20050209, 04:50  #1 
Sep 2002
89 Posts 
ARE THE ODDS CORRECT..Please help
According to NY state lottery http://www.nylottery.org/ny/nyStore/...avRoot_300.htm
the odds for winning are 1 in 1:45,057,474 for picking 6 numbers 159. I've figured that out and its ok but since they give you 2 plays for a dollar they say the odds of winning on a dollar are 1 in 1:22,528,737 am I wrong in assuming this is wrong I feel the odds are 2 in 45,057,474 which I don't believe is the same thing. I was never strong in probability but i feel the statement they are making is wrong. Could someone please help? L.P.Murray Last fiddled with by lpmurray on 20050209 at 04:55 
20050209, 07:31  #2 
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
5^{2}×17 Posts 
The odds should be okay as long as you make sure that the two bets are not identical.
OTOH if you pick numbers at random, there is always a (small) chance that the two plays will be identical, and the probability will be slightly different. 
20050209, 07:50  #3 
Sep 2002
59_{16} Posts 
Its just that with their way of thinking if you pick 4 numbers the odds go down to 1 in 11 million if you pick 8 it goes to 1 in 6.5 millon and so on. With their way of thinking if you pick 22million numbers you have a 1 in 2 chance to win but the way I figure it you have a 22million out of 45million chance of winning which is completely different. Am I missing something here? We both can't be right????
If their wrong I would like to point it out to them....... 
20050209, 10:30  #4 
Sep 2002
Oeiras, Portugal
11×131 Posts 
The chance of winning is given by the ratio between the number of plays and the total number of possibilities (for 6 out of 59 numbers and just one play that is equal to 1/combin(59,6)=1/45,057,474 ). So, if you play 2 times 6 numbers, the odds are indeed the double (2/combin(59,6), assuming the numbers are different and havenĀ“t been chosen at random, as Patrick pointed out. Therefore, if you play 22528737 million times 6 numbers your odds will indeed be 1 in 2.
When in your post you write "pick 2 (or 4 or 8 or 22 millions) numbers", I assume that you are referring to number of plays, 6 numbers each. In fact if you were picking in a single play more than 6 numbers that would correspond to a larger number of plays. In general, if you play with x numbers (x>6) the probability of winning is given by combin(x,6) / combin(59,6). Last fiddled with by lycorn on 20050209 at 10:42 
20050209, 10:38  #5  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
As long as there is only one prize and you select DIFFERENT plays, the odds add just as fractions: 1/N + 1/N = 2/N = 1/(N/2) In fact, if you were to purchase 45,057,474 DIFFERENT plays, you would be certain to win. (An Australian syndicate did that once  The lottery was changed to make it logistically impossible for them to do it again.) 

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