

Subject:
probability of winning in a local raffle ticket draw
Category: Science > Math Asked by: pepperosityga List Price: $40.00 
Posted:
27 Oct 2005 04:48 PDT
Expires: 26 Nov 2005 03:48 PST Question ID: 585592 

Subject:
Re: probability of winning in a local raffle ticket draw
Answered By: elmartoga on 27 Oct 2005 18:21 PDT Rated: 
Hello pepperosity! The answer given by txags911 is quite close, but not exactly correct. According to the lottery rules shown in the web page you mentioned, the procedure for awarding prizes is: 1) Draw 1 ticket from the 85,000. That's the winner of one prize 2) Return ticket to pool 3) Repeat (1) and (2) 4,255 times, one for each prize. As txags911 mentioned, the easiest way to calculate the probability of winning at least one prize is to calculate the probability of winning no prizes at all, and subtract that value from 1: Prob. of winning 1 or more prizes = 1  (Prob of winning 0 prizes) Now, for each ticket draw, since you have 5 tickets, the probability of NOT winning the prize for that draw is 84,955/85,000 = 0.999941... In order for you to win no prizes, you would have to lose each of 4,255 draws. The probability of this happening is 0.999941^4255, which gives 0.778566... So now we plug this value in the previous formula to get: Prob. of winning 1 or more prizes = 1  0.77856... = 0.221433... Therefore, if you purchased 5 tickets, the probability of winning at least 1 prize is approximately 22.1433% Additional Information The answer for this question is an application of the binomial distribution. The binomial distribution gives the probability of obtaining a number X of "successes" out of an experiment that is repeated N times, with the probability of success held constant across the experiments. In this lotterty case, the experiment is repeated 4,255 times, and your chance of winning each of the times is 5/85000= 0.0000588235. Following is a link to a binomial probability calculator, which you will be able to use to find out, if you are interested, the probability of winning exactly 1 prize, exactly 100 prizes, more than 3 prizes, etc. http://www.swogstat.org/stat/public/binomial_calculator.htm [Enter 4255 for N, and 0.0000588235 for p] I hope this helps! If you have any questions regarding my answer, please don't hesitate to request a clarification. Otherwise I await your rating and final comments. Best wishes! elmarto 
pepperosityga
rated this answer:
Thank you Elmarto. I had to read through the question a few times s l o w l y in order to fully grasp the laws of probability but I got there in the end thanks to your very informative answer. Thanks again. 

Subject:
Re: probability of winning in a local raffle ticket draw
From: txags911ga on 27 Oct 2005 08:59 PDT 
With one ticket, the odds of losing are (850004255)/85000 = 94.99% Second ticket, just remove one of the overall tickets to get (849994255)/84999 = 94.99% Three more times yield percentages of 94.99% as well. So many tickets, that the odds don't change as you go down too much. So the odds of losing four times is (.9499)^5 = 77.34% chance. That leaves a 22.66% chance of winning at least once. 
Subject:
Re: probability of winning in a local raffle ticket draw
From: kmcleanga on 27 Oct 2005 10:38 PDT 
Txags991 is close, and the question is not well defined; however, but you are failing to account for wining multiple prizes. Winning on 2,3,4 or all 5 prizes at one time should only be counted as 1 successful try. Assuming the "winning a prize" is to mean winning at least one prize. If the statement "winning a prize" is means winning ONLY ONE prize, then you would have deduct all the multiple winnings (2,3,4 or 5) from the successful try list and the probability would decrease further. Since this sounds like a homework problem I will leave it at that. kmcleanga 
Subject:
Re: probability of winning in a local raffle ticket draw
From: txags911ga on 27 Oct 2005 12:42 PDT 
No, I think my answer is exactly what the question is looking for. If you win five times, then you have "won a prize." And I absolutely accounted for the possibility of winning multiple prizes. W\Essentially, rather than figuring out the odds of winning exactly one prize, adding the odds of winning exactly two, and so on, I just figured the odds of winning no prizes. There are six cases that can occur: 1. Win exactly 0 prizes. 2. Win exactly 1 prizes. 3. Win exactly 2 prizes. 4. Win exactly 3 prizes. 5. Win exactly 4 prizes. 6. Win exactly 5 prizes. Each has some probability of occurring, and assuming the question is asking for the probability of winning at least one prize, all you have to do is figure out the probability of the one negative case (#1 above), and subtract it from 100%. So the probability of losing 5 times in a row is 77.34% (.9499^5). That means the other 5 cases combined have a 22.66% chance of occurring. 
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