Re: the probability theory has holes!
Kook Spotter
2) gmail address.
3) total fuckwit.
Bart Goddard
521e6a...@m25g2000yqc.googlegroups.com:
> i have a simple question for you?
>
> If the probability that a number has as its factor
> 2 is 1/2
>
> 2 and 3 is 1/(2*3)
>
> 2 and 3 and 5 is 1/(2*3*5)...
>
> If we cover every event the probabilities add up and give 1!
>
> So 1=1/2+1/(2*3)+1/(2*3*5)+.....
If you've covered every event, then please tell us
which fraction in your series accounts for 1/14.
> are you ok?
I'm OK. What's wrong with you?
B.
--
Cheerfully resisting change since 1959.
eestath
power
case 1: a number has as its factor 2 : probability 1/2
case 2: a number has as its factor 2 and 3 : probability 1/
(2*3)
case 3: a number has as its factor 2 and 3 and 5 : probability 1/
(2*3*5)
...
if we cover every case probabilities add up to 1!
So 1=1/2+1/(2*3)+1/(2*3*5)+.....
are you ok? you get it? Is simple is elementary ok?
1=1+1/2+1/(2*3)+1/(2*3*5)+.....<1/2+1/4+1/8.....=1
So 1<1
plus 1+1/2+1/(2*3)+1/(2*3*5)+.... is irrational!
Henry
> i have a simple question for you?
>
> If the probability that a number has as its factor
> 2 is 1/2
>
> 2 and 3 is 1/(2*3)
>
> 2 and 3 and 5 is 1/(2*3*5)...
>
>
> So 1=1/2+1/(2*3)+1/(2*3*5)+.....
>
> are you ok? you get it? Is simple is elementary ok?
>
> 1=1+1/2+1/(2*3)+1/(2*3*5)+.....<1/2+1/4+1/8.....=1
>
> So 1<1
>
> plus 1+1/2+1/(2*3)+1/(2*3*5)+.... is irrational!
>
> stathopoulo...@hotmail.com
OK, so you are a troll.
What you want is:
The probability that a number has as its factor
2 is 1/2
3 but not 2 is (1-1/2)/3 = 1/6
5 but not 2 or 3 is (1-1/2-1/6)/5 = 1/15...
If we cover every event the probabilities add up and give 1!
So 1 = 1/2 + 1/6 + 1/15 + 4/105 + 8/385 + 16/1001 + .....
(see http://www.research.att.com/~njas/sequences/A038110 and
http://www.research.att.com/~njas/sequences/A038111)
Bart Goddard
66d54c...@v25g2000yqk.googlegroups.com:
> are you ok?
I answered your question, why won't you answer mine?
eestath
for 2: we exlude the first 1/2
for 3:then we exclude 1/3 but half of them have factor 2 so we
actually exclude 1/2*3
for 5:then we exclude 1/5 but half of them have factor 2 and 1/3
have factor 3 so we actually exclude 1/2*3*5
So the Series is 1=1/2 + 1/(2*3) + 1/(2*3*5)+....
You answer but you are actually wrong!!!!!!!!!!
James Burns
> What i had in mind is Eratosthenes Sieve:
>
> for 2: we exlude the first 1/2
>
> for 3:then we exclude 1/3 but half of them
> have factor 2 so we actually exclude 1/2*3
>
> for 5:then we exclude 1/5 but half of them
> have factor 2 and 1/3 have factor 3 so we
> actually exclude 1/2*3*5
Right here is the problem: "1/3 have factor 3".
What you want is "2/3 DO NOT have factor 3".
The 1/3 with factor 3 have already been excluded.
Another 1/5 of the fraction ( 1 - 1/2 )*( 1 - 1/3 )
that do not have either 2 or 3 as a factor is
what you want to exclude for the next step.
And ( 1 - 1/2 )*( 1 - 1/3 )*1/5 = 1/15.
> So the Series is 1=1/2 + 1/(2*3) + 1/(2*3*5)+....
Another way to look at this:
(fraction not div by 2)*(fraction not div by 3)*
(fraction not div by 5)* ...
= (fraction not div by any prime)
which means
( 1 - 1/2 )*( 1 - 1/3 )*( 1 - 1/5 )* ... = 0
If you are not already familiar with the Riemann
zeta function, then I think you would find this
very interesting:
http://en.wikipedia.org/wiki/Riemann_zeta_function#Euler_product_formula
> You answer but you are actually wrong!!!!!!!!!!
One of the more pleasant surprises on the Internet
is finding someone who will admit that they were wrong
(when they finally understand that they were wrong).
I wonder, are you going to surprise us?
Jim Burns
Bart Goddard
3190a5...@c3g2000yqd.googlegroups.com:
How can I be wrong when I only asked a question?
Which of your fractions counts the exclusion of
the integer 14?
If you can't answer the question, then you need
to re-think your assertion.
