So you keep fliping a coin. 50/50 chances

[DyslexicHobo]

Originally posted by: zerogear
IMO there is no 'randomness' in flipping a coin. Its simply that we haven't been able to determine all the actual factors, when we can't calculate every single factors of the chance, it becomes 'random'

Of course there is no randomness. But I believe it is impossible to condition your body to be able to flip the coin the exact amount of times and have it land on a (hard) surface the exact same way and bounce the same way every time.

You might get lucky a couple times in a row, but I doubt more than that.

 

[Who Me]

Randomness cannot exist in a deterministic universe.
And to my mind, the universe we inhabit is deterministic.
Ergo - yes, there is no randomness.

 

[firebyyrd]

according to the theory of evolution.......

given time, the coins will turn into a human being!

 

[Aisengard]

I thought there was actually randomness in quantum mechanics, hence the Einstein quote "God doesn't play dice with the universe" when discussing his dislike of the topic.

 

[CycloWizard]

Originally posted by: Who Me
Randomness cannot exist in a deterministic universe.
And to my mind, the universe we inhabit is deterministic.
Ergo - yes, there is no randomness.

Begging the question. "If A, then not B. I assume A, therefore, not B." Randomness either exists or it doesn't. Whether it does is still an open question. However, that has little to do with this topic, which is likely better described by chaos theory than true randomness.

 

[thesurge]

Originally posted by: Matthias99

Originally posted by: Skotty
Also, can you mathematically proove that .000...infinity...001 is equal to 0? I know it's easy to prove .9 repeating is equal to 1 by simply multiplying by 10 and cancelling out the .9 repeating. Through that you could claim that .9 repeating + .0...1 is equal to 1, and since .9 repeating is equal to 1, .0...1 must equal 0, but I'm not sure that is mathematically sound.

Mathematically, "0.0000...0001" is not a number (at least not a real or complex number). There's no way to actually define it. Logically, an "infinite number of zeroes followed by a one" cannot exist, since if there is a 1 at the end the zeroes would not actually be infinite.

"0.9999..." is actually "the limit as x goes to infinity of (the sum of i from 1 to x of (9 * (1/(10^i)))" -- that is, 0.9 + 0.09 + 0.009 + 0.0009 + ..., which is a well-defined infinite arithmetic series. This series is exactly equal to one.

The best you can do for "0.000...0001" is "the limit as x goes to infinity of (1/(10^x))". This limit is actually what you get when you take 1 and subtract (0.9 + 0.09 + 0.009 + ...) from it. That limit is exactly equal to 0 -- that's the way limits are defined.

geometric

 

[Soccerman06]

Doesnt quantom physics say both iterations are possible at the same time, ie its always going to happen the same everytime (superposition).

 

[darkhorror]

Seems like a simple problem to me, just take your basic formula to get the number of heads at any number N, then take the limit as N goes twards infinity.

I think the problem is that people don't really know how to think of infinity and they try to limit it to something. Try and give an exact answer to what the exact rolls in order would be with out one that ends up repeating. If you look at this like you flip a coin then flip again, and keep on going then you arn't looking at infinity because no matter how many times you flip a coin you will never "reach infinity"

 

[Unitary]

Assuming idealized coin flips, the correct language would be to say that the coin flips will "almost surely" not repeat forever. The outcomes that repeat forever form a set of measure zero within our probablility space, meaning it's not likely at all (probability zero), but sill possible for that event to happen. The idea is that we find the probability by integrating (summing) over all the events that we want and the result would be the probability of one of the events happening. Since the events we are intersted in have measure zero our integral, more specifically the probability, is 0 meaning it won't happen almost surely.

 

[Fallen Kell]

Originally posted by: Molondo
What is the chance of something repeating forever?

Very simple. Go read up on the "Poisson distribution", as this is exactly the answer you are looking for. The Poisson distribution will tell you the probability of getting a specific result from a number of repeated events that have a set probability outcome.

So for your answer, you will want to look at the Limit of the Possion Distribution as the k approaches infinity.

lim->k,infinity f(k,lamda) = (e^(-lamda)*lamda^(k))/k!
where k is the unmber of occurrences of the event (in your case you are looking at infinity)
lamda is the probability of receiving the result desired result from your test (in this case, the probability of getting "heads" in a coin flip is ".5")

So as you can see as k approaches infinity, the above formula approaches 0 VERY quickly, as you are effectively dividing (e^(-.5)*(.5^infinity) by infinity! (the factorial of infinity), which is an extremely high order level of infinity as you are esseitially multiplying infinity by itself infinite number of times!!!

