# Odds of winning a prize in a weighted, random raffle

There is a raffle a state holds annually to assign salmon fishing licenses to fisherman; in a specific stetch of river, to control harvest and limit access, to ensure resource management and a high quality recreational experience.

There are 772 licenses available up for grabs in the draw. There are 1,256 applicants in the draw. The draw is random and sequential. However, there is a catch: applicants can apply as a 'party' of up 5 members. If a 'party' is drawn and sufficient licenses remain at that trial of the drawing, all members of the party get licenses. Then, the amount of licenses available for the next round of the drawing is reduced by the number awarded; and so on, until all 772 licenses are awarded. If on the last trial less licenses remain than the number in the party, the state simply awards tags to all in that party and the draw is over. There is no way of knowing a priori how many of the 1,272 applicants are individual applicants or parties consisting of 2, 3, 4 or 5 members.

How would the odds of drawing a license be calculated for this problem? Many thanks, in advance!

• What is the context? (e.g. is this from a class where you just learned certain facts/methods/similar examples?) What are your thoughts about the problem? What have you tried/where did you get stuck? All of this information could help people help you. – Mark S. Mar 17 at 18:16
• I think you must make some assumption about the parties to make a sensible attempt at the problem. If everyone is an individual, my chances of getting a license are $772/1256\approx 61\%$ If everyone but me is a party of $5,$ my chances are $1/155<1\%$ – saulspatz Mar 17 at 18:23
• Its recreational problem solving; I work in a technical field and have advanced degrees in the sciences where I use mathematics daily; some friends asked me how to solve this problem for them and I was getting stuck at the part of the uknown quantity of party applications within the applicant pool. It seems the overall odds of drawing are 772/1,256 but the aspect of party applications and drawing without replacement could makes the odds far worse; and hence, an interesting problem. I wonder if there is some way to at least gets bounds on the best and worse case odds? – Paleo73 Mar 17 at 18:23
• SaulSpatz- that was sort of where I was headed :) – Paleo73 Mar 17 at 18:27
• When you said there are $1256$ applicants, what is an "applicant"? Is it a party (including possibly party-of-1)? Or is it a single person? In the former case there can be as many as $1256 \times 5$ people involved, while in the latter case there can be as few as $252$ parties involved. Which do you mean? – antkam Mar 18 at 21:13