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March 16th, 2007, 04:10 PM  #1 
Joined: Mar 2007 Posts: 1 Thanks: 0  Probability of Winning
Hello, OK, there is this home and car lottery in my area. They state that the odds of winning are 1 in 15. I assume that all they are doing is using the total number of prizes over total number of tickets to generate this result. For example 1,000 prizes to be won with 15,000 tickets being sold. This would equal a 1 in 15 chance of winning. Now my question is if I buy 3 tickets what are my odds and probability of winning. I thought that I could just use 1/15 + 1/15 + 1/15 = 3/15 or 1 in 5 chance. But then by that logic if I bought 15 tickets I should have odds of 1:1 which I know is not possible. What am I doing wrong???? 
March 16th, 2007, 04:24 PM  #2 
Joined: Dec 2006 Posts: 1,111 Thanks: 0 
That's right; If you buy 15 tickets, you are theoretically guaranteed to win. However, since theory isn't always reality, it's possible that by some accident you still won't get any prize, and somebody else who bought 15 tickets will wind up with two prizes. But, yes, you do theoretically have a 1:1 chance of winning if you buy 15 tickets. Of course, you still usually end up losing, because probability games are based on the idea that, say, each ticket costs $1, and the chance of winning is 1 in 15, but the prize is $10. Thus, even if you win lots of prizes, you still loose. It's just another example of the economic law of TANSTAAFL: There Ain't No Such Thing As A Free Lunch. 
March 16th, 2007, 05:15 PM  #3 
Joined: Mar 2007 From: UK Posts: 21 Thanks: 0 
If the chances of a ticket winning are 1/15, then the chances of it not winning are 14/15 Assuming that more one ticket can win at the same time (if there are 1000 prizes this should be possible), first calculate the probability that none of your tickets wins. This is (14/15)(14/15)(14/15) = 2744/3375. The probability that at least one of your tickets wins is therefore 1  2744/3375 = 631/3375. This is not equal to 1/15 + 1/15 + 1/15. If you buy 15 tickets, the chances of winning (using the above method) turn out to be 1  (14/15)^15 = approximately 0.645. And this is not equal to 1. :P (However it is higher than 631/3375, which is approximately 0.187. Therefore buying more tickets does give you more chances of winning, which goes well with common sense. ) 
March 17th, 2007, 04:18 AM  #4 
Joined: Dec 2006 Posts: 1,111 Thanks: 0 
OK, but if we simplify the game using the same ratio, this time saying that there are only 15 tickets and 1 prize, then what you're saying is that buying 15 tickets would only give me a chance of winning of 0.645. It's pretty easy to see that that's false, since I am actually guaranteed to win. Of course, with more tickets and prizes, you are no longer guaranteed to win, but the probability of winning remains the same for a proportionate number of tickets. The probability of not winning is 1(number of tickets bought)/15. Thus, if we buy 15 tickets, we have a 1  1 = 0 probability of not winning. 
March 17th, 2007, 06:53 AM  #5  
Joined: Dec 2006 From: St. Paul MN USA Posts: 37 Thanks: 0  Quote:
 
March 17th, 2007, 09:44 AM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,019 Thanks: 917 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Let's say that exactly 15,000 tickets will be sold* and that 1,000 distinct tickets will get prizes. Then buying one ticket gives a you the advertised 1/15 chance of winning. If you buy 15, your expected number of prizes is 1. Your odds of winning at least one prize is 1(14/15)^15, or about 64.47%. This is made up for by your roughly 26.41% chance of winning more than once. To guarantee at least one win you must buy at least 14,001 tickets. * That is, if you buy more tickets someone else buys less to keep the total the same. 
March 17th, 2007, 10:44 AM  #7  
Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
By the way, what are the odds of that actually occuring?  
March 17th, 2007, 10:54 AM  #8  
Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
 
March 17th, 2007, 03:10 PM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,019 Thanks: 917 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Otherwise, there are 14000 places the single winning ticket could go, so there are an additional 14000!/(15000^13999 * 1!) * 14000 possibilities, for a total chance of about 1 in 6.5e6488.  
March 17th, 2007, 03:16 PM  #10 
Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
In other words, in order for only one of your tickets to win, you would have to be the most unlucky person in the universe .


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