yet, yes, I think there one could (in a certain formalization) show that one needs at least exponential decrease of expected win to make this work.
A: Probably you can easily adapt the THE PACHIRA LOTTREE (proposed by Douceur & Moscibroda) to your needs. The authors state the postulate of the unprofitable Sybil attack which seems in agreement with your requirement of no gain from buying more tickets. 
A: I think that The simple lottery is an answer to your question. Since I have no background in the game theory, I certainly omit some details and the COMMENTs section below might be far from correct.
On second thought, also The unique-is-the-winner lottery below would be a solution, too.
=THE SIMPLE LOTTERY=
For the purpose of discussion I present more variants of the lottery.
THE GAME. Suppose we have $n=100$ players each of which chooses to buy $x=x_i$ tickets, each of which costs 1 dollar, where


*

*(v3) $x\in \{0,1,2,3\}$

*(vI) $x\in \{0,1,2,\dots\}$

*(vR) $x$ is a non-negative real number.


The lottery is made with each ticked having equal probability to win the price of $P=202.0202 \dots > 1$ dolars so that $i$-th player wins with probability $x_i/\sum_j x_j$.
ASSUMPTIONS. The number of players is well known. Some game-theory related assumptions might be needed like that the players are inteligent, their goal is to maximize gain expectation (average over probability, in the linear fashion),
the actual move of each player is secret while other information is well known etc. In particular I consider the player's goals uniformity and their public knowledge to be sensitive. (If there are players with different goals like keep-what-I-have-with-certainty or get-maximum-or-noghing, the game might possibly get disbalanced).
THE STRATEGY.
A game-theoretical result like a Nash equilibrium theorem provides us
(I hope; at least in the variant (v3)) with an optimal strategy for each player.
Case (vR) lacks compactness and might be a problem. (vI) is somewhat more compact.
The optimal strategy will be generally a mixed (i.e., probabilistic) strategy.
Because of the symmetry, the mixed strategies of individual players will be the same. 
COMMENT.
Though this is not important, I choosed the particular values so that, I guess, (some of) the optimal mixed stragies of the players are actually pure strategies of bying $x=2$ tickets. I think so because assuming(!) the other players play pure strategy of buing $a$ tickets and if I buy $x$ tickets, the outcome expectations is $E(x)=e(x)-x$ where $e(x) = \frac{P x}{(n-1)a + x}$ is win expectation and $x$ is the investment in the tickets.
I consider the variant (vR). The graph of $e(x)$ is an increasing hyperbola with $e(0)=0$ and $\lim_{x\to \infty} e(x) = P = 202.0202\dots$.
If $e'(0)\le 1$ (that is, $a \ge P/(n-1)$) I will buy $x=0$ tickets.
But this is not the equilibrium since the strategy is not optimal to other players
(their expactations are $E_i=P a/((n-1)a) - a = P/(n-1) -a \le 0$ and each of them should consider to buy e.g. $0$ tickets instead).
Otherwise the best pure strategy is to buy $x$ tickets where $e'(x)=1$ ($E'(x)=0$),
which means $x=x_0 = \sqrt{(n-1)a}(\sqrt{P}-\sqrt{(n-1)a})$. For the equilibrium strategy
we have $x_0=a$, which leads to $x_0=a=P(n-1)/n^2$.
If players play like that, the lottery maker gets $P(n-1)/n$ dolars and gives away $P$ dolars, thus paying price $P/n$ for attracting playres to participate in the procedure.
BTW, if $n=1$ and I am the only player, I invest one cent in the tickets and get full $P$.
Now back the mixed strategies: If think of changing my pure $x_0$ ticket strategy to some mixed strategy, I see the utility function is concave which makes the pure $x_0$ strategy
best among all mixed strategies. And more generally, if the strategy of other players is fixed (arbitrarily, mixed), then the utility function is strictly concave (because an average of strictly concave functions), making some pure strategy the optimal strategy. (This even sounds as a proof that the equilibrium strategies for (vR) are pure.) For variants (v3) and (vI), this suggest the optimal would be to mix the two integer values that are next the calculated $x_0$. 
=THE ZERO LOTTERY.=
(For the sake of completness I have to note that also the zero lottery seems to be a solution to your problem. The rules are "No one wins." Then, obviously, the chances are equal. That is to clear that you wish to get positive probability that someone wins. May be you wich that probability to be $1$.)
=THE UNIQUE-IS-THE-WINNER LOTTERY=
This lottery has nonzero probabity of no-winner, but it is simpler and more accessible.
Moreover, it can be repeated until a winner is found.
We have $n$ players which have option to by $0$ or $1$ tickets, which costs one dolar. A player wins $P=100$ dolars if he is the only one who bought the ticket. Of no one boght the
ticket, no one wins. If at least two tickets were sould, no one wins.
COMMENT.
Silly player like me chooses to buy a ticket with probability $1/n$. More clever one
perhaps buys with probability proportional to $P$. Even clever player will estimate  the price of the game to be about $P/n$, hence, for large $n$, he buys with probability smaller than $P/n$. For smaller $n$ he only observers the lottery is likely to repeat several times and he has no idea what the optimal strategy is. 
A: One can perhaps pick the winner by elimination. The algorithm would be :


*

*Pick a ticket at random and eliminate the holder. 

*Repeat (1) until only one ticket is left. 


In this scheme, players who buy more tickets would also have a higher chance of being eliminated. An issue with the scheme could be to determine the identity of the holder of the ticket which is picked for elimination in each step. The holder does not have an incentive to identify himself/herself. 
This scheme is similar to method is Peter Rindal's answer except that you do not eliminate based on personal characteristics.
A: How about something simple: make the tickets expensive (say 1000 Euros) and set the prize to be worth slightly less than two tickets (say 1950 Euros). There is incentive to participate and there is no incentive to buy more than one ticket.
