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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: acinod on April 05, 2013, 10:52:29 pm

Title: Oz Lotto Probability
Post by: acinod on April 05, 2013, 10:52:29 pm: In the game of Oz Lotto, you pick 7 numbers.

Every Tuesday night, 7 main winning numbers and 2 supplementary numbers are chosen from the numbers 1 to 45.

You win a prize from a division if the numbers you chose match any of the following:

Division 1: All 7 main winning numbers.
Division 2: Any 6 main winning numbers and either supplementary number.
Division 3: Any 6 main winning numbers.
Division 4: Any 5 main winning numbers and either supplementary number.
Division 5: Any 5 main winning numbers.
Division 6: Any 4 main winning numbers.
Division 7: Any 3 main winning numbers and either supplementary number.

What is the probability on winning each division?

I know how to calculate Division 1; that's simple because it's 1/(45C7). But what about the other divisions? I don't really understand the formulas on the Oz Lotto site either (http://tatts.com/tattersalls/games/oz-lotto/game-rules-and-odds)
Title: Re: Oz Lotto Probability
Post by: b^3 on April 05, 2013, 11:43:40 pm: I remember doing this question a couple of years ago now, looking back at my old textbook they've updated the question as they changed the rules slightly for the game. Anyways...

Lets look at division 2.
What you need to do to find the probability is to flip their formula, but now to explain where the formula comes from.

We have $45$ numbers, of which we want to pick $7$ from. The possible number of ways to pick $7$ numbers from $45$ is $nCr(45,7)=\frac{45!}{7!38!}$ , so that will be our denominator.
Now to win division 2, we need at least $6$ of those winning numbers and at least $1$ supp number. So how many ways can we pick $6$ winning numbers from those $7$ ? $nCr(7)=\frac{7!}{6!1!}$
Now for the second part the key here is that the supplementary numbers can't be any of those $7$ winning numbers. That is we only have $45-7=38$ numbers left that can be the supp numbers. Now the $nCr(38,1)-nCr(36,1)$ part is basically the number of options left we have, take away the number of options left that are wrong (that is that don't give us either of the two supps). So there is $38$ numbers besides the winning numbers, and of those we are picking $1$ , since we only have one number left to make our $7$ , take away the number of ways we can pick $1$ from the $36$ remaining numbers that aren't supps.
Then the number of ways those two situations work together is the product of the two, which becomes our numerator.

The logic works through for the other options, I'll explain say division 5 & 7 though.

Division 5:
Total number of ways we can pick the $7$ numbers $=nCr(45,7)=\frac{45}{7!38!}$ . The number of ways we can pick $5$ correct numbers from the $7$ correct numbers is $nCr(7,5)=\frac{7!}{5!2!}$ , and then we can pick any two other incorrect numbers that are not supps as well, that is we have $45-7-2=36$ numbers remaining to pick that are basically useless, and from those we want $7-5=2$ numbers, i.e. $nCr(36,2)=\frac{36!}{2!34!}$ . Multiplying those together and the dividing by the total number of ways that we found previously, gives the probability.

Division 7:
$nCr(45,7)=\frac{45!}{7!38!}$ is the number of options we can have picking $7$ numbers from the $45$ . The $nCr(7,3)=\frac{7!}{3!4!}$ is the number of ways we can pick $3$ correct numbers from the $7$ correct numbers. $nCr(38,4)=\frac{38!}{4!34!}$ is the number of ways we can pick $4$ numbers from the remaining $38$ , but to get the amount of options that are correct, we need to take away the number of options of picking the numbers that aren't supps, there are $36$ of those left, and we have $4$ spots to fill. Then again to get the combination of both, we multiply them together, and then divide by the total number of options.

I hope that helps, and makes sense....
Title: Re: Oz Lotto Probability
Post by: acinod on April 06, 2013, 11:21:43 am: Thanks for the help!

One thing I don't understand is how come to win Division 6, the website says the formula is (45c7 / (7c4 x 38c3))? Shouldn't it be (45c7 / (7c4 x 36c3))?
Title: Re: Oz Lotto Probability
Post by: b^3 on April 06, 2013, 02:01:44 pm: Once we've picked 4 of the correct numbers, we have to pick another 3 from the remaining numbers. For division 6, it does not matter if these numbers are supps or not, so supps will be included in these. So 45-7 (the amount of correct numbers) gives 38 remaining numbers (which includes the supps). Hence $\frac{nCr(45,7)}{nCr(7,4)\times nCr(38,3)}&=\frac{45!}{7!38!}\div\left(\frac{7!}{4!3!}\times\frac{38!}{3!35!}\right)$