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Probability Question about Lottery

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Originally posted by: konichiwa
haha well I think my math was right in the situation I had assumed. And i think for big lotteries (powerball and the like) both of those cases are true... but who knows 😉

Still wrong. In Powerball, order does not matter, and numbers can not be repeated. There IS, however, a special 'powerball' for the last number. This number is drawn from a different pool, and can be a repeat number.
 
The odds are really easy to calculate actually.

It's just 43!/49! which is approx. 1 in 10 billion assuming that the same number can't be picked twice.
If numbers can be picked more than once, it's 1/49^6 which is roughly 1 in 13.8 billion

Ooops I forgot about the order thing; spazzychicken is right, you have to multiply your odds by 6! to get 1 in 14 million, I knew I was forgetting something 😱

Both of these are for the jackpot.

To calculate the lower prizes is a little more complex. I can show you all the steps of it if anyone really would like to know.
 
Originally posted by: Xiety
Originally posted by: konichiwa
ok i still don't know if i really understand your question but:

assuming you need to pick six numbers from 1 to 49, that gives 49 possibilities per "slot."

that means there are 49^6 possible combinations. that's 13841287201 ways of picking six numbers, 1-49.

if i'm not mistaken (been a long time since i took statistics too) the probability you pick the right numbers in the right order is 1/13,841,287,201

assuming you are correct, maybe i should save my money :Q

maybe if my chances were in millions, i would try, but not if in billions 😱

But you can't win if you don't play 🙂

Cheers,
Aquaman
 
$60? That's an office pool! LoL

I drop a buck on the Cali lotto every so often, but that's about it.

I do scratch offs more often, and generally am ahead by about $400 on average. But it's never more than $3 at a time.
 
For the first number you can pick any one of 49 numbers.

For the second, you can choose one of 48 numbers, since one has already been chosen.

For the third, one of 47...
For the fourth, one of 46...
For the fifth, one of 45...
For the sixth, one of 44...

49*48*47*46*45*44 = 10,068,347,520 possible combinations.

If you play 100 times (choosing a different combination of numbers each time) your odds go up to one in 1,006,834,752.

Edit: Hm. Looks like SpazzyChicken has thought this through more than I have...
 
Actually the odds are really not known until it is over. Teh chances of getting the numbers right is known, but to win you have to be the ONLY one with ALL the numbers matching. If someone else has a match then you each win half, etc...

But even then you only get about 20-40% of the winnings, not that I would complain. 🙂
 
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