I'm a grade 12 student, in Toronto, Ontario, taking the course MGM4U. As part of an assignment, we have to determine the different number of tickets that could win each of the prizes, and then determine the expected loss given average amounts won (this involves first finding the probabilities). Up to this point I've been fine with counting strategies, and I got near-perfect on the test. Unfortunately, this assignment's a real bugger. I found a site that's helped me A LOT:
http://icarus.mcmaster.ca/fred/Lotto/
But it's still not enough, as it doesn't quite explain everything as clearly and fully as I think it should. Could someone please help me, and try to help make more sense of that site?
This is what I have so far:
n(different tickets that can win 1st prize) = 1 x 1 x 1 x 1 x 1 x 1 x 43 = 43 different tickets
n(different tickets that can win 2nd prize) = (1 x 1 x 1 x 1 x 1 x 42 x 1) x 6 = 2158 different tickets
n(different tickets that can win 3rd prize) = (1 x 1 x 1 x 1 x 1 x 42 x 42) x 6 = 11352
n(different tickets that can win 4th prize) = (1 x 1 x 1 x 1 x 45 x 44 x 43) x 6P2 = 2554200
n(different tickets that can win 5th prize) = (1 x 1 x 1 x 46 x 45 x 44 x 43) x 6P3 =
469972800
6P2 means permutation (6,2), or 6!/(6-2)!= 6!/4!. It's the same thing with 6P3.
Although, I know I have to fix the last few because only very recently did I remember that the sets aren't disjoint in this case (in other words, some of the combinations I have listed could actually win more than one possible prize, which isn't possible--and I still have to fix that, among other things).
Any help would be much appreciated.
http://icarus.mcmaster.ca/fred/Lotto/
But it's still not enough, as it doesn't quite explain everything as clearly and fully as I think it should. Could someone please help me, and try to help make more sense of that site?
This is what I have so far:
n(different tickets that can win 1st prize) = 1 x 1 x 1 x 1 x 1 x 1 x 43 = 43 different tickets
n(different tickets that can win 2nd prize) = (1 x 1 x 1 x 1 x 1 x 42 x 1) x 6 = 2158 different tickets
n(different tickets that can win 3rd prize) = (1 x 1 x 1 x 1 x 1 x 42 x 42) x 6 = 11352
n(different tickets that can win 4th prize) = (1 x 1 x 1 x 1 x 45 x 44 x 43) x 6P2 = 2554200
n(different tickets that can win 5th prize) = (1 x 1 x 1 x 46 x 45 x 44 x 43) x 6P3 =
469972800
6P2 means permutation (6,2), or 6!/(6-2)!= 6!/4!. It's the same thing with 6P3.
Although, I know I have to fix the last few because only very recently did I remember that the sets aren't disjoint in this case (in other words, some of the combinations I have listed could actually win more than one possible prize, which isn't possible--and I still have to fix that, among other things).
Any help would be much appreciated.
