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So you keep fliping a coin. 50/50 chances

Originally posted by: Molondo
What is the chance of something repeating forever?
Click to expand...

There are many scientists researching that very same thing. A law or theory that there is nothing random,etc....
 
Originally posted by: Molondo
What is the chance of something repeating forever?
Click to expand...
In this case, extremely low. The chance of the next flip screwing up a repeating series is 50% 😉

ps - I will bet on the monkey's recreating the works of Shakespeare first.

 
Originally posted by: Molondo
What is the chance of something repeating forever?
Click to expand...

Statistically, the chances of it repeating forever would be the same as any other pattern. The reality is, however, that unless the coin was tossed in exactly the same way and under the exact same conditions, the chances of heads versus tails is not going to be the same each time. So the chances would be pretty low for you to predict any given pattern.
 
I think the probability would be (1/2)^n, where 'n' is the number of iterations. Generally, for a 1:m odds occurrence, I think it would be (1/m)^n. According to this formula, the chances of flipping 1000 heads in a row is about 9.3326e-302.
 
I think it kind of works like this.


One flip 50% chance of getting heads

two flips 33% chance of getting both two flips heads
(Possible choices, (1st flip/2nd flip) heads/heads, heads/tails, tails/heads, tails/tails))

three flips 16.66% chance of getting all three flips heads
(possible choices (1st flip/2nd flip/3rd flip) (heads/heads/heads, heads/tails/heads, heads/heads/tails, tails/heads/heads, tails/tails/heads, tails/tails/tails)


I mean, already at 4 flips you are talking about a very minute chance of being able to flip all four coines heads, but of course it is possible considering that there is always that pinhole small .000000000000-infiti-0001% chance that the flip could be landing on heads every single time and never stop.


 
Originally posted by: mozirry
I think it kind of works like this.


One flip 50% chance of getting heads

two flips 33% chance of getting both two flips heads
(Possible choices, (1st flip/2nd flip) heads/heads, heads/tails, tails/heads, tails/tails))

three flips 16.66% chance of getting all three flips heads
(possible choices (1st flip/2nd flip/3rd flip) (heads/heads/heads, heads/tails/heads, heads/heads/tails, tails/heads/heads, tails/tails/heads, tails/tails/tails)
Click to expand...

Except that ALL combinations of heads and tails have the same chance of happening assuming the toss happens under the same circumstances each time. So you are no more likely to get 10 heads and 10 tails in 20 tosses than you are of getting all 20 heads or all 20 tails. You have the same chance each time of getting heads or tails, regardless of what came up in previous tosses.
 
Originally posted by: HeartView
So you are no more likely to get 10 heads and 10 tails in 20 tosses than you are of getting all 20 heads or all 20 tails. You have the same chance each time of getting heads or tails, regardless of what came up in previous tosses.
Click to expand...

Yes, you are. You are 184,756 times likely to get 10 heads and 10 tails, than 20 tails.

However, it is equally likely to get 10 heads then 10 tails, as 20 tails.
 
What CycloWizard said: (1/2)^n. If n goes towards infinity, the resulting probability approximates to zero. Or step by step:

50% probability for 1 flip
25% for 2 flips
12.5% for 3 flips
...
0% for an infinite amount of flips.
 
Originally posted by: mozirry
I think it kind of works like this.


One flip 50% chance of getting heads

two flips 33% chance of getting both two flips heads
(Possible choices, (1st flip/2nd flip) heads/heads, heads/tails, tails/heads, tails/tails))

three flips 16.66% chance of getting all three flips heads
(possible choices (1st flip/2nd flip/3rd flip) (heads/heads/heads, heads/tails/heads, heads/heads/tails, tails/heads/heads, tails/tails/heads, tails/tails/tails)


I mean, already at 4 flips you are talking about a very minute chance of being able to flip all four coines heads, but of course it is possible considering that there is always that pinhole small .000000000000-infiti-0001% chance that the flip could be landing on heads every single time and never stop.
Click to expand...

Actually, that works out to 25% for 2 heads in a row. (You listed the 4 possibilities; it's one out of those 4) This would be (1/2)^2
The probability of 50 heads in a row is (1/2)^50

edit: That's really weird. I'd swear chcarnage's post wasn't there when I was viewing the thread... Now it shows up with a "new" icon on the top of the post. It was posted longer ago than I came online (dial-up)... no other posts show as "new" in the thread. Weird.
 
Originally posted by: chcarnage
What CycloWizard said: (1/2)^n. If n goes towards infinity, the resulting probability approximates to zero. Or step by step:

50% probability for 1 flip
25% for 2 flips
12.5% for 3 flips
...
0% for an infinite amount of flips.
Click to expand...

bingo
 
So is it zero? Or as close to zero as possible without actually hitting it?

So we've determined that it's extremely close to 0% chance, but is it exactly 0%? Or is it ".000000000000-infiti-0001%"?
 
Originally posted by: DyslexicHobo
So is it zero? Or as close to zero as possible without actually hitting it?

So we've determined that it's extremely close to 0% chance, but is it exactly 0%? Or is it ".000000000000-infiti-0001%"?
Click to expand...

".000000000000-infiti-0001%" is zero. (This one has been around a few times in ATOT)
 
Originally posted by: chcarnage
Originally posted by: DyslexicHobo
So is it zero? Or as close to zero as possible without actually hitting it?

So we've determined that it's extremely close to 0% chance, but is it exactly 0%? Or is it ".000000000000-infiti-0001%"?
Click to expand...

".000000000000-infiti-0001%" is zero. (This one has been around a few times in ATOT)
Click to expand...

This comes up in 1=.999... You can't have infinite zeros with a 1 at the end. The zeros would NOT be infinite if they ended with a 1!
 
Originally posted by: chcarnage
Originally posted by: DyslexicHobo
So is it zero? Or as close to zero as possible without actually hitting it?

So we've determined that it's extremely close to 0% chance, but is it exactly 0%? Or is it ".000000000000-infiti-0001%"?
Click to expand...

".000000000000-infiti-0001%" is zero. (This one has been around a few times in ATOT)
Click to expand...

You can't even have a 1 at the "end" if you have already specified that it is an infinite sequence of 0's.

 
The standard answer is that its, informally, 2 to the power of -infinity, or zero. The formal answer is its the limit of 2^-n as n approaches infinity. Which is also zero.

In non-standard analysis, the answer is not zero ; it differs by a computable amount. The answer is 2^-H, where H is an enormous integer that is much larger than any real number.
 
A number of years ago I wrote a couple of Basic programs to test randomness on a computer. The first one was the coin toss with the 2nd one a pair of dice. As you would expect the more rolls in a run the closer the results came to what you would expect. With the dice program I generated a graph at the end of a run showing the distribution. The resulting shape would fit a normal distribution curve (bell shapped) with a large number of rolls. It has been over 20 years since I did that, so I can't give exact results, but it was as one would expect.

In all the runs I never saw it repeat for a run. There was always variation. Others have given formulas that calculate the chance of this happening and it very quickly goes to near infinity.
 
Just to confuse the issue and make some people ponder...

What's the probability that the universe would exist as it is now? With seemingly infinte possibilities over an unmeasured timeframe, would the chance be 0%? So the world does exist as it is today...was the chance of that 0%? Would the chance for any state be 0%? If all those 0%'s didn't add up to more than 0%, it would impossible for the universe to exist in any state.

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.
 
Also, if you consider some of the ideas in the randomness thread or the radioactive decay thread involving randomness, you could argue that there is no randomness in the universe and the chance it exists as it does now is 100%, and the chance of all other states is 0%. Though I would hate to throw away my free will so casually.
 
To clarify, P({Event}) = 0 in no way shape or form implies that {event} is impossible. Think of a cdf.

If all the 0's didn't add to exactly 1, then it makes no sense to talk about it in terms of probability, as you're not dealing with a probability distribution at all (this is one part of the definition of a pdf).

Randomness is an explanation for factors that cannot be estimated. There are a lot fo those.

.000...1 is not a number that exists. Infinitessimal is a concept, not a number, much like it's antonym, infinite.
 
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