Re: A challenge for Pitman-math

Seanpit wrote:
On Jan 12, 9:38 pm, Rusty Sites <SpameYou...@xxxxxxxxxx> wrote:
My calculation does give an exact, rational result. You are off by
way more than approximation error and are exposed as a bullshitter once
more.
You're quite mistaken. See if you can get anyone else to agree with
you here . . .
Choice 1

Use your formula to produce a number for a small example like matching 3
aa in a 4 aa chain. I can do some simple math that even you can
understand and you can see that your way doesn't work.

Ok, consider a simple coin flip example. Say you flip a quarter 10
times. Given that the coin tosses are "fair", what are the odds that
your flip exactly 9 heads and 1 tail in a given try?

The formula is:

(e^-5)(5^9) / 9! = ~0.0362


So the chance of getting one head in two tosses would be

(e^-1)(1^1)/1! = 0.368

In actuality, there are 4 equally probable cases. HH HT TH TT. Two of the cases have exactly one head so the probability is actually 0.5. This is about as close as your method ever gets. Now back to your example with one tail in 10 tosses. There are 1024 cases (2^10, can you see that?) so I can't list them all, but perhaps you can see that there are 10 cases that have exactly one tail. The one tail can occur on any of the ten tosses so the chance of getting one tail and nine heads is

10/1024 = 0.0098

and you aren't even close. Anybody who has had a decent introductory course in probability will tell you that the number of heads in a given number of tosses is governed by a binomial distribution. For one tail in ten tosses the binomial distribution yields

(10 1)*(1/2)^9*(1/2)^1 = 10/2^10 = 0.0098




In other words, a little more than 3% of the time. That's the right
answer.

Well, no. It's actually less than 1%.


The odds of getting exactly 5 heads and 5 tails in a given trail is
about 0.17546 or around 17.5% of the time.

In the real world the answer is

(10 5)*(1/2)^5*(1/2)^5 = 0.246

That's pretty close for you.


Clearly, your "math" is just a bit off base.

You mean "clearly" because it doesn't agree with the results of Sean Pitman?


The solution to the problem at hand goes as follows: If the expected
number of occurrences in this interval is λ, then the probability that
there are exactly k occurrences (k being a non-negative integer, k =
0, 1, 2, ...) is equal to:

(e^-λ) (λ^k) / k!

where

e is the base of the natural logarithm (e = 2.71828...)
k is the number of occurrences of an event - the probability of which
is given by the function
k! is the factorial of k
λ is a positive real number, equal to the expected number of
occurrences that occur during the given interval. For instance, if the
events occur on average 4 times per minute, and you are interested in
the number of events occurring in a 10 minute interval, you would use
as your model a Poisson distribution with λ = 10*4 = 40.

http://en.wikipedia.org/wiki/Poisson_distribution

That's the correct formula.

You are a classic crackpot. They are never wrong. If it means the whole world is wrong, then it is. When you finally have to admit you are wrong about something, you will change things around, perhaps drastically, but you will always get the same result.


Choice 2

Go see how it is really done here.

http://en.wikipedia.org/wiki/Binomial_distribution

It will be interesting to see if you can figure out how to apply this to
the problem at hand and verify that my result is correct. I solved the
problem directly but this is the formal way to do it.

This is a link to an explanation of the binomial distribution. The
Poisson distribution can be derived as a limiting case of the binomial
distribution.

So the Poisson distribution is the one to use because it can approximate a binomial distribution, but using the binomial distribution is wrong. Is that about it?

The Poisson distribution can be applied to systems with
a large number of possible events, each of which is rare. A classic
example is the nuclear decay of atoms. The Poisson distribution is
sometimes called a Poissonian, analogous to the term Gaussian for a
Gauss or normal distribution.

Thanks for the math lesson. You are so erudite.


I have no idea what formula you used.

Because it's the correct one.

Whatever it was, you messed it
up.


Would you like those words with or without salt?

.

Loading