# >Let's see, if I flip a coin 10000 times in one session and then flip it >10000 times in a

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>Let's see, if I flip a coin 10000 times in one session and then flip it
>10000 times in another session, what are the odds that I will have the
>exact same run of heads and tails for both sessions? It would be
>1/2^10000, which is a pretty small number. Considering that the odds of
>duplicating two exact runs of 10000 tosses is so small, it must mean
>that the God is directing the coin throws right?

Let's turn the screw so it bites some more for these people.

let's say I'm flying along in an airplane, and fire a gun. Now,
at 10 km up, I might have a maximum range of 20 kilometers till the
bullet hits the ground. [ I don't know, no flames on this! ]  The possible
landing zone area is Pi R^2, or 1256 square km. Or, 1.25 X 10^9 m^2.
If the bullet is axp 1/2 cm in radius, this gives an axproximate probability
of hitting any square centimeter of 8 x 10^-14!

Awe hell, let's get more specific. The change that the very tip of
the bullet will hit a certain area of 1 Ao^2 is ~ 1 x 10^-27.

Nope, that's not specific enough. We have to account the probability
of the alignment of the very atoms in the bullet. Let's say it is 200 gm of
lead, or one mole 6.02 x 10^23 atoms. Over my head, as I do not know the
lattice structure of Pb.

Let's make it simple:
First atom: 6 x 10 ^23
Second atom: 6 X 10^23 -1
Etc...
last atom: 1

Whoops, my calculator just gave out...
What the hell is the (6 X 10 ^23)!

okay, okay, let's get out the CrC
*grumble* *grumble* *grumble*

n*(n-1)*(n-2)...(1)   where n=6 x 10^23

erg. can't find it for a number that large.

Okay, let's get real sloppy.
There will be 5 x 10^23 of 10^23 terms
There will be 9 X 10^22 of 10^22 terms
etc.
9 x 10^0  of 10^0  terms

so, the total would be of the order
10 ^(4.64 x 10 ^42 )

Yowzzers! I probably made a mistake there, but you get the picture.

So 10^(-4.64 x 10^42) is the probability of the formation of the
atoms.
The total probabilty ( so far ) of the structure, and its landing
zone is just of the order: 10^((-5. x 10^ 42) -27)    Ouch!

Now of course, we need the structure of each atom....
Then of course, we'll have to take into the account of the exact
trajectory, location, speed, direction, etc! In addtion, we need to know the
exact moment of impact!

Awe hell, we all know we can keep making the gradient finer and
finer [ to the limit of quantum mechanics at least }

Let's just make the probabilty a nice round number of
10^(-10^1000)

Or 1/1000.... (a gougleplex 0's)

Clearly, the chance of the bullet landing where it does, in the
manner that it did, is so astronomically

( considering that this is magnitudes far far above the total
number of subatomic particles, let alone neutrinos and photons
in the universe, its far beyond any astronomic term!)

small, it MUST have been guided by an all powerful and all knowing god!
Logic dictates that it must be! There is no way we can even imagine a number
a hundred magnitudes larger, let alone comprehend the number.

If you still don't, let's add in the probability for the structure
and behavior of the entire universe! Tack on another couple of 10^1000 terms
there. Get absolutely downright silly. The chance of the bullet travelling
as it did, with spacial coordinates in reference to every other single
particle in the universe would be something like

10^(-1000000^10000000^1000000000^100000000 )

Unfortunately, the said bullet hit the Pope right between the eyes
and he was killed instantly.
Clearly, it was an act of God when speaking ex cathedra.

How could you possibly deny it?

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E-Mail Fredric L. Rice / The Skeptic Tank