merkin

ill get them to start turning their standard deviation models to the lottery

MrDeference

For the benefit of the thickos can you explain this in more detail please Mark?

MarkO

Okay. Most people assume that with the odds of winning the lottery with one ticket being 14m to 1, by buying another ticket they are halving the odds (i.e., making themselves twice as likely to win). They're not. The odds of winning are now 14m to 2.

Crap explanation, I know. Perhaps one of the better mathemeticians (carl, or merkin's bods, by proxy, might want to give a proper explanation)?

MrDeference

OK. I'll write why I though it did, and you can point out the error.
Lets say there are only 10 combinations of numbers. Pick the winning number from 1-10.
I pay £1 and pick "1". I have 1/10 chance of winning. 9/10 chance of loosing.
I then pay another £1, and pick "2". I have doubled my stake, and selected two numbers : 1/10 + 1/10 chance of winning. This is 2/10, or .2, and double the chance. Odds of me winning have doubled, odds of me not winning have fallen by 1/10.

What's wrong with my thinking? Surely it's exactly the same with the lottery?

merkin

Mr deference, that is true, but if you buy 2 lottery tickets, the odds become 2/14,000,000 , NOT 2,000,000/14,000,000 - i.e a change so small its almost irrelevant. if however you bought 2 million tickets, your odds would improve somewhat, however, rather you than me

MarkO

Sounds plausible. I've had this discussion before, and somebody explained to me why it's kindof different with the lottery. Hmmmmm.

MrDeference

Hell, I don't disagree that a probability POV 1/14M is not much worse than 1/7M, however, it is still twice less probable.

Bollox. I am going to type the above sentence for the third time, and this time see if I can get convey some meaning:

Hell, I don't disagree that from a probability point of view 1/7million is not much better odds than 1/14million. It is however twice as probable.


[Edited by MrDeference - 11/6/2002 3:43:47 PM]

merkin

2 in 14m is absolutely not the same as 1 in 7m!!!!

if you had a pot with 14 million ***** in, and you picked 1, and then had another go, how many ***** would be left in the pot??

MrDeference

Ahh, but what difference does that make?
1/14000000 + 1/13999999 is near as damnit 2/14000000.

merkin

yes, it would be if you could add the 2 together as a probability, but you cant, they are unconnected events, its the same effect, picking twice from a pot of 14m ***** doesnt half your odds of picking out the right one, it merely increases it by 1 in 14m.

or expressed mathematically non conditional probability.

its not an exponential relationship

i.e. by your maths you would only have to buy about 25 tickets, and your odds of winning the jackpot would be virtually guaranteed!

MarkO

Oi, that's just what I was going to say (in a roundabout way)!

carl

I think we're getting out of sync here. The odds that are important are the odds of the ***** coming out the same as you have on your ticket. If you buy two tickets with different sets of numbers you have halved the odds from 14 million:1 to 7 million:1. But 1/14x10^6 is so small that 2/14x10^6 isn't a lot bigger.

If you buy two tickets with the same numbers on, you have the same chance of winning as if you bought one ticket (14 million:1) because that's the chance of those numbers coming out (it's not like a tote in that it doesn't depend on the number of tickets sold). Of course you (a) increase the jackpot funds available by a miniscule amount and (b) double your share of the jackpot should you win.

HTH

MrDeference

To recap, you are saying that an analog between picking ***** out of a bag applies. I disagree. They are linked events where the existance of the act of buying a ticket decreases the problem space. This is a linked event.

I am saying that buying a ticket has a 1/14m chance of winning. Each additional purchase will add those same odds of winning.
so, to have a prob of 1, you would need to buy 14million tickets.
My maths makes it plain that is isn't exponential, and each purchase of a ticket increases your probability of winning by 1/14m.

It is your bad understanding that is deducing that I am saying each ticket doubles the chance of winning. That is arrant nonsense of course. It doesn't change the fact that the underlying purchase of two tickets over one doubles the chances.

merkin

linked events or not, if you have a 1 in 14m chance of picking out the correct 6 number combination , then if you had a pot with 14m ***** in, chances are the same, think of each ball in that pot as a combination of numbers.

(havent done the math on this, im assuming the correct odds of picking the correct 6 numbers are indeed 1 in 14m

MarkO

Yes, but your implication is that by doubling your chances, you're significantly reducing the likelihood of not winning. This isn't the case. Yes, you double your chances, but the chances are so insignificant that doubling them as near as dammit makes no difference.

What you appear to be trying to imply is that by buying a 2nd ticket, you halve the odds of winning. If this was the case, as merkin says, then buying a 3rd ticket would half the subsequent odds again, and so on - meaning that buying 25 or so tickets would effectively guarantee a jackpot win. Clearly that isn't the case - you'd simply have a 25 in 14,000,000 chance of winning, which is still a ludicrously insigificant chance of hitting the jackpot.

GaryK

guys,

this is gonna run and run, we dont need mathematicians we need clairvoyants and tell us what the numbers will actually be

I always wondered if mystic meg is so fecking good how come she hasn't won???

Gary

carl

No it doesn't, because two people are allowed to have the same numbers. The act of a ball dropping out of the machine reduces the problem space, but the act of buying a ticket doesn't.

The analog with removing ***** from a bag would only hold if you returned each ball to the bag after it was picked, thus ensuring that it could be picked again.

[Edited by carl - 11/6/2002 4:10:40 PM]

merkin

o.k this will hopefully help explain.... how many people do you have to get in a room before two have the same birthday?

merkin

p.s. Mr Deference, i suggest you pick a new user name

MrDeference

I was thinking the same.

MarkO

Hang on. That statement of mine that you say we're all in disagreement with? Well, I'm not. I still agree with it!

merkin

LOL, so how many people do you have to have in 1 room before you get two with the same birthday?

MrDeference

MarkO,
No you don't: We disagree.
We agree:
367. FFS

carl

It's 26 people in the room to have a 50% chance of two having the same birthday.

If you want it to be 100%, it has to be 367 people.

merkin

nope, the answer is actually 22 or 23!

MrDeference

Now your just trying to wind me up.

MrDeference

Really - I though probability and maths would have featured in Financial software...

carl

Birthday Paradox

merkin

LOL @ you two

I made the fatal mistake of not putting my question properly (i.e. missing the 50% bit) due to being busy.

YOu are right, its 23 to have a 50% chance,

I was trying to relate this to the lottery question, i.e. birthdays can match any of the people in the room, lottery you have to match a set of numbers that CANNOT be the same as another set of numbers.

carl

...but it can be the same as someone else's ticket. What was the question again?