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 $j_l=i$
Best Answer
(10*.3 + 100 * .7 + 50 *.8) / 160 = 113/160 = 70.625 %
that said, there are many unknowns perhaps that can help us model the question better.