There are $m$ people. The $i^{th}$ person buys $t_i$ tickets. All tickets are put in a box.
A ticket is drawn uniformly at random. The person who brought that ticket wins the $1^{st}$ prize. Henceforth this person is ineligible to win additional prizes; if any tickets owned by this person are drawn in the future, they will be discarded. This procedure is repeated till all $m$ people get a prize.
What is the probability that the $i^{th}$ person will get the $n^{th}$ prize? I can only think of the naive solution where you look at all $m!$ possible outcomes and calculate the probabilities using that information. But is there a smarter way to calculate this, possibly by computer? What if we need this information for all $i$ and all $n$? Will the naive approach be the most efficient in this case?
The following three questions have discussed this before, but are not general enough, since they are concerned only with winning any prize, but not with the prize number:
I doubt that there's a very compact formula. Let $T$ be the total number of tickets sold. Then the probability that person $p$ wins the second prize is $$\sum_{k=1 \\k \ne p}^n{\frac{t_p}{T-t_k}}\cdot\frac{t_k}{T} = \frac{t_p}{T}\sum_{k=1 \\k \ne p}^n{\frac{t_k}{T-t_k}}$$
This doesn't look so bad, because one can define $$P_1 =\sum_{k=1}^n{\frac{t_k}{T-t_k}},$$ and then the above probability becomes $$\frac{t_p}{T}\left(P_1- \frac{t_p}{T-T_p}\right)$$
To get the probability that person $p$ wins third prize, you'll have to condition on which pair of other other people won the first two prizes, and while you may be able to write a compact formula, it will still involve a lot of computation. It will just get worse as you go on.
NOTE: If someone who reads this knows how to typeset the sum, so that $k \ne p$ is set below $k=1,$ instead of alongside, please do so, so that I can learn how. I tried to do it with \underset, but I couldn't make it work.