Chain Reactions Within An Equilibrium Of Chaos And Entropy

[Corey...@gmail.com]

Roulette is a 50/50 game if you are playing black vs red, except for
the one green slip of the wheel, but the odds of hitting that are very
remote.

So here is a scam you can test out at home.

Get 21 pennies, and a quarter.

Now get 3 pennies and the quarter ready. Put 2 pennies in a pile, and
one by itself.

If the quarter comes up heads move the penny by itself into the pile of
2, and tails you do the opposite.

The odds are 50% higher that you will win at roulette if you have the
pile of 2 pennies. And here is why. Because the extra pennie is a
freebie that your pretty girlfriend scored from some stooge playing
poker who was probably on a winning streak.

Now you can keep playing this game by adding a penny to each pile, and
the odds of you taking the whole pot continue to go down logically.
When you are playing 3 on 2, the odds of winning aren't 50% higher,
they are only 25% higher. And with each penny you add the odds go down
by half. Or something like that. You do the math.

But the more money your girlfriend hustles the greater your odds of
winning at roulette are. Or you could bring in fake chips yourself.
But the great thing is the more free chips your playing with, the more
of the casinos pot you have a good chance of raking in.

It doesn't matter, just always bet on black.

There are lots of other ways to get that "free penny" as well. Being
able to read good body language at a poker table gets you a free penny,
being a pretty girl gets you that extra, I'm going to overbet my bad
hand penny. You know

But the idea is that if you play the 50/50 game, and you always trying
to win just half of what you are willing to gamble, your odds improve
greatly.

2 against 1 , you have a 75% chance of winning

You win that pot

3 against 1, you have better than 75% chance of winning

4 against 2, you have a 75% chance of winning.

So every time you win a pot you get a free penny!

And once you go up on a streak of luck like this, even when you start
to come down, you can continue going back up, and find a happy medium.
You just have to know when to quit, and know how to score those free
pennies. Just look at the whole pot that you have at the time as
something you are willing to risk to get whatever 'goal' you have at
the time.

Can you imagine what the odds would look like if you only go after one
penny as your goal? As the pile of pennies on your side keeps growing
larger, your odds go up exponentially after you reach each goal. But
obviously there is still a chance you will lose everything because you
are risking all of your pennies to win.

[Calcas]

You're joking, right?

[Corey...@gmail.com]

Now that this post is featured on JREF let me explain the math a little
bit more.

If I have 2 pennies and I want to start a chain reaction, in a 50/50
game, then I only go after one pennie at a time.

There are 3 ways the game can go, I can lose a penny, get it back, and
then score the penny that is my goal. Or I can win the penny right
off. Or I can lose completely.

So I have a 66.6% chance of winning

And now that I have 3 pennies and I go after one more penny, the odds
are 75% that I will win, because there are 3 ways I can win and 1 way I
can lose.

If I take the chance of playing these two games in a row, There are 7
ways I can win and 5 ways I can lose. So the chances of winning are
50/50. 2/3 + 3/4 = 5/7. And that's
71.42857%

Which is a whole lot better than starting with 2 pennies and playing
against 2 pennies!!

[Corey...@gmail.com]

So here is some further results of my theory: Corey's Punctuated
Theory On Gravitational Chain Reactions Within Equilibriums Of Chaos
And Entropy...

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.

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. 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. But my question was, how do I
calculate the average number of coin flips before the larger pile
wins.

And the answer is k(n-k)

That's right, k(n-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. 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. 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. For example
if we look at the earth as being a mass of 10 pennies, and we look as
the signle penny as being a distance of 4.9 meters, then if we follow
this equation.

t = sqrt( ( 2(4.9 m) ) / ( 9.8 m/s^2 ) ) = 1 s

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. Now obviously with correct
preportions of pennies, and more sophisticated mathematics, and a
better understanding of the physical formulas for gravity. We could do
a lot more. And be far more precise.

So here is my gravity theory. We are using the quadratic formula to
solve: 2*n/9.8 = k(n-k) , for k
k=(1/14) (7n +- sqrt(49 n^2 - 40 n)).

So now an example...

We are dropping a ball from 10 meters above the ground. So we plug 10
meters into n to solve for k.

k=(1/14) (7n +- sqrt(49 n^2 - 40 n))
k=9.791574237

My question to calculate the average number of coin flips in my game is
k(n-k), so we plug in k & n:

k*(10-k) = 2.040816327 = average number of coin flips

Now we take the square root of the average number of flips to get the
actual time it takes to land:

sqrt(avg flips) = 1.428571429 = number of seconds to land.

Now finally to factor in a problem with my equation we say that if k is
9.791574327, that means our large gravity pile is that many pennies.
And our small gravity pile is exactly 0.208425673 pennies!