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After reading this paragraph:

A simpler version of this distinction might be more palatable: flip a coin infinitely many times. The probability that you flip heads every time is zero, but it isn't impossible (at least, it isn't more impossible than flipping a coin infinitely many times to begin with!).

From here: Is it generally accepted that if you throw a dart at a number line you will NEVER hit a rational number?

I thought to myself, why is the probability of flipping heads every time zero? If we have a perfect coin and we flip it twice we get the probability to equal this:

$$\frac 12 \times \frac 12 = \frac 14$$

or for four flips we get this:

$$\frac 12 \times \frac 12 \times \frac 12 \times \frac 12 = \frac 1{16}$$

So as we approach an infinite amount of coin flips, the probability gets smaller and smaller, but it shouldn't ever reach zero, so the probability shouldn't be zero, it should be $1/+\infty$. Does this mean that $1/+\infty$ equals zero or have I misunderstood the question?

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Everyone below is trying to be rigorous, but for all intents and purposes, yes $1/\infty=0$, and this naturally occurs by simply taking your real numbers $\mathbb{R}$ and tacking on the additional element $\infty$... the real problem comes in when you want to consider $\infty/\infty$ or $0/0$. –  Chris Gerig Jun 10 '12 at 22:44
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Should this be closed as a duplicate of this question? –  Zev Chonoles Jun 11 '12 at 4:16
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6 Answers

Some of the answers so far have claimed that it is not possible to make sense of flipping a coin infinitely many times. In mathematics, this is not true; there is a perfectly well-defined mathematical model (a probability space) of what it means to flip a coin infinitely many times. This model has the following properties:

  • It assigns a probability, which is a non-negative real number, to various sets of possible infinite sequences of coin flips. (There are technicalities and subtleties around the issue of what sets have well-defined probabilities that are not relevant to this question.)
  • This assignment $\mu$ has the property that the probability that an infinite sequence of coin flips starts with any particular sequence of coin flips of length $N$ is $\frac{1}{2^N}$.
  • This assignment $\mu$ also has the property that if $S$ is a set of sequences of coin flips and $T$ a set of sequences of coin flips which contains $S$, then $\mu(S) \le \mu(T)$. (This is a general property of measures.)

From this the probability of only flipping heads is uniquely determined: it is at most the probability of flipping $N$ heads first for every positive integer $N$ (so at most $\frac{1}{2^N}$), and it is a non-negative real. The only non-negative real with this property is $0$ by the Archimedean property of the reals. So the probability of only flipping heads must be $0$.

One can think of this argument as computing a limit via the monotone convergence theorem for sets, but it is computing something slightly more basic, namely an infimum. In any case, $\frac{1}{\infty}$ is not a probability because it is not a real number.

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"So the probability of only flipping heads must be 0." Given the assumption that every probability has to come as a non-negative real number. The catch lies in asking "is that assumption really correct"? I don't see how you can just write off the notion of non-real probabilities such as infinitesimal probabilities or fuzzy probabilities... though you can certainly focus on real-valued probabilities to see where such a theory goes. –  Doug Spoonwood Jun 10 '12 at 22:56
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@Doug: I'm not writing it off. That simply isn't how the standard Kolmogorov-style axiomatization of probability works, which is what I'm presenting ("this model has the following properties..."). Of course mathematicians are free to develop other axiomatizations of probability. I am sure someone has thought about nonstandard probability spaces which have probabilities taking values in the nonstandard reals. –  Qiaochu Yuan Jun 10 '12 at 23:10
    
Glad you made that clear. One might get the wrong idea from the term "properties". –  Doug Spoonwood Jun 11 '12 at 0:43
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You can't flip a coin an infinite number of times. The probability of such a sequence of coin flips always yielding 0, comes as an extrapolation from flipping the coin a finite number of times. The probability of "0" here indicates a limit not an actual frequency. I agree that such might work out better as (1/+inf), but that does NOT imply that (1/+inf)=0, since (1/+inf) is NOT a number. Which positive infinity do you mean when you write "+inf"? We could just say that such a probability equals an infinitesimal, and I doubt that any real objection to that will exist (only people who object to non-zero infinitesimals would object, I think).

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What do you mean, when you ask "which positive infinity do you mean"? I thought there were only two infinites? –  Ell Jun 10 '12 at 14:03
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I like non-zero infinitesimals, and I would object on the grounds that in every number system I know that has an element named $+\infty$, we have $1/(+\infty) = 0$. –  Hurkyl Jun 10 '12 at 14:04
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@Ell, I think you might want to look at transfinite numbers. $\infty$ isn't a real number, and so when you try to talk about infinite as more than a limiting process you run into some very surprising and interesting conclusions. Infinities often require you to get very nit-picky when working with them, since they often come to results that don't hold in finite cases. –  Robert Mastragostino Jun 10 '12 at 14:18
    
@Hurkyl In a hyperreal number system, (1/+oo) equals a positive infinitesimal, where +oo indicates an infinite positive number. –  Doug Spoonwood Jun 10 '12 at 22:44
    
@Ell You might also want to take a look at Jerome Keisler's Elementary Calculus book: math.wisc.edu/~keisler/calc.html –  Doug Spoonwood Jun 10 '12 at 22:45
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The answer to your question is available in this wiki page. In particular look at the section on tossing a coin.

