Funny how you just double the first number and leave the second alone...

ho hum.

So you thought there was a special lottery to see which day of the year they hold the DAILY NUMBER lottery?

:1orglaugh

Sorry, too funny.

LOL!
This thread rules!
Holy shit I'm tired!

That is fucking creepy!! And same as the dollar bill.

The Law of Probability says that the odds of a specific series of numbers being drawn on a specific day are not the same as the odds of any set of numbers being drawn on any day. Anyone who has taken statistics in college should know that it's not as simple as it seems if you are matching the date with numbers.

Do a Google search or read this, for those of you that care:
http://www.peterwebb.co.uk/probability.htm


Periodic events

So far we have discussed probabilities in terms of outcomes either occurring or not occurring, but sometimes the gambler will wish to know the probability of an outcome occurring within a given time or during a given sequence of events. If a given event must occur exactly twice in 2 years, then if one year is chosen as a unit of time, and the continuous occurrences of the event are called E1, and E2, there are four possibilities:-

Year A Year B
1.E1,E2 ------
2.E1 E2
3.E2 E1
4.----- E1, E2

Note that these possibilities are similar to the tossing of two coins, and that care has been taken not to reduce the possibilities to three by combining 2 and 3 as one possibility.

It follows that if an event must occur exactly twice only in two years, the probability of it occurring at least once in either year is 3/4. There are three times within the table above in either year that the event occurs only once and one occurrence where it would not occur. Only in possibility 4 does the event not occur in year A. There is a formula to work out these probabilities.

If an event must occur exactly x times in a period divided equally into N smaller periods, then the probability of the event occurring at least once in any small period is

=(power(n,x)-(power((n-1),x))/power(n,x)

Thus, if an event must occur four times in ten years, the probability of it occurring in the first two years is

=(power(5,4)-(power((5-1),4))/power(5,4)

=((625-256))/625 = (369/325) = 0.5904 or much better than 50%.

The Chevalier de Mere, a rich Frenchman who liked gambling, was responsible for inviting the philosopher and mathematician Blaise Pascal to carry out some of the earliest work on probability theory.

De Mere played a gambling game in which he bet that he could throw a six in four throws of a die. De Mere progressed from this game to betting that with two dice he could throw a double in 24 throws. It was known that the odds were in his favour with the first game, and gamblers of the time reckoned that as four is to six (the numbers of ways a die can fall) as 24 is to 36 (the ways two dice can fall), the second game should be favourable. The Chevalier de Mere was not satisfied with this assumption and asked Pascal to work out the true probabilities.

To work out these two problems it is necessary to work out the converse probabilities. The probability of not throwing a 6 in four throws is:-

=power((5/6),4)=625/1296

Therefore the probability of throwing at least one 6 is :-

=1-power((5/6),4) or 0.5177

As an even money proposition, therefore a bet to throw a six in four throws of a die is slightly in favour of the thrower. The probability of not throwing a double six in 24 throws is:-

=power((35/36),24)) = 0.5087

This means that the bet is (slightly) against the thrower.

Various formulae have been put forward to determine the number of throws necessary to make the throwing of a double-6 a better than even chance. The old gamblers' rule in operation in de Mere?s time relied on knowing the answer to the lowest number of throws necessary to give a probability of throwing 6 with a single die of more than 1/2. As has been said, this was known to be 4, which might be called the break-even number. According to the rule the break-even number for the double event (N2) is the probability of success on the single event (P1), known to be 1/6 (the probability of a 6 with one die), divided by the probability of success on the double event (P2), known to be 1/36 (the probability of a double-6 with two dice), multiplied by the break-even number of the single event (N1), known to be 4. Thus

n2=(p1/p2)*n1 by substitution n2=((1/6)/(1/36))x4, or n2=((36/6)x4)

In this equation n2=24, guessed correctly by the Chevalier de Mere to be wrong. Abraham de Moivre, in a book The Doctrine of Chances published in 1716, set out a formula for discovering the approximate break-even point (n) as:-

n=(0.6931/p)

(0.6931 is the ?Natural? logarithm of 2). So this gives the answer to our problem as 24.9516. An approximate break even number of 25, which is correct.

This gives an effective break number of 25, which is correct, although de Moivre?s formula gives too high a number, which is only slightly too high when the probability is small, but may be critically too high when the probability is larger. For example, de Moivre's formula for the break-even number for throwing a 6 with one die gives n = 0.6931 x 6 = 4.1586, which gives an effective break-even number of 5, which is too high, 4 being correct, as we have found.

