Hi Tony F.,
A fairly standard probability problem. The difficult part is deciding exactly which tickets to use as you calculate along.
For the first part, you need the probability of NOT drawing all three of your tickets BEFORE the last 4 remain. [That's fairly standard for these types of probability problems; you have to go at them "backwards", so to speak, b/c the calculations for drawing various combinations of cards including some of yours over a long series of choices becomes otherwise mathematically too cumbersome.] But it's OK for this if 2 of your tickets ARE drawn before that, b/c you would still have 1 left. So on the first 2 drawings, you don't care if your first 2 tickets go! But then, you have to keep your remaining ticket out of the pot! The combined probability is a sequential multiplication of the odds at each drawing step:
So that is: (19/19)*(18/18)*(16/17)*(15/16)* .... *(4/5). :There are now 4 tickets left, one of which is one of yours. Total odds are 4/17 (all the other numbers cancel each other out).
You should be able from the reasoning above, to continue the multiplication series over the next 3 terms, to answer the second part. There is NO drawing for the last bottle per se (although you might think, from the progression of terms, that it should be *(0/1)), you have the last ticket, and it is the only one left!
So if chance favors you (and the laws of the state permit), enjoy your prize(s).
-- Cheers, -- Mr. d.