wrote:
I have been working on a theory of gravity, and would like help
modeling everything else that we know about gravity into the theory. So
I can self publish my work in a journal or on a website like Wikipedia.
Gravity is represented as F= G( (m1*m2) / r^2 ) , and I would like to
change this theory, but make sure that the math still works exactly the
same. The Force Of Gravity is equal to the Gravitiational Constant
multiplide by the masses of two objects, and divided by their distance
appart.
Make that "divided by the square of their distance apart."
So in my game where you form a circle of 10 pennies that represent the
gravitational pull of one object, and my individual penny which sits
outside of the circle entirely solitary. The odds still remain 10/11,
when you are flipping a fair coin to decide which pile wins each round.
What determines a 'win' for a pile of pennies? Which coin are you
flipping? How is a 'fair coin' giving you anything other than 50/50
odds?
As the piles move there is a .09765625% of flipping 10 wins in a row
for the individual penny, and a 50% chance that the pile of 10 pennies
will win on the first round.
That number '.09765625%' ( = 1/2^10th) doesn't look relevant, but until
you tell us what a 'win' is we can't check your work.
But my question was, how do I calculate
the average number of coin flips before the larger pile wins.
You could use theoretical statistics, but you would need to define your
problem more carefully, first.
And the answer is k(n-k)
What are n and k?
That's right, k(n-k).
What are n and k?
So in my illistration, you can see that the
circle of 10 pennies attracts to lonely penny into its gravitational
field after 10 coin flips on average.
How does 'winning' a coin flip relate to gravitational field or
attraction?
But theoretically the number of
rounds in the game could come close to infinity. And in practice you
win after the first round or too.
I still don't understant what you are trying to say.
And I think you can see how this example illistrates a basic
understanding of gravity. If we assume that gravity accelerates
everything on earth at 9.8 m/s^2.
Why should we make that assumption? We know it is wrong.
For example if we look at the earth as being a mass of 10 pennies,
Pretty massive pennies...
and
we look as the signle penny as being a distance of 4.9 meters,
Pretty low altitude for such a massive penny.
then if
we follow this equation.
t = sqrt( ( 2(4.9 m) ) / ( 9.8 m/s^2 ) ) = 1 s
Galileo's gravity...
And if we say the average number of coin flips it takes to produce this
effect is 10, then each coin flip represents 1/10th of a second. So on
average it takes just 1 second.
You are beyond confused, now.
Now obviously with correct preportions of pennies, and more
sophisticated mathematics, and a better understanding of the physical
formulas for gravity.
That isn't even a sentence. It is illucid.
We could do a lot more. And be far more precise.
Try "reading." I'm sure you've heard about it. It's been in all the
papers.
Everything we know about Galileo's gravity can be found in a high
schools physics book.
Everything we know about Newton's gravity can be found in an
undergraduate college physics book.
Everything we know about Einstein's gravity can be found in a graduate
level physics book.
Given your demonstrated knowledge of statistics, I would like to have
you over on a poker night sometime soon.
Tom Davidson
Richmond, VA
.
