<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] I don't think you'll find a closed formula for this. If you're OK with only specifying the odds up to a couple of decimal places, the easiest way to do this would be to simulate the draw a couple of billion times and use the resulting relative frequencies. If you want the exact probabilities instead, you can keep track of all possibilities of who already won and work through one draw at a time. For instance, if $A$, $B$ and $C$ bought $1$, $2$ and $3$ tickets, respectively, and $2$ prizes are being drawn, in the first step $A$ would be drawn with probability $\frac16$, $B$ with probability $\frac26=\frac13$ and $C$ with probability $\frac36=\frac12$. Then in the second step, if $A$ was drawn in the first step, there would be $5$ tickets left, and $B$ would be drawn with probability $\frac25$ and $C$ with probability $\frac35$, and so on. The total probability of being drawn would be $\frac16+\frac26\cdot\frac14+\frac36\cdot\frac13=\frac5{12}$ for $A$, $\frac16\cdot\frac25+\frac26+\frac36\cdot\frac23=\frac{11}{15}$ for $B$ and $\frac16\cdot\frac35+\frac26\cdot\frac34+\frac36=\frac{51}{60}$ for $C$. For debugging your code, you can check to see that they add up to $2$ as they should.
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