It depends on whether it’s possible for you to buy two tickets with the same number(s). If the numbers must be different, then you add odds; if there’s a possibility of duplication (though the nubers are still random and chosen independently of each other), then you multiply the odds – and you also have a change of winning 2 or more first prizes.
Assuming all numbers are different, your chance of winning is equal to the number of tickets you purchase divided by the number of possible combinations. So 10/195M, 100/195M, 1/195K, and 1/195.
Your ‘odds of losing’ line is incorrect because you’re buying tickets for the same draw.
Except that I don’t know of any lottery that works that way.
Has anyone heard of a case where someone filled out a slip for some specific numbers, and then the computer rejected it because someone else already bought a ticket with those numbers?
Actually, I don’t think such a thing is even possible (except for scratch-off cards and the like, where the customer cannot choose his numbers). The reason is that the computers are not fast enough to check and see if someone might have bought those numbers elsewhere a half-second ago.
(Here’s my proof that the computers are not fast enough: If they were fast enough, then a half-second after the numbers are drawn, the computer could tell us whether or not anyone won. But they aren’t that fast, and we have to wait several hours until the computers look at all the numbers and tell us if there was a winner or not.)
Ooops. I just realized a much simpler way to prove that duplicate numbers are indeed sold: Sometimes several people all win and share the jackpot!