Pubkeybreaker
> eestath <stathopoulo...@gmail.com> wrote in news:e38c715c-4ef2-4d72-9463-
> 3190a5550...@c3g2000yqd.googlegroups.com:
>
> > What i had in mind is Eratosthenes Sieve:
>
> > for 2: we exlude the first 1/2
>
> > for 3:then we exclude 1/3 but half of them have factor 2 so we
> > actually exclude 1/2*3
>
> > for 5:then we exclude 1/5 but half of them have factor 2 and 1/3
> > have factor 3 so we actually exclude 1/2*3*5
>
> > So the Series is 1=1/2 + 1/(2*3) + 1/(2*3*5)+....
>
> > You answer but you are actually wrong!!!!!!!!!!
>
> How can I be wrong when I only asked a question?
> Which of your fractions counts the exclusion of
> the integer 14?
>
> If you can't answer the question, then you need
> to re-think your assertion.
He also needs to think about the proposition:
The probability that a number has 2 as a factor is 1/2.
This is, quite simply, wrong. There is no way to select
an integer UNIFORMLY AT RANDOM from Z+. Once you discuss
probability, you must consider the underlying density function
from which you are drawing the sample.
If what he means is that 1/2 of the integers less than some given
integer N have 2 as a factor, this is wrong as well, since it
depends on N mod 2. (and so forth for 3,5,7, etc.)
If the OP really wants to probe further, he will get introduced
to the wonderful world of sieve methods! Welcome to the parity
problem!
Welcome to the fundamental lemma of the sieve! Welcome to Mertens'
theorem! He needs to study this theorem. In particular, he needs to
learn why the constant is exp(gamma) and not 1.
I'm not telling you (Bart) anything you don't know.
BTW, we met once, at the West Coast Number Theory Conference at
Asilomar.
Bart Goddard
86ab-278...@k19g2000yqc.googlegroups.com:
> I'm not telling you (Bart) anything you don't know.
> BTW, we met once, at the West Coast Number Theory Conference at
> Asilomar.
>
I remember. I think it was more than once. Maybe also
at another WCNT site(?) I think I still have that
Xmas card you were passing out (the holes in
the cover "decrypted" the message hidden in the text
on the inside.)
B.
BURT
> Pubkeybreaker <pubkeybrea...@aol.com> wrote in news:c04b37bb-de78-4e27-
> 86ab-278a8dec6...@k19g2000yqc.googlegroups.com:
>
> > I'm not telling you (Bart) anything you don't know.
> > BTW, we met once, at the West Coast Number Theory Conference at
> > Asilomar.
>
> I remember. I think it was more than once. Maybe also
> at another WCNT site(?) I think I still have that
> Xmas card you were passing out (the holes in
> the cover "decrypted" the message hidden in the text
> on the inside.)
>
> B.
>
> --
> Cheerfully resisting change since 1959.
If you think reality is statistical in nature then I consider you a
statistic.
Mitch Raemsch
Gerry Myerson
Bart Goddard <godd...@netscape.net> wrote:
You know, we WCNTers need a secret handshake
we can use on Usenet to identify each other.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Bart Goddard
E2F5D8.123...@news.eternal-september.org:
Ooh! That would have been much cheaper than the
T-shirt. Hindsight....
B.
PS. If you pronounce WCNT, it sounds like "WackNut"...
gives me an idea for a logo....
eestath
If you model Eratosthenes sieve:
first you exclude 1/2 of the numbers: 2,4,6,8,10,...then
you exclude 1/3 of them but 1/2 of them are allready excluded, so you
actually exclude 1/(2*3) then
you exclude 1/5 of them but 1/(2*3) of them are allready excluded so
you actually exclude 1/(2*3*5)
....
1.There is no way a number to be left out
2.I didn' t count anything twice
So the series 1 = 1/2 + 1/(2*3) + 1/(2*3*5)+....
But it isn't 1= 1/2 + 1/(2*3) + 1/(2*3*5)+....<1/2 + 1/4
+1/8+......=1
1<1.
plus 1/2 + 1/(2*3) + 1/(2*3*5)+.... is irratinal so it can be 1
eestath
Dik T. Winter
> every integer greater than 2 is a product of primes to an integer
> power
>
> case 1: a number has as its factor 2 : probability 1/2
> case 2: a number has as its factor 2 and 3 : probability 1/
> (2*3)
> case 3: a number has as its factor 2 and 3 and 5 : probability 1/
> (2*3*5)
> ...
>
> if we cover every case probabilities add up to 1!
>
> So 1=1/2+1/(2*3)+1/(2*3*5)+.....
No, it is not 1 because it can have only factors 3 without any other
factors.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> try to understand what i' am saying not what you are taught!
>
> If you model Eratosthenes sieve:
>
> first you exclude 1/2 of the numbers: 2,4,6,8,10,...then
> you exclude 1/3 of them but 1/2 of them are allready excluded, so you
> actually exclude 1/(2*3) then
> you exclude 1/5 of them but 1/(2*3) of them are allready excluded so
No it is not 1/(2*3) that is already excluded. It is 1/2 + 1/(2*3) that is
already excluded.
