Probability of winning raffle with multiple tickets (max one win each)

Question

I have 90 tickets in a raffle.
20,000 people have bought tickets.
Each person has bought between between 60 and 100 tickets.
The total number of tickets sold is 1,360,000 giving an average of 68 tickets per person.
There are 5,000 prizes.
Each person can only win once, so their tickets are removed from the raffle as soon as they win a prize.

What are my chances of winning a prize given I have 90 tickets.
(The exact distribution of the ticket allocation is not known, so an assumption that everyone else but me has 68 tickets would be a close enough estimation)?

Answer 1

NAIVELY,

 

Probability of 1 win = 1 - Probability of zero wins

 

That is, the probability of winning the prize equals 1 minus the probability of losing.

 

Since you have 90 tickets, the probability of losing on the first pick is 1359910/1360000

 

Since the winning tickets are disqualified from winning again, the numerator and denominator

are reduced by an average of 68 after each pick.

 

So we are interested in the sum of the infinite series:

 

  1 -  SUM [90/(1360000-68*N)] ,   N=0,1,2,3,.....,5000

 

which is a finite arithmetic sequence

 

The formula is Sn = N( A1+An)/2 where N is the number of terms, A1 is the first term, and An is the last term

 

The first term is 90/1360000 = 9/136000

The 5000th term is 90/(1360000-68*5000) = 90/1020000 = 9/102000 = 3/34000 = 12/136000

 

The partial sum is then Sn = 5000*(9/136000 + 12/136000)/2 = 2500*21/136000 = 0.386029411764705882359....