So you have e^(-.5) * (.5^infinity) in the numerator. Well, what is .5^infinity? .5*.5*.5*.5... -> approximates 0.

Now you have (~0)/(infinity!)

So you have a numerator which approaches 0, divided by an extremely high magnitude infinity on the denominator ( infinity! -> infinity^(infinity) ). We know that any number divided by another number which approaches infinity approaches 0 for the answer.

In this equation we have 2 distinctly separete events which are forcing the resulting value to be 0. The value of the numerator itself is approaching 0 as k approaches infinity. So by the rule that 0 divided by any number is also zero, the answer is 0.

We also have the fact that as any value divided by another number that is approaching infinity makes the answer approach 0.

We have a value in the numerator approaching 0 and a value in the denominator approaching infinity, the resulting answer is 0.

So the answer to your question is "0".

 

[DyslexicHobo]

Originally posted by: Fallen Kell

Originally posted by: Molondo
What is the chance of something repeating forever?

Very simple. Go read up on the "Poisson distribution", as this is exactly the answer you are looking for. The Poisson distribution will tell you the probability of getting a specific result from a number of repeated events that have a set probability outcome.

So for your answer, you will want to look at the Limit of the Possion Distribution as the k approaches infinity.

lim->k,infinity f(k,lamda) = (e^(-lamda)*lamda^(k))/k!
where k is the unmber of occurrences of the event (in your case you are looking at infinity)
lamda is the probability of receiving the result desired result from your test (in this case, the probability of getting "heads" in a coin flip is ".5")

So as you can see as k approaches infinity, the above formula approaches 0 VERY quickly, as you are effectively dividing (e^(-.5)*(.5^infinity) by infinity! (the factorial of infinity), which is an extremely high order level of infinity as you are esseitially multiplying infinity by itself infinite number of times!!!

So you have e^(-.5) * (.5^infinity) in the numerator. Well, what is .5^infinity? .5*.5*.5*.5... -> approximates 0.

Now you have (~0)/(infinity!)

So you have a numerator which approaches 0, divided by an extremely high magnitude infinity on the denominator ( infinity! -> infinity^(infinity) ). We know that any number divided by another number which approaches infinity approaches 0 for the answer.

In this equation we have 2 distinctly separete events which are forcing the resulting value to be 0. The value of the numerator itself is approaching 0 as k approaches infinity. So by the rule that 0 divided by any number is also zero, the answer is 0.

We also have the fact that as any value divided by another number that is approaching infinity makes the answer approach 0.

We have a value in the numerator approaching 0 and a value in the denominator approaching infinity, the resulting answer is 0.

So the answer to your question is "0".

But using this answer we reach an anomaly. If each of the possible sequences of coin flips have an equal chance to occur, every outcome has a 0% chance of being true. If we add up all of these sequences, we should come up with 100% (such as if we flipped one coin once, 50% tails + 50% heads = 100%). However, 0+0+...+0+0 = 0!

 

[Unitary]

Originally posted by: DyslexicHobo

Originally posted by: Fallen Kell

Originally posted by: Molondo
What is the chance of something repeating forever?

Very simple. Go read up on the "Poisson distribution", as this is exactly the answer you are looking for. The Poisson distribution will tell you the probability of getting a specific result from a number of repeated events that have a set probability outcome.

So for your answer, you will want to look at the Limit of the Possion Distribution as the k approaches infinity.

lim->k,infinity f(k,lamda) = (e^(-lamda)*lamda^(k))/k!
where k is the unmber of occurrences of the event (in your case you are looking at infinity)
lamda is the probability of receiving the result desired result from your test (in this case, the probability of getting "heads" in a coin flip is ".5")

So as you can see as k approaches infinity, the above formula approaches 0 VERY quickly, as you are effectively dividing (e^(-.5)*(.5^infinity) by infinity! (the factorial of infinity), which is an extremely high order level of infinity as you are esseitially multiplying infinity by itself infinite number of times!!!

So you have e^(-.5) * (.5^infinity) in the numerator. Well, what is .5^infinity? .5*.5*.5*.5... -> approximates 0.

Now you have (~0)/(infinity!)

So you have a numerator which approaches 0, divided by an extremely high magnitude infinity on the denominator ( infinity! -> infinity^(infinity) ). We know that any number divided by another number which approaches infinity approaches 0 for the answer.

In this equation we have 2 distinctly separete events which are forcing the resulting value to be 0. The value of the numerator itself is approaching 0 as k approaches infinity. So by the rule that 0 divided by any number is also zero, the answer is 0.

We also have the fact that as any value divided by another number that is approaching infinity makes the answer approach 0.

We have a value in the numerator approaching 0 and a value in the denominator approaching infinity, the resulting answer is 0.

So the answer to your question is "0".

But using this answer we reach an anomaly. If each of the possible sequences of coin flips have an equal chance to occur, every outcome has a 0% chance of being true. If we add up all of these sequences, we should come up with 100% (such as if we flipped one coin once, 50% tails + 50% heads = 100%). However, 0+0+...+0+0 = 0!

It certainly can! It's another time our intuition fails when we deal with infinite things. Take, for example, the Riemann integral. We estimate the area under the curve by approximating it by rectangles. In the limit, the width of the rectangles goes to zero. So the area under the curve would effectively be "0+0+0+....", but the sum, the area under the curve, is most likely not 0.

 

[DyslexicHobo]

Originally posted by: Unitary
It certainly can! It's another time our intuition fails when we deal with infinite things. Take, for example, the Riemann integral. We estimate the area under the curve by approximating it by rectangles. In the limit, the width of the rectangles goes to zero. So the area under the curve would effectively be "0+0+0+....", but the sum, the area under the curve, is most likely not 0.

Oh, good analogy. I give up with infinity. This is the second time today where I've looked at what seem to be by-products of a faulty number system and have gotten completely confused.

I don't know who's to blame: my simple-minded human brain, or a number system that my brain can't comprehend!

 

[DarkThinker]

Originally posted by: Born2bwire

Originally posted by: Molondo
What is the chance of something repeating forever?

We're still working on it. /flip

bahahaaa :beer:

 

[darkhorror]

Originally posted by: DyslexicHobo

Originally posted by: Fallen Kell

Originally posted by: Molondo
What is the chance of something repeating forever?

Very simple. Go read up on the "Poisson distribution", as this is exactly the answer you are looking for. The Poisson distribution will tell you the probability of getting a specific result from a number of repeated events that have a set probability outcome.

So for your answer, you will want to look at the Limit of the Possion Distribution as the k approaches infinity.

lim->k,infinity f(k,lamda) = (e^(-lamda)*lamda^(k))/k!
where k is the unmber of occurrences of the event (in your case you are looking at infinity)
lamda is the probability of receiving the result desired result from your test (in this case, the probability of getting "heads" in a coin flip is ".5")

So as you can see as k approaches infinity, the above formula approaches 0 VERY quickly, as you are effectively dividing (e^(-.5)*(.5^infinity) by infinity! (the factorial of infinity), which is an extremely high order level of infinity as you are esseitially multiplying infinity by itself infinite number of times!!!

So you have e^(-.5) * (.5^infinity) in the numerator. Well, what is .5^infinity? .5*.5*.5*.5... -> approximates 0.

Now you have (~0)/(infinity!)

So you have a numerator which approaches 0, divided by an extremely high magnitude infinity on the denominator ( infinity! -> infinity^(infinity) ). We know that any number divided by another number which approaches infinity approaches 0 for the answer.

In this equation we have 2 distinctly separete events which are forcing the resulting value to be 0. The value of the numerator itself is approaching 0 as k approaches infinity. So by the rule that 0 divided by any number is also zero, the answer is 0.

We also have the fact that as any value divided by another number that is approaching infinity makes the answer approach 0.

We have a value in the numerator approaching 0 and a value in the denominator approaching infinity, the resulting answer is 0.

So the answer to your question is "0".

But using this answer we reach an anomaly. If each of the possible sequences of coin flips have an equal chance to occur, every outcome has a 0% chance of being true. If we add up all of these sequences, we should come up with 100% (such as if we flipped one coin once, 50% tails + 50% heads = 100%). However, 0+0+...+0+0 = 0!

You are trying to take the limit and apply it as the value this is not how it works.

 

[BassBomb]

Interesting discussions!

 

[CycloWizard]

Originally posted by: DyslexicHobo
Oh, good analogy. I give up with infinity. This is the second time today where I've looked at what seem to be by-products of a faulty number system and have gotten completely confused.

I don't know who's to blame: my simple-minded human brain, or a number system that my brain can't comprehend!

If it makes you feel better, +/- infinity isn't generally included in the set of all real numbers - they must be made up!

 

[QuantumPion]

I think given an infinite number of flips, it is possible to achieve an arbitrarily large but finite number of repeats in a row. But the chance of getting infinite repeats in infinite time is undefined, because it boils down to infinities divided by infinities and such.