The explanation to your question is convergence. We aren't really considering $1/\infty$ but rather the sequence $1/n$ as $n \rightarrow \infty$.

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I think you are confusing "equal to" with the concept of the limit. In mathematics one defines the limit of a sequence to be a certain number, if the result is "closer" then any deviation from that number.

In your example, the probability of throwing heads $N$ times is $2^{-N}$. For any small deviation from zero $\epsilon$ we can find an $N$ such than $2^{-N} < \epsilon$. It is therefore correct to state:

The limit of probability of throwing heads $N$ times as $N$ goes to infinity is zero

This is often incorrectly shortened to:

The probability of flipping heads every time is zero

The reason this last statement is incorrect, is simply because the probability of flipping heads "every time" is simply undefined without first defining what you mean by "every time". This is exactly why we need the more formal definition of the limit.

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There is nothing incorrect about what I said. There is a completely well-defined probability space (en.wikipedia.org/wiki/Probability_space) whose points are countable sequences of coin flips, there is a completely well-defined probability measure on this space with the obvious properties, and relative to this measure the probability of any particular infinite sequence of coin flips is exactly zero. (Formally the relevant probability measure is normalized Haar measure (en.wikipedia.org/wiki/Haar_measure) on the compact group $(\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}}$.) –  Qiaochu Yuan Jun 10 '12 at 19:49
    
@QiaochuYuan: You have certainly produced a completely well-defined concept, but I would be inclined to disagree that it's a concept that answers the question "what happens when I flip infinitely many coins?", because the only real answer to that question is "you don't". I suppose you are providing the definition which the question asked for of "every time", but I don't know that it's fair to say that yours is manifestly the "correct" definition, or even that the notion of a "correct" definition makes any sense, since there is no experiment for it to be correct to. –  Ben Millwood Jun 24 '12 at 17:31
    
@ben: do we agree that it makes sense to flip $n$ coins for any positive integer $n$? The construction I describe can be regarded as a convenient way to encapsulate these scenarios, each of which make sense, into a single mathematical object. Ultimately it is a matter of language whether one deals with this single infinite object or with an infinite family of finite objects. Terence Tao's remarks on equivalences between finitary and infinitary mathematics are relevant here: terrytao.wordpress.com/2010/03/19/… –  Qiaochu Yuan Jun 24 '12 at 21:05
    
@QiaochuYuan: absolutely. I don't doubt that your construction is a good one, and it works great for the finite case, and there is an interpretation of the infinite case such that it works there too. It's probably true that that's the only good interpretation of the infinite case. But I still think it's important to recognise that the interpretation is going on, and you're not really talking about an endless line of coins being tossed into the air. (That said, that article is lovely!) –  Ben Millwood Jun 24 '12 at 21:46
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If you are interested in extending the number system (the real numbers, complex numbers), then you may want to think about the surreal numbers. In the real numbers we have the following statement: "The limit of $1/x$ as $x$ approaches infinity is zero". This is not the same as $1/\infty=0$, for the reasons outlined in the above answers. However, in the surreal numbers, we have many more numbers: the real numbers, various infinite cardinalities, and infinitesimal and even square roots of these. In the surreal numbers the question needs to be revised: is their an infinite number whose reciprocal is zero. The answer to this is no. The reciprocal of an infinite number is an infinitesimal.

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However, just as for every other ordered field, there is a corresponding extended surreal number $+\infty$, and $1/(+\infty)=0$ in that number system. –  Hurkyl Jun 10 '12 at 14:19
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While others have illustrated ways in which you can reason about infinite coin flips, I think I can show more directly why your reasoning is incorrect:

So as we approach an infinite amount of coin flips, the probability gets smaller and smaller, but it shouldn't ever reach zero, so the probability shouldn't be zero, it should be 1/+∞.

But just as the probability never "reaches" zero, so the denominator never "reaches" $+\infty$ (even if you come up with a firm idea of what that symbol means, and how you could reach it). All this tells you is that looking exclusively at what happens for any finite number of coin flips doesn't tell you anything for sure about what happens when you conduct an infinite number of coin flips. The infinite is altogether more exotic and strange than the finite.

Here's an illustrative example: suppose I start with the question "If I flip $N$ coins, what is the probability I get $N$ heads?" and decide I'm interested in a version of this question with $N=\infty$. Either of the following sound like sensible ways forward:

  1. If I flip infinitely many coins, what is the probability that infinitely many of them are heads?
  2. If I flip infinitely many coins, what is the probability that all of them are heads?

They're both essentially the result of thinking about $\infty$ instead of $N$. But a moment's thought shows they are completely different questions! In particular, the answer (in the usual probability model, as described by Qiaochu Yuan) is $1$ for the first question and $0$ for the second – they couldn't possibly be more different