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
Best Answer
Let $ x_i $ be the number of tickets bought be each of the m people. Suppose n tickets are sold.
Person i has probability $P_{i,1} = x_i/n$ of being the first winner.
Person i has probability $P_{i,2}= (\sum_{j \neq i} P_{j,1}\frac{x_i}{n-x_j})=(\sum_{j \neq i} \frac{x_j}{n}\frac{x_i}{n-x_j}) $ of being the second winner.
$$P_{i,3}=(\sum_{j_p \neq i} \frac{x_{j_1}}{n}\frac{x_{j_2}}{n-x_{j_1}}\frac{x_i}{n-x_{j_1}-x_{j_2}}).$$
For i to be the third winner, you are multiplying the probabilities that i did not win the first 2 times. For each other person, $j_1$ you compute the chance that they were the first winner (already done in part 1,$\frac{x_{j_1}}{n}$ ), then you compute the probability that a second person, $j_2$, won in the second round, given $j_1$ won in the first ($\frac{x_{j_2}}{n-x_{j_1}}$). Finally you multiply by the chance that i wins in the third round, given $j_1$ won in the first and $j_2$ won in the second.
So $$P_{i,k}=(\sum_{j_p \neq i}\prod_{l=1}^{k-1}\bigg[\frac{ x_{j_l}}{n-\sum_{t=1}^{l-1}x_{j_t}}\bigg]\frac{x_i}{n-\sum_{t=1}^{k-1}x_{j_t}})$$
So for person i you must add together these probabilities from 1 to 10 (for 10 winners).
To break this down, you go over each case, multiplying the probabilities that all combinations of previous people won the first round and for each combination you multiply by the chance that person i wins this time, given that particular combination of people won . You add up all theses probabilities, and as I mentioned before multiply by the chance that person i was not selected before.
This formula can be written in a more intuitive way:
P(i wins)=$ \sum_{k=1}^{10} P_{i,k}$
$$P_{i,k}=(\text{probability i loses all prior draws}) \times \sum_{\text{all combinations of k-1 winners from all people except i}} P(\text{i wins kth round} | j_1,j_2,...,j_{k-1} \text{won the first k-1 rounds} ) \times P(j_1,j_2,...,j_{k-1} \text{won the first k-1 rounds}))$$
Let's do an example with 4 people, who bought 1,2,3 and 4 tickets respectively. 3 winners will be drawn.
$P_{i,1}=i/10$ (since person i bought i of the ten tickets).
$P_{1,1}=1/10=.1$
$P_{1,2}=(2/10)(1/8)+(3/10)*(1/7)+(4/10)*(1/6)=.1345$ (probability each other person won round 1 times the probability person 1 wins round 2 given the other person won round 1)
$P_{1,3}=(2/10)[(3/8)(1/5)+(4/8)(1/4)]+(3/10)[(2/7)(1/5)+(4/7)(1/3)]+(4/10)[(2/6)(1/4)+(3/6)(1/3)].2143$
So the probability that person 1 is one of the 3 winners is .1+.1345+.2143=.4488.
Alternatively, it may be easier to compute the probability of each scenario:
Let $W_{a,b,c}$ mean that the winners were, in order a, then b, then c.
So $P(W_{1,2,3})=(1/10)(2/9)(3/7)$.
For a game with m players, and k winners, in order to compute P(player 1 wins), using this method you compute every combination of $P(W_{j_1,j_2,\ldots j_k})$ that contains player 1, and add together those probabilities to find the probability player 1 wins. This seems easier than the method I proposed earlier. The number of computations this takes is: for $P(W_{j_1,j_2,\ldots j_k})$ takes 2k multiplications and n(n+1)/2 additions, and you must add together ${m \choose k-1}$ of them.
In general to compute solve the problem this way,
$P(W_{j_1,j_2,\ldots j_k})=\frac{x_{j_1}}{n}\frac{x_{j_2}}{n-x_{j_1}}\frac{x_{j_3}}{n-x_{j_1}-x_{j_2}}\cdots\frac{x_{j_k}}{n-\sum_{t=1}^{k-1}x_{j_t}}$
P(player i wins) = $\sum P(W_{j_1,j_2,\ldots j_k})$, where at least 1 of the $j_l=i$