combinatorics - x tickets, y prizes, can only win once - what is probability given distribution of tickets

Given this problem:

There is a draw for x prizes with y tickets sold. Each person can own 1 or more tickets but can only win 1 prize (i.e. Once 1 of their tickets wins all of their other tickets are void).

I have seen a similar question asked before but have not really seen a good answer.

If I know the number of people and the number of tickets that each person has how would I calculate the probability of winning a prize for each person?

For example:

6 prizes to be won
25 tickets sold
2 people have 4 tickets
2 people have 2 tickets
13 people have 1 ticket each

How do I calculate the probability that 1 of the people with 4 tickets will win a prize, given that each person can only win once.

My starting point was:

$1 - \frac{\binom{21}{6}}{\binom{25}{6}}$

which would tell me the probability of the person who has 4 tickets winning. However this is naive and is only correct when there are 21 people each with 1 ticket (the numerator in this fraction) whereas I have 13 people with 1, 2 people with 2 and 1 other person with 4 .

Because there are other people who have more than 1 ticket I need to calculate how many of the $\binom{21}{6}$ selections are invalid (i.e. contain more than 1 ticket for a player)

To do this I found all of the selections that didn't have more than 1 ticket for a player:

$\binom{13}{6} + \binom{13}{5}\binom{2}{1} + \binom{13}{5}\binom{2}{1} + \binom{13}{5}\binom{4}{1} + \binom{13}{4}\binom{2}{1}\binom{2}{1} + \binom{13}{4}\binom{4}{1}\binom{2}{1} + \binom{13}{4}\binom{4}{1}\binom{2}{1} + \binom{13}{3}\binom{4}{1}\binom{2}{1}\binom{2}{1} =30,888$

Then I found the selections which did have multiple tickets for the same player:

Selections with multiple player tickets = $\binom{21}{6} - 30,888 = 23,376$

and then I used this in my initial example:

$1 - \frac{30,888}{(\binom{26}{6} - 23,376)}$

which gives me a probability of 79.99% that one of the players with 4 tickets wins one of the 6 prizes.

I am not 100% sure I haven't made a mistake and Im also sure there must be a better way to calculate this.

Would be interested to know peoples ideas.

EDIT
My suggested solution is incorrect as it assumes that a selection containing more than 1 ticket for a person results in no prizes awarded (i.e. is invalid) but actually if there are 4 tickets drawn owned by different people and then the final 2 tickets are for the same person there are still 5 prizes awarded but there would be an extra ticket(s) selected to award the 6th prize. As comment below states the order is therefore important.