If a coin is tossed an infinite amount of times, is it possible for it to land on heads every flip for infinity? 

Or must it eventually succumb to infinity, and land on tails?

I don't think that would be possible, since it is really a 50/50 matter and having heads for infinity would mean being really lucky with the tosses.

It's possible yes

I don't think that would be possible, since it is really a 50/50 matter and having heads for infinity would mean being really lucky with the tosses.

lol. I wasn't asking how "lucky" you would have to be for it to happen, I was asking strictly if it was possible. It dosen't matter that the chance of it happening is less than .00000000000000000000000000000000000000000000000000000000000000000000001. As long as the chance is there, it is possible.

On the other hand, in my view, it seems equally rational to assume that since it is infinity, there also is the inevitability of it landing on tails, since there are unlimited chances.

So idk.

 

I don't think that would be possible, since it is really a 50/50 matter and having heads for infinity would mean being really lucky with the tosses.

lol. I wasn't asking how "lucky" you would have to be for it to happen, I was asking strictly if it was possible. It dosen't matter that the chance of it happening is less than .00000000000000000000000000000000000000000000000000000000000000000000001. As long as the chance is there, it is possible.

On the other hand, in my view, it seems equally rational to assume that since it is infinity, there also is the inevitability of it landing on tails, since there are unlimited chances.

So idk.

 

Oh, well then ya, it can be possible

This is called the St. Petersburg paradox. It also relates to the Infinite Monkey Theorem. I will try my best to answer this question based off of those, since my knowledge relates to that.

 

"The infinite monkey theorem states that a monkey hitting keys at random on a typewriter keyboard for an infinite amount of time will almost surely type a given text, such as the complete works of William Shakespeare.

 

In this context, "almost surely" is a mathematical term with a precise meaning, and the "monkey" is not an actual monkey, but a metaphor for an abstract device that produces an endless random sequence of letters and symbols. The relevance of the theorem is questionable—the probability of a monkey exactly typing a complete work such as Shakespeare's Hamlet is so tiny that the chance of it occurring during a period of time even a hundred thousand orders of magnitude longer than the age of the universe is extremely low (but not zero).

 

There is a straightforward proof of this theorem. As an introduction, recall that if two events are statistically independent, then the probability of both happening equals the product of the probabilities of each one happening independently. For example, if the chance of rain in Moscow on a particular day in the future is 0.4 and the chance of an earthquake in San Francisco on that same day is 0.00003, then the chance of both happening on that day is 0.4 × 0.00003 = 0.000012, assuming that they are indeed independent.

Suppose the typewriter has 50 keys, and the word to be typed is banana. If the keys are pressed randomly and independently, it means that each key has an equal chance of being pressed. Then, the chance that the first letter typed is 'b' is 1/50, and the chance that the second letter typed is a is also 1/50, and so on. Therefore, the chance of the first six letters spelling banana is

(1/50) × (1/50) × (1/50) × (1/50) × (1/50) × (1/50) = (1/50)6 = 1/15 625 000 000 ,

less than one in 15 billion, but not zero, hence a possible outcome.

 

From the above, the chance of not typing banana in a given block of 6 letters is 1 − (1/50)6. Because each block is typed independently, the chance Xn of not typing banana in any of the first n blocks of 6 letters is

Xn = (1 - 1 / 50^6)^n

 

As n grows, Xn gets smaller. For an n of a million, Xn is roughly 0.9999, but for an n of 10 billion Xn is roughly 0.53 and for an n of 100 billion it is roughly 0.0017. As n approaches infinity, the probability Xn approaches zero; that is, by making n large enough, Xn can be made as small as is desired, and the chance of typing banana approaches 100%.

The same argument shows why at least one of infinitely many monkeys will produce a text as quickly as it would be produced by a perfectly accurate human typist copying it from the original. In this case Xn = (1 − (1/50)6)n where Xn represents the probability that none of the first n monkeys types banana correctly on their first try. When we consider 100 billion monkeys, the probability falls to 0.17%, and as the number of monkeys n increases, the value of Xn – the probability of the monkeys failing to reproduce the given text – approaches zero arbitrarily closely. The limit, for n going to infinity, is zero.

However, for physically meaningful numbers of monkeys typing for physically meaningful lengths of time the results are reversed. If there are as many monkeys as there are particles in the observable universe (1080), and each types 1,000 keystrokes per second for 100 times the life of the universe (1020 seconds), the probability of the monkeys replicating even a short book is nearly zero.

 

Ignoring punctuation, spacing, and capitalization, a monkey typing letters uniformly at random has a chance of one in 26 of correctly typing the first letter of Hamlet. It has a chance of one in 676 (26 × 26) of typing the first two letters. Because the probability shrinks exponentially, at 20 letters it already has only a chance of one in 2620 = 19,928,148,895,209,409,152,340,197,376 (almost 2 × 1028). In the case of the entire text of Hamlet, the probabilities are so vanishingly small they can barely be conceived in human terms. The text of Hamlet contains approximately 130,000 letters.Thus there is a probability of one in 3.4 × 10183,946 to get the text right at the first trial. The average number of letters that needs to be typed until the text appears is also 3.4 × 10183,946, or including punctuation, 4.4 × 10360,783.