Dik T. Winter
> What i had in mind is Eratosthenes Sieve:
>
> for 2: we exlude the first 1/2
What remains is 1/2
> for 3:then we exclude 1/3 but half of them have factor 2 so we
> actually exclude 1/2*3
What remains is 1/2 - 1/2*3 = 1/3
> for 5:then we exclude 1/5 but half of them have factor 2 and 1/3
> have factor 3 so we actually exclude 1/2*3*5
Actually you exclude 1/3*5
Bart Goddard
e2207c...@r5g2000yqb.googlegroups.com:
> try to understand what i' am saying not what you are taught!
I asked a question. How is that not trying to understand?
None of your fractions seems to exclude 14, yet you say
you've counted all possible cases. Please explain.
eestath
14!
Bart Goddard
8a9260...@d10g2000yqh.googlegroups.com:
> if you exclude the numbers witch have as factor 2 you do not exclude
> 14!
Exactly. So when DO you exclude 14? That is, which fraction accounts
for excluding 14?
eestath
> eestath <stathopoulo...@gmail.com> wrote in news:4e58d51a-aa4e-4013-b2fc-
> 8a92605b6...@d10g2000yqh.googlegroups.com:
>
> > if you exclude the numbers witch have as factor 2 you do not exclude
> > 14!
>
> Exactly. So when DO you exclude 14? That is, which fraction accounts
> for excluding 14?
1/2 the first...read more...
Richard Tobin
eestath <stathop...@gmail.com> wrote:
>What i had in mind is Eratosthenes Sieve:
>
>for 2: we exlude the first 1/2
Yes.
>for 3:then we exclude 1/3 but half of them have factor 2 so we
>actually exclude 1/2*3
Yes.
>for 5:then we exclude 1/5 but half of them have factor 2 and 1/3
>have factor 3 so we actually exclude 1/2*3*5
No.
1/3 have factor 3, so you now exclude the 2/3 that don't. So you
now exclude 1/5 * 1/2 * 2/3.
-- Richard
--
Please remember to mention me / in tapes you leave behind.
Bart Goddard
news:41851288-d73e-44f3...@p8g2000yqb.googlegroups.com:
You said that 1/2 counted excluding numbers divisible by 2.
Then you said it did not. Now you say it did. I sure don't
see how reading more direct contradictions is going to help
here.
But at least you finally answered a simple question. Let's
try another: If you take the first 1,000,000 integers and
exclude all multiples of 2 and all multiples of 2 and 3 and
all multiples of 2 and 3 and 5, how many integers will be
left, according to your formula?
I'm guessing that it should be 1,000,000 x (1-1/2-1/6-1/30),
but I'd rather you answer.
Ostap S. B. M. Bender Jr.
I have a surprise for him: the uniform measure on Z+ (or any other
infinite countable set) doesn't exist, so there can be no valid
probability arguments of this sort.
James Burns
I suppose I understand what you're saying here:
There is no function f: Z+ -> R such that both
f(m) = f(n) for all m,n in Z+
and
Sum_{n in Z+} f(n) = 1
That's easy enough to prove. Certainly, I am not
disputing that.
However, it looks to me as though a good argument
very much like the quick-and-dirty arguments
being used in this thread is possible with
an expanded definition of a measure on Z+,
a definition much like that for generalized
functions, with the Dirac delta function as a
particular example.
As I understand it, the Dirac delta is described
as a function that is 0 everywhere except one
point, but, when it is integrated across that
point, we get 1. This is obviously (and provably)
impossible, as it is stated. The work-around that
I am familiar with uses a sequence of
functions < d_n > such that
d_n(x) -> 0, as n -> infinity, for x <> 0,
and
INT{-inf, +inf} d_n(x) dx = 1, all n
There are a lot of choices for such a sequence.
A rigorous argument involving Dirac deltas
would solve some problem using a convenient
choice for the d_n, get an answer involving
n, and then take the limit as n -> infinity
as the "true" answer. The operations behind
what we call the Dirac delta are completely
unexceptional, but they have the same effect
as the provably impossible function.
My question is: why can't something similar
be done to create a uniform measure on Z+,
or, at least, something that would give us
the answers a uniform measure would, if such
a thing existed?
For concreteness, suppose
f_n : { m in Z+ | m =< n } -> R
where
f_n(m) = 1/n
Suppose we ask what the probability that a
"random number" is not divisible by 2, 3, or 5
(for example), and get a constant plus some
bounds that -> 0 as n -> infinity. Without
doing very much in the way of calculation, I
feel confident that the constant would be
(1/2)(2/3)(4/5).
What is wrong with this argument?
Jim Burns
FredJeffries
Edwin T. Jaynes in "Probability theory: the logic of science"
http://omega.albany.edu:8008/JaynesBook.html
Offers this advice:
"Apply the ordinary processes of arithmetic and analysis only to
expressions with a finite number n of terms. Then after the
calculation is done, observe how the resulting finite expressions
behave as the parameter n increases indefinitely."
See especially chapter 15 "Paradoxes of Probability Theory". He also
discusses the error of "Jumping directly into an infinite set at the
beginning of a problem" in his discussion of "Finite vs Countable
Additivity" in that chapter. And in Appendix B he discusses "extend
[ing] probability theory to infinite sets".