ARE THE ODDS CORRECT..Please help

[lpmurray]

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 1-59. 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

[patrik]

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.

[lpmurray]

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.......

[lycorn]

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 haven´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).

[Wacky]

Originally Posted by lpmurray

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.

Believe what you want, but they are the same thing. "Odds" are just a ratio and, like common fractions, are normally reduced to relatively prime integers.

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.)