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General Science / The odds of a coin landing vertically? + 51/49 theory
« on: 22/12/2016 12:17:53 »
Is it possible or easy at all to calculate mathematically?
If you considered only the initial landing position and ignored the physics that happened thereafter, would it just be surface area of the sides X and surface area of the width Y where the odds would be 1 in (X:Y)?
Of course, if you included actual physics, that would reduce the odds as there is a considerable chance that the coin would flip over after landing vertically due to momentum, angle, etc.
Also, while I'm at it, another question (but pls answer the first):
I've read a few times a supposedly true fact that the coin has a 2% higher chance of landing on the upper side due to that side being upwards the same or one more amount of times compared to the lower side.
However, this does not make sense to me as no one would consider 0 flips to be a valid coin toss, obviously. Neither would they accept 1 flip as valid, most likely.
So it depends on the minimum amount of flips one deems to validate the coin toss. I would think that number would be 2, if I had to guess, so that would bring us again to the upper side having the advantage; but wouldn't it have a higher advantage than 2% then?
Surely, if you consider the minimum amount of flips that makes a coin toss valid + the realistic amount of possible coin flips in a real toss, wouldn't it be hard to arrive at the value of +2%? If I'm wrong, please explain why that specific value.
I have zero education in mathematics, so forgive me if I'm missing some major, well known equations. Odds and probabilities fascinate me and I like to think of them and try to apply them in games/uncertain situations etc. so answers to questions like these are most welcome.
If you considered only the initial landing position and ignored the physics that happened thereafter, would it just be surface area of the sides X and surface area of the width Y where the odds would be 1 in (X:Y)?
Of course, if you included actual physics, that would reduce the odds as there is a considerable chance that the coin would flip over after landing vertically due to momentum, angle, etc.
Also, while I'm at it, another question (but pls answer the first):
I've read a few times a supposedly true fact that the coin has a 2% higher chance of landing on the upper side due to that side being upwards the same or one more amount of times compared to the lower side.
However, this does not make sense to me as no one would consider 0 flips to be a valid coin toss, obviously. Neither would they accept 1 flip as valid, most likely.
So it depends on the minimum amount of flips one deems to validate the coin toss. I would think that number would be 2, if I had to guess, so that would bring us again to the upper side having the advantage; but wouldn't it have a higher advantage than 2% then?
Surely, if you consider the minimum amount of flips that makes a coin toss valid + the realistic amount of possible coin flips in a real toss, wouldn't it be hard to arrive at the value of +2%? If I'm wrong, please explain why that specific value.
I have zero education in mathematics, so forgive me if I'm missing some major, well known equations. Odds and probabilities fascinate me and I like to think of them and try to apply them in games/uncertain situations etc. so answers to questions like these are most welcome.