Suppose there are 10 different people, each holding 1 lottery ticket, and I give each of them 10% chance of becoming the 'winner'.
Now let's say one of those 10 people (I dont know which one) decided to buy a second ticket.
I could choose at that point to lower every ticket's chances to 5% (effectively increasing the chance that nobody will win). Since that person now holds two tickets, which both give 5% chance, whoever it was didn't increase their chance of winning, since he still has 10% total chance, like before. Now suppose another ticket gets sold, I can lower every ticket's chances to 3.33%, and still the person who may hold 3 tickets, gained absolutely nothing from doing so.
The problem with this solution (although it works) is that very soon I have to lower the chances to such a tiny percentage, that its almost garantueed that nobody will win.
Is there a simple scheme possible to prevent this, or is this unavoidable?
After the OP's clarifications, I will answer the question for the following specific set up: There are $n$ persons and $n-1+k$ tickets, $k=1,...$ (up to some finite number). Initially $k=1$, and all persons buy one ticket. The price of each ticket is fixed at $v$, and it doesn't change. The winning ticket is decided by a "random draw", namely, all tickets always have the same probability of winning. All tickets take part in the draw, irrespective of whether they have been bought or not. Therefore the probability of a ticket winning is $p_k= \frac 1{n-1+k}$. The winner collects an amount of $W$ which can be variable.
Question: how can we decide on the value of $W_k$, as a function of the number of tickets, so that there is no incentive for somebody to buy an additional ticket, if one is issued?
For $k=1$ the expected profit of each one of the participants (we are ignoring issues of risk aversion and so we deal in terms of expected profits and not of expected utility) is the expected value of collecting the prize, minus the sure cost of buying one ticket: $$EP_{11} = p_1\cdot W_1-1\cdot v \qquad [1]$$
where the double index on $EP$ refers the first to the value of $k$ and the second to the number of tickets somebody holds.
Assume now that $k=2$. If somebody buys this ticket, his expected profit will now be $$EP_{22} = 2p_2\cdot W_2-2\cdot v \qquad [2]$$ while the expected profit of all others will be $$EP_{21} = p_2\cdot W_2-1\cdot v \qquad [3]$$
We want the prospective buyer of the 2nd ticket to have the same expected profit by buying the extra ticket as the expected profit he had when he had bought the first ticket, so we want to determine $W_2$ by equalizing equations $[1]$ and $[2]$:
$$EP_{11} = EP_{22}\Rightarrow p_1\cdot W_1-1\cdot v = 2p_2\cdot W_2-2\cdot v \Rightarrow W_2 = \frac {p_1W_1+v}{2p_2} \qquad [4]$$
So assume that we issue this extra ticket and we announce that now the prize is $W_2$ as determined by $[4]$. The buyers will do the calculation and they will find that they won't gain anything in terms of expected profit if they buy the extra ticket... does this mean that they won't buy it? No, it does not imply that. Because, the buyers have already invested in the lottery by buying the first ticket. So there is no point in comparing the expected profit if they buy the 2nd ticket with the expected profit they had before the extra ticket was issued, because it is no more relevant. What they will do is to compare the profit the will now have, in this new set of rules, if they buy the extra ticket and if they don't. Namely, being rational, they will compare $EP_{22}$ with $EP_{21}$. So if we as organizers calculate $W_2$ as above, then we have that
$$EP_{21} = p_2\cdot W_2-1\cdot v = p_2\frac {p_1W_1+v}{2p_2} - v = \frac {p_1W_1-v}{2} - =\frac 12EP_{11} = \frac 12EP_{22} < EP_{22}$$
So we see that, if we calculate $W_2: EP_{22} = EP_{11}$ we do not make the buyers indifferent in buying the extra ticket or not - on the contrary, buying the ticket is the only way to maintain the same expected profit as before - because if they don't, they will have expected profit $EP_{21}$ which is lower. So expect a lively bidding for this extra ticket! (and think how high you can set its price so that the buyers would still want to buy it).
If we want to make the players truly indifferent in buying the extra ticket we must equalize the expected profits
$$EP_{21} = EP_{22} \Rightarrow 2p_2\cdot W_2-2\cdot v = p_2\cdot W_2-1\cdot v$$ which is obviously possible only if
$$W_2 = v$$
...but if you do set $W_2$ like that, expect a lively beating instead of a lively bidding...
ADDENDUM
Following conversation with the OP in the comments, if we don't play around with the value of the prize and keep it fixed, then the "no-incentive-to-buy twice" result can materialize if a priori the announced rules of the lottery are that (denoting now $n$ the number of persons and $k$ the number of additional tickets above $n$):
a) The draw is "random" (equal probability per ticket)
b) Only bought tickets take part in the lottery
c) That the probability of winning depending on the number of tickets is such that
$$p_n=2p_{n+1} =3p_{n+2}...=(k+1)p_{n+k} \Rightarrow p_{n+k} = \frac {p_n}{(k+1)}$$ What is the consequence of such a rule? Since $p_n = 1/n$ we have that
$$p_{n+k} = \frac {1}{n(k+1)}$$
But the numbers of tickets taking part is $n+k$ so the total probability allocated to ticket buyers is $$\frac {n+k}{n(k+1)} < 1$$ In other words, except in the initial state with $n$ tickets, a "house probability arises", i.e. a probability that no-ticket wins. So for $k>0$, each ticket has the same probability of winning compared to the other tickets, but the "house" acquires necessarily a probability of winning too, and this probability is increasing in $k$ and tends to $(n-1)/n$ ($n$ is fixed here). This is a formalization of the initial idea described in the question.