In the current issue of Automobile magazine, there is an ad for a car raffle to benefit the National Humane Society. In the raffle, 2399 tickets are available at $100 each. The grand prize is a 2003 Porsche Carrera, Chevrolet Corvette Z06, a “Jag”, or $50,000 cash.
There’s an offer to “buy 3 get 3 free” and the odds of winning are “399-1”.
This doesn’t sound right.
Well, if there are only 2399 tickets being sold, and one of them has to win, and you have six of them, I figure the odds to be six times 1/2399 which is 6/2399, which is 1/399.833…
So I would have rounded up to 1/400, but it’s pretty close.
Well, 1 in 400 is 399-1; i.e 399 chances of losing for each chance of winning.
Okay, still, it just doesn’t sound right, but it’s been a long time since my high school permutation & odds math class.
Are you saying the odds sound too likely or too unlikely?
If you have six tickets, you have 6/2399 of all the tickets out there. 6/2399= 1/399.833… as Scammer said.
Personally, those are long odds for me to pony up $600.
Seems like the odds should be a bit higher.