No you are wrong. You need to study statistics mate. The thing is theree are 18 teams. So if we had to rank them up 18 teams can be on first place. Thus only 17 teams can be on second place. Thus 16 for 3 rd place. Likewise to the end. Therefore 18! = 6.40237e^15. And only one specific order can be correct, thus 1\6.402..... which equals .0000000000000001. Thus The probability that a person would win the Wiliam Hill Prize is 0..0000000000000001. That means if they played this predicting game 10,000,000,000,000,000 times they would win megabucks one time.
maths problem
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OK, after 2 hours of trying to work out the captachas for this ^&% site:
http://mathhelpforum.com/math-help-forum.php
I gave up and decided to post my question here. I had also solved the captacha once to find a sudden mission impossible style message of this captcha lasts for 2 minutes and had thought oh &*&^%, pressed the register button only to find "no way hosea you have not pressed agree to forum rules".
Seriously folks. You wanna make a million bucks. Solve the above &*^ing crap. It ain't rocket science.
Back to my maths question.
In the Australian Football League there are 18 teams. William Hill has been saying that if someone picks all 18 in the correct end of season finishing order then you get megabucks.
Now, I am not a betting person. So out of cynicism, and assuming that they hired a math head to tell the the odds, let's work the odds out now.
18! is obvious for the chance. It's a massive number:
6402373705728000
That isn't that interesting. But what interests me more is if we take the pop of Australia. It's approx 23 million. So let's call it that. Say everyone in the 23 million had a go once and used a different combination of the 18!
Now, I'm not sure how to go about this. So I tried out
6402373705728000/23000000 =
278364074.16208695652173913043478
So does that mean that there is a 1/278364074 chance that anyone in Australia will win the William Hill prize? I am not sure. If so it shows what a con it is!