# 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.

• It's gonna be hard, because the probability of getting a particular set of winners differs depending on the order, so your sample space is really all ordered sequence of winners, and the formula for the probability is gross. For example, if $T$ is the total tickets, and persons $i=1,2,\dots,x$ have $t_i$ tickets, the probability that the drawn winners will be persons $1$ to $x$ in order is $$\frac{t_1}{T}\cdot \frac{t_2}{T-t_1}\cdots\frac{t_x}{T-t_1-t_2-\cdots -t_{x-1}}$$ The denominator is going to be gross and depends on the order. The numerator do not. – Thomas Andrews Feb 2 '18 at 17:13
• With the EDIT, you have confused the problem. I have given a solution assuming that if you hold $4$ prize winning tickets, say, only $1$ prize is given to you, and $3$ prizes are cancelled. – true blue anil Feb 3 '18 at 6:29
• In continuation of previous comment, suppose $3$ tickets have to be added to restore the cancelled prizes, how will they be distributed ? Further, the redistribution could again result in someone holding more than one prize winning ticket, making it a horribly complicated for both the organizers and the players ! – true blue anil Feb 3 '18 at 6:41
• As per question, there are 25 tickets, 6 prizes and the distribution is I have 4, another person has 4, 2 people have 2 and the rest have 1 ticket. If we select 6 unique tickets first time then that is it however if 1 person has 2 winning tickets in the first 6 then we would keep drawing until we get another winner. That is why the order is important. In my solution above it would be correct if everyone else had 1 ticket but because some have more than 1 some of the combinations will require drawing extra tickets and so I actually may win more times. – user527452 Feb 4 '18 at 6:57
• e.g. in an extreme case (but a valid outcome) if the first 6 tickets all belong to 2 people (i.e. the person with 4 and 1 of the people with 2) then we only have 2 prizes distributed and would need to continue drawing for the final 4 prizes. In this case we would need to draw (at least) another 4 tickets and in some of those cases our ticket would be drawn however we are not counting those cases in my solution because I am assuming that only 6 tickets are ever drawn and that in the case of duplicates the whole draw is invalid. – user527452 Feb 4 '18 at 7:01

It isn't difficult to find out whether you get a prize, holding $4$ out of $25$ tickets.
and the only way you don't get a prize is by drawing all $4$ from the "non-winning" $19$ tickets,
thus P(you get a prize) $= 1 - \frac{\binom{19}4}{\binom{25}4} \approx 0.6936$
• I hold 4 tickets however there are 6 prizes so I am not sure your answer is correct is it? <br> Isn't it P(I win) = 1 - $\frac{\binom{21}{6}}{\binom{26}{6}}$ i.e. 1 minus the number of ways of picking 6 from all the tickets that are not mine over the number of ways of picking 6 tickets totally. However, I don't think it is correct as the order matters. However, the issue is that other people have more than 1 ticket and so I will actually have more chances because in some cases they will need to pick 7 ,8,9...tickets to get 6 unique winners. – user527452 Feb 4 '18 at 6:50
• Sorry, typo in above but can't edit anymore. should be: P(I win) = 1 - $\frac{\binom{21}{6}}{\binom{25}{6}}$ i.e. All the ways of picking 6 from 21 tickets (i.e. not including mine) over the total ways of picking 6 from all the tickets. Gives probability that I don't win and then 1 minus that gives me the probability I win. As above, I believe there are issues with this method though due to the fact that others also have multiple tickets and the fact that when there is a duplicate draw we keep drawing more. I therefore think this underestimates my chances. – user527452 Feb 4 '18 at 7:04
• OK, I follow your answer. I think we have both got to the same place just via a different approach: $\frac{\binom{21}{6}}{\binom{25}{6}} == \frac{\binom{19}{4}}{\binom{25}{4}}$ You either have 25 tickets with 6 total winners and you draw our 4 tickets from it or you have 25 tickets including our 4 winners and draw 6 winning tickets – user527452 Feb 5 '18 at 12:22
• Timeout on the comment again... OK, I follow your answer. I think we have both got to the same place just via a different approach: $\frac{\binom{21}{6}}{\binom{25}{6}} == \frac{\binom{19}{4}}{\binom{25}{4}}$ You either have 25 tickets including 6 winners and draw our 4 tickets from it (as you have done) or you have 25 tickets including our 4 tickets and draw 6 winning tickets from that. – user527452 Feb 5 '18 at 12:28