A simple question that many don't know the answer to

How can you tell if a coin will be on heads or tails?

I was just thinking what if I could always know the outcome of a coin toss with 100% accuracy?

I decided to see if I could figure it out.

So I started with the mass of the coin.

I will denote this as M.

Then I decided to figure out the amount of force that is applied upward on the object.(In Newtons)

I will denote this as F.


After I needed to know at what angle the coin is placed on the finger.

I will denote this as A.

I also needed to map what the air pressure is from the finger to about 7 feet from the ground. (Has to be equal throughout the area)

I will denote this as AP.
I will denote the increase or decrease in air pressure by n-10% or n+10%


So now I will put this in to an equation to see what the outcome is.

Ok the coin we are choosing is a quarter.

Now we divide the mass of the coin by the force applied on it to get the acceleration.

This looks like...

M/F=Ac(Acceleration is Ac)

Now this is where the angle comes into play. Whichever side the quarter is lowest then it will move toward that side. But the tilt decreases with the increase in air pressure, So we have to decrease the speed by a certain percent due to the pressure.

This looks like...

Ac - n-10% n-20% n-30%

(And so forth until acceleration reaches zero.)

You do the same process but increase the pressure as the object gets lower until it goes to a stop.

This can show you how much rotations the object has gone through.

If it started at heads and ends up being 10.5 it will be at tails.

But if it becomes 10 complete rotations then it will have been at heads.

Basically if it is a number and .5 it will not be what you began with.

But if it is a whole number then it will be what you began with.

So if you have any questions, comments, concerns, or conundrums please post them here.

Here I will tell you

Flip a coin ten times

Here are the results

head
head
tail
head
tail
tail
tail
head
tail
head

Simple question aye?
second line.

reply to post by calstorm


If you think about it.
2nd line

You're missing important part here ... what if coin drops on the edge ... 3rd possibility?

reply to post by valhala


It would be at .75

reply to post by Gentill Abdulla


So basically I have .33333~ to hit tails or head?

Edited to fix typos (sorry english not my PL)

[edit on 28.6.2010 by valhala]

Originally posted by valhala
reply to post by Gentill Abdulla

 

So basicly I have .33333~ to hit tails or head?

A .74999999999~ percent chance of heads or tails.

If everything is EXACTLY aligned then you still have a 24.00000000000000~ chance of getting it halfway.

[edit on 28-6-2010 by Gentill Abdulla]

reply to post by Gentill Abdulla


OK ... I have 3 possibilities to hit ... head/tail/edge ... that's .3333~

Originally posted by valhala
reply to post by Gentill Abdulla

 

OK ... I have 3 possibilities to hit ... head/tail/edge ... that's .3333~

But there are different variables that change the percentage of each possible action.

Which is why there is a .749999999~ that you can get either heads or tails.

I have a book called " Numbers" In it there was a piece where it talked about Pascal and Fermat gambling on by flipping a coin. If the coin landed heads up Fermat won a point, tails up Pascal won a point. 1st to 10 points win. Both put down 50 francs, so the total would be 100 francs. They played the game until it was 8 points for Fermat and 7 points for Pascal.(Also keep in mind that they did this threw sending letters of there results to each other back and forth) They had to stop playing tho because Fermat received a message that a friend was very ill and he had to leave immediately. Here is a part from the book on how they decided to split the pot.

Dear Blaise.
As to the problem of how to divide the 100 francs, I think I have found a solution that you will find fair. Seeing as I needed only 2 points to win the game, and you needed 3,I think we can establish that after four more tosses of the coin, the game would have been over. For, in those four tosses, if you did not get the necessary 3 points for your victory, this would imply that I had in fact gained the necessary 2 points for my victory. In a similar manner, if I had not achieved the necessary 2 points for my victory, this would imply that you had in fact achieved at least 3 points and had therefore won the game. Thus, I believe the following list of possible ending to the game is exhaustive. I have denoted "heads" by an "h" and tails by a "t". I have starred the outcomes that indicate a win for myself.

hhhh* hhht* hhth* hhtt*

hthh* htht* htth* httt

thhh* thht* thth* thtt

tthh* ttht ttth tttt

I think you will agree that all of these outcomes are equally likely. Thus I believe that we should divide the stakes by the ration 11:5 in my favor, that is, I should receive (11/16)*100=68.75 francs, while you should receive 31.25 francs

I hope all is well in Paris. Your friend and colleague
Pierre

reply to post by Gentill Abdulla


Can't agree ... there is equal chance to get either head, tail or edge ... mathematicly speaking.
Variables are figure of mind (or physics on this planet) (sorry for being blunt)

Originally posted by valhala
reply to post by Gentill Abdulla

 

Can't agree ... there is equal chance to get either head, tail or edge ... mathematicly speaking.
Variables are figure of mind (or physics on this planet)

(sorry for being blunt)

Mathematically speaking a variable can change the mathematical probability of any outcome.

Originally posted by Gentill Abdulla

Originally posted by valhala
reply to post by Gentill Abdulla

 

Can't agree ... there is equal chance to get either head, tail or edge ... mathematicly speaking.
Variables are figure of mind (or physics on this planet)

(sorry for being blunt)

Mathematically speaking a variable can change the mathematical probability of any outcome.

Why not write a computer input/output program of the angles, what it starts on, air pressure, etc.?

reply to post by Gentill Abdulla


Let's take simple formula a * b = c.
I think we both agree we have only (3) variables here.

But only 3 numbers fit into this equation to make it right.

reply to post by Phlynx


Give me all input variables you think we need and I'll write a proggy

The mass of the coin is not equal all around. That means your equation would only work in a vacuum? The force up is not equal to the force down with normal G.

Since there are only two sides to a coin. Its logic that a even number would be the right answer. There are only two possibilities.

But how would you know beforehand which side the coin would land on before you tossed it?

Your equation does not tell you that.

[edit on 27.06.08 by spy66]

reply to post by spy66


There are 3 sides of coin!

I wouldn't mind working on this but first I have to find a quarter.


Oh almost forgot - then I have to find my Math skills!

Originally posted by valhala
reply to post by spy66

 

There are 3 sides of coin!

Well than you have to add the odds for that ever happening after you have tossed it in the air.

But if the bet is heads or tail. The third side wont count, its not a part of the bet.

[edit on 27.06.08 by spy66]