It is interesting that John Scarne, in his famous book Scarne's Complete Guide to Gambling, published in 1961, states that the odds to one should be multiplied by 0-6931. Thus his approximate answer to de Mere's problem is n = 0-6931 x 35 = 24.2585. This answer is too low, although it gives the correct effective break-even number as 25. Scarne's formula applied to throwing a 6 with a single die gives the answer n = 0-6931 x 5 = 3.4655, again slightly low, but giving the correct break-even number of 4. The third formula to find the break-even number is as follows:-

If the probability of the outcome is (1/a) then :-

n=(log 2/((log a)-(log(a-1))))

In this case, the logarithms are the ordinary logarithms, to base 10.
The break-even number for throwing a double 6 thus becomes:-

n = (log 2/(log 36 -log 35))

which, with the use of four figure log tables, gives n a value of 24.6721, thus giving 25 as the effective break-even number. For throwing a 6 with one die, n = 3.8005, confirming 4 as the effective number.

When the Chevalier de Mere asked Pascal to help with this problem Pascal provided him with his answer, and began a correspondence about probabilities with Pierre de Fermat, another French mathematician, which established probability theory as a new branch of mathematics.

question: what is the probabilty for the numbers 9,1,1 to come up on september 11th?

answer: 1/1000




i will be only laughing if you think there is a different answer to this question. :1orglaugh maybe you should have done your homeworks when you took statistics classes (if you ever did)

The numbers I was thinking related to if there was a 365 day lottery, as opposed to 1.

Er, er, er.

1/1000 is the odds that 9 1 1 will be drawn on any day, not the probability that it will be drawn on a specific day.

i posted this msg... i didnt notice Equinox was logged in on my PC, fucker..

fuck my head hurts

This thread is too much to read. But to back myself up and to say that Voodoo is right, all I was saying is that it depends on what you are talking about. If you are talking about just that one number than it is 1:1000. If you are talking about everything that lead up to it including the creation of the universe than it is almost infinite. (I know, "almost infinite" makes no sense, but you get the point.)

*sigh*

Then please tell us all what the exact odds for 911 to come up in NY's lottery drawing last night (9/11) were.

And I will make you the same $$ offer I made Voodoo when I scared him away.

it is 1/1000 that it will be drawn on one specific day... you specify the day, there are 1000 possible numbers for that day, each have a chance of 1/1000...

the chance that 911 will be drawn on september 12th is 1/1000 or any other day its the same... every day its 1/1000 chance...


an important rule of statistics:
in an event, if you add all probabilities for each outcome, it has to add up to 1 .. here the event is a drawing, the drawing on 9/11 .. there are 1000 possible outcomes of this single event.. since each number has the same probability, they all have the odds of 1/1000 each..
1000 x 1/1000 = 1


whatever date and whatever number you pick, the number to be drawn on that day has the odds of 1/1000...

jeez, how else can i explain this..... if you still dont get it, go back to college or something..

Lane - how am I ever going to make any money off of these guys if you keep explaining this to them so clearly? I'd much rather teach them the hard way, when it hurts their bank accounts.

Then again, it's been explained about 10 different way now and some people STILL don't seem to get it, so maybe there is money to be made yet.

I can only hope on of these guys who thinks the odds are greater than 1000:1 will step up with some cash. Come on smart guys, let's do it! If you are so sure, let's play.........

According to Prof.Gerard Ben Arous of New York University's Courant Institute of Mathematical Sciences its 1 in 1000

http://uk.news.yahoo.com/020912/80/d9hnp.html

Gosh. That's a shocker.

I guess Prof.Gerard Ben Arous of New York University's Courant Institute of Mathematical Sciences has at least half a brain.

I'm not a fucking mathematician. :)

I'm just saying that the simple odds of something happening are different than the probability of specific events happening in a specific order, or at a specific time or place.

I'm not making this up, it's statistics stuff. Not that I paid enough attention to know all the formulas, but I know that it's more complex than it first seems if you look at a bigger picture.

That's exactly what I said - on any given day the odds are 1/1000.

"on any given day"

the given day is 9/11 and the odds are still 1/1000

So even though we have explained and proved this to you numerous different ways, you are convinced that you are right, even though you can't prove or explain it?

Ok, one more time... as simply as I can state it:

There are only 1000 different combinations. Don't believe me? Count them:

001
002
003
...
998
999
000

Tada - 1000.

When drawing any ONE of those 1000 combinations (even 911), the odds are OBVIOUSLY 1:1000.

The numbers do not give a shit what the current date is, what number came up last, how many posts are in this thread, or if you wacked off at 3:47 that day.

100.

I wonder how that will affect tonight's drawing. I guess the odds of 1-0-0 coming up tonight would be astronomical, since there are now 100 replies to this post.

:1orglaugh

Exactly :thumbsup

To the morons who think that the odds of drawing 9-1-1 on 9/11 are greater than 1:1000.

You are looking to deep into this. You are actually trying to figure out the probability of 9-1-1 being drawn ONLY on 9/11 through out the enitre year and only in NY. Which has nothing to do with the odds/chance of 9-1-1 being drawn on any given day at any given place. In the state of NY on 9/11 there was a 1:1000 chance that 9-1-1 would be the winning numbers period, nothing you say can change that.

You've been served.. move on

Hey, Tera... You didn't scare me off.... I just found it pointless to try to teach someone who lacks the capacity to learn.

Back at ya boy! :ak47:

Lol it was fucking rigged.

I understand.

Kinda like how I just don't get it when you say over and over that 2+2=5 without knowing why or even trying to prove it.

If you are so sure, I'm not sure why you won't take me up on my bet. You stand to make a LOT of money. Here is my offer again, in case you forgot:

----------
Ok, dude.... if you are so sure about this, why don't you put your money where your mouth is?

You say the odds of 911 coming up in a single drawing yesterday (on 9/11) were 1:720,000, right? So by your logic, the odds of 912 coming up in a random three digit drawing today would be the same - 1:720,000.

So, here is what I propose. You be the "house" and I'll be the "gambler." If 912 should only come up once in 720,000 drawings today, you should make a KILLING if you pay me at 100,000:1 odds. Right? I mean, theoretically, after 720,000 drawings, you would win 719,999 of them and lose just one. So, if we bet $1 pert drawing, you would win $719,999 and pay out $100,000 (on the one you lost), netting you nearly $620,000.

Damn! You're gonna be rich!!!!!!

So, let's do it. I must be CRAZY to take this bet at only 100,000:1 odds, but I'm gonna do it anyway.

So, to make sure everything is fair, we will use a random number generator.... maybe www.random.org if that's cool with you. Whatever. Doesn't matter to me.

And then we will start drawing numbers. Every time 912 doesn't come up, I will pay you $1. And then we will draw again. And again, and again, and again. I'll play all day long, as long as you want, and I will keep GLADLY handing you $1 bills every time 912 doesn't come up. But as soon as 912 does come up - yes, today, on 9/12 (odds - 720,000:1 according to you), you have to pay me $100,000.

Sound good? Ready to play? Let's do it! Daddy needs a new pair of shoes!
----------

****

I'm watching CNN right now, and they just went to commercial. They are doing a story on last night's drawing after the break.

****

TERA....
Your bet is absolutely rediculous!!!!! Here is a TRUE bet that I will take you up on.... which pertains to the topic of this thread....
YOU bet me $720,000.00 that 912 will be drawn in the NY State pick 3 Lottery TODAY. And I will bet you DOUBLE, that it WON'T come up today in the NY State Pick 3 Lottery!

Back at ya again! Now... Put YOUR money where YOUR mouth is!

By YOUR odds, YOU'VE got a 1:1000 chance to win!!!... and MY odds are 1:720,000.

So, let's roll!

:eek7

What the fuck are you talking about? How does that have anything to do with anything?

Dude... I'm giving up on you. You're not even making any sense any more. My offer is EXACTLY what we've been talking about. You are just talking crazy now.

And I'll bet you there is a 1:1000 chance of any number you choose to get pulled tonite. I think that pertains to what we are talking about here wanna bet? :Graucho

Have a good day then Tera! :) Enjoy your bliss.

its 1/1000 dammit

Damn... your post disappeared just as I hit reply. What a shame.

I take it you woke up and smelled the roses?

haha... whatever dude.

Discussing this with you is like discussing it with a 5 year old, who I wouldn't expect to understand. Why do I even try?

voodoo, you cant prove the wrong.. insisting wont make it right.. this is math..
it can cost you a lot if something similar involves you and your money one day..

You are all in a senseless argument over the definition of the question, not the answer. That mathematician in that link was NOT wrong. Read closer what he says.

Of course he wasn't wrong. He said the odds of 911 coming up in last night's NY Pick 3 lottery were 1 in 1000.... which is EXACTLY what we (the smart ones) have been saying all along.

Qet Quiet in here and have him settle this.

He already did, and proved himself to be one of the smart ones. Go back a page or so.

Just curious why everyone besides the first poster, thinks they're a badass for having posted the clause of 1:1000?

Tera, Romeo, all you other guys, you aren't badasses and ya didn't serve anyone...

Thanks, move on, I already admitted my ignorance.