<s>[INST] Formula for calculating the odds per user of winning in a raffle each player can win once Am trying to write a program which gives each user in a raffle contest their odds so far at winning.
The rules of the game are simple: 
A predefined number of tickets are sold for example 1000 each user can at most buy 50 tickets and a user can only win once in the game after that all his tickets become invalid or pulled out and a new draw begins giving the other users a better chance of winning. The game will have 10 winners only. 
With the data collected from the software I would then know how many tickets were sold, the users that bought them and the amount of tickets that each user has. Now I just need a formula or pseudocode that would give each user their statistic probability of winning based on the data acquired, so that it be can used before and after each draw in the game to show each user their odds so far.
I have looked at similar questions asked here, but no one seems to want to address the part that if a person wins the rest of their tickets become invalid. Am not good with probability or understand those fancy notations, so I don't understand is such a thing possible to calculate per user.
Thanks for the help
Update
Testing my understanding of joriki second method:
lets say 10 tickets were sold to 4 users each bought A: 1, B: 2, C: 4, D: 3 
and there will be 3 prizes given to users.
I calculated the total probability of being drawn for each user to be 
A = $\frac{1}{10} + \frac{2}{10}*\frac{1}{8}*\frac{1}{6} + \frac{4}{10}*\frac{1}{6}*\frac{1}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.1482
B = $\frac{1}{10}*\frac{2}{9}*\frac{2}{8} + \frac{2}{10} + \frac{4}{10}*\frac{2}{6}*\frac{2}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.3817
C=  $\frac{1}{10}*\frac{4}{9}*\frac{4}{8} + \frac{2}{10}*\frac{4}{8}*\frac{4}{6} + \frac{4}{10} + \frac{3}{10}*\frac{4}{7}*\frac{4}{4}$ = 0.6603
D=  $\frac{1}{10}*\frac{3}{9}*\frac{3}{8} + \frac{2}{10}*\frac{3}{8}*\frac{3}{6} + \frac{4}{10}*\frac{3}{6}*\frac{3}{2} + \frac{3}{10}$ = 0.6500
Their total sum is 1.8403 and not 3 ? also is this considered the total probability of being drawn for the 3 draws or just for the first round of the game with the tickets becoming invalid
 [/INST] Let $N_i$ be the number of tickets that the $i$-th player bought.  Then the total number of possible draws of $m$ tickets is the coefficient of $x^m$ in the polynomial
$$
P(x)=\prod_i\left(1-N_i x\right);
$$
the number of possible draws where the $j$-th player doesn't win is the coefficient of $x^m$ in
$$
Q_j(x)=\frac{P(x)}{1-N_j x}=\prod_{i\neq j}\left(1-N_i x\right);
$$
and the winning probability for the $j$-th player is $1$ minus the ratio of the two coefficients.  Because you only need to keep coefficients out to the $m$-th power of $x$ during the calculation, you can compute the exact winning probabilities for all players in time proportional to the number of players and the square of the number of tickets to be drawn.

To demonstrate this (generating function) method for OP's example, where players A,B,C,D buy $1,2,4,3$ tickets respectively:
$$
\begin{eqnarray}
P(x)&=&(1-x)(1-2x)(1-4x)(1-3x) \\ &=&1 - 10x + 35x^2 - 50x^3 + 24 x^4; \\
Q_A(x)&=&\frac{P(x)}{1-x}=1-9x+26x^2-24x^3; \\
Q_B(x)&=&\frac{P(x)}{1-2x}=1-8x+19x^2-12x^3; \\
Q_C(x)&=&\frac{P(x)}{1-4x}=1-6x+11x^2-6x^3; \\
Q_D(x)&=&\frac{P(x)}{1-3x}=1-7x+14x^2-8x^3.
\end{eqnarray}
$$
What's nice is that these polynomials encode the probabilities of winning for every possible number of tickets.  If one ticket is drawn, the probabilities come from the coefficients of $x^1$:
$$
p_A = 1-(-9)/(-10) = 1/10, p_B = 1-(-8)/(-10)=1/5, \\p_C=1-(-6)/(-10)=2/5, p_D=1-(-7)/(-10)=3/10,
$$
which are obviously correct and sum to $1$.  If two tickets are drawn, the probabilities come from the coefficients of $x^2$:
$$
p_A = 1-26/35=9/35, p_B=1-19/35=16/35, \\p_C=1-11/35=24/35, p_D=1-14/35=3/5;
$$
these sum to $2$, as they should.  Finally from the coefficients of $x^3$, we find the probabilities of winning when three tickets are drawn:
$$
p_A=1-(-24)/(-50)=13/25, p_B=1-(-12)/(-50)=19/25, \\p_C=1-(-6)/(-50)=22/25, p_D=1-(-8)/(-50)=21/25,
$$
which sum to $3$.  (And from the coefficients of $x^4$, when four tickets are drawn, each player wins with probability $1-0/24=1$.)
In an actual application, where the number of players was much larger than the number of tickets to be drawn, each polynomial multiplication would be truncated after the $x^m$ term.
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