Even if the observable universe were filled with monkeys the size of atoms typing from now until the end of the universe, their total probability to produce a single instance of Hamlet would still be a great many orders of magnitude less than one in 10183,800. As Kittel and Kroemer put it, "The probability of Hamlet is therefore zero in any operational sense of an event...", and the statement that the monkeys must eventually succeed "gives a misleading conclusion about very, very large numbers." This is from their textbook on thermodynamics, the field whose statistical foundations motivated the first known expositions of typing monkeys."

 

So to answer your question, yes, it is possible, but it almost surely will not happen. But you can also argue that it is not possible, because since you are flipping for an infinite amount of time, you will never know the outcome.

 

If it is flipped infinity amount of times, it has to land on tails at least once, right? 

 

It almost surely will, but not certainly

Lets do a little math (or atleast I'll make up some math, and hope it makes sense)

 

What's the chance of flipping heads once in a row? 1/2

What's the chance of flipping heads twice in a row? 1/4

What's the chance of flipping heads three times in a row? 1/8

What's the chance of flipping heads four times in a row? 1/16

 

As you can see, this follows a simple pattern of 1/2^n

 

We can express this as a product series where x_i = 1/2. For those of you unfamiliar with a product series, here is a picture http://gyazo.com/a5cd47e6f6244c345f0c4213378242d4

 

A product series is denoted by a pi symbol with a lower bound below the pi symbol, denoted by i, and an upper bound above the pi symbol, denoted by n. This is followed by an argument x_i, which in our case is 1/2.

 

If we are flipping a coin infinite times, our product series will go from i=0 to n=infinity. Which is basically just (1/2)*(1/2)*(1/2)...over and over again infinitely many times.

 

The limit of this series (i.e, the number which it converges to) is 0, because any number divided by infinity = 0. 

Therefore the probability of flipping a coin and it landing on heads infinitely many times is exactly 0.

 

Note: You cannot actually divide by infinity, I added this just for simplicity. Instead of dividing by infinity, you would take the limit of 1/x as x goes to infinity. For those who do not yet know what limits are, basically if x becomes really really really really big, what number does 1/x tend towards?

 

TL;DR: The probability of getting heads again approaches 0 as you flip the coin more times. If you flip it infinitely many times, the probability becomes 0, because that's how math works.

Edited September 29, 201312 yr by ohhungry

i'd say it's possible, but near impossible.

i'd say it's possible, but near impossible.

I just proved it's impossible! 0% chance, as in it cannot happen :P

Edited September 29, 201312 yr by ohhungry

 

i'd say it's possible, but near impossible.

I just proved it's impossible! 0% chance, as in it cannot happen

 

Your whole case seems to rest on your belief which you've "simplified" as "any number divided by infinity = 0. "

Which, you pulled out of your ass- It is not factually based, at all. I don't know where you got that from.

You're going to have to provide actual evidence that it is a 0% chance. Provide, for example, the number of times in which the coin is flipped where, from there on out it can reach heads no longer. I would be very interested, as that makes no sense at all.

Read Zappa's post.

Edited September 29, 201312 yr by StolenLogic

Lets do a little math (or atleast I'll make up some math, and hope it makes sense)

 

What's the chance of flipping heads once in a row? 1/2

What's the chance of flipping heads twice in a row? 1/4

What's the chance of flipping heads three times in a row? 1/8

What's the chance of flipping heads four times in a row? 1/16

 

As you can see, this follows a simple pattern of 1/2^n

 

We can express this as a product series where x_i = 1/2. For those of you unfamiliar with a product series, here is a picture http://gyazo.com/a5cd47e6f6244c345f0c4213378242d4

 

A product series is denoted by a pi symbol with a lower bound below the pi symbol, denoted by i, and an upper bound above the pi symbol, denoted by n. This is followed by an argument x_i, which in our case is 1/2.

 

If we are flipping a coin infinite times, our product series will go from i=0 to n=infinity. Which is basically just (1/2)*(1/2)*(1/2)...over and over again infinitely many times.

 

The limit of this series (i.e, the number which it converges to) is 0, because any number divided by infinity = 0. 

Therefore the probability of flipping a coin and it landing on heads infinitely many times is exactly 0.

 

Note: You cannot actually divide by infinity, I added this just for simplicity. Instead of dividing by infinity, you would take the limit of 1/x as x goes to infinity. For those who do not yet know what limits are, basically if x becomes really really really really big, what number does 1/x tend towards?

 

TL;DR: The probability of getting heads again approaches 0 as you flip the coin more times. If you flip it infinitely many times, the probability becomes 0, because that's how math works.

 

I am sorry, but you cannot divide by infinity, because infinity is not a number, it is a concept.

Of course its possible, just that infinity is not a number.

Just like Zappa said infinity is a concept, not a number. The chances of a coin landing on heads or tails is 50/50 and it will stay consistent no matter how many times the coin is flipped. The probability that the coin will land on one side for every flip an infinite amount of times is extremely small, but still possible.

 

Look at it this way. Since infinity represents something without a limit, in this case a number for the amount of flips, that number can always go up. No matter how high your total flip count may be, you can always go higher. This works vice versa for the chance of it landing on a certain side. First flip is 1/2, next is 1/4, then 1/8, and will continue the same way your amount of flips can increase, which is infinitely.

 

 

So in short yes, it is possible to land on one particular side an infinite amount of times, despite the infinitely decreasing chance.

 

If a coin is tossed an infinite amount of times, is it possible for it to land on heads every flip for infinity? 

Or must it eventually succumb to infinity, and land on tails?

 

It is possible. The chances of that happening is very low. The more flips you do, the lower the chances of it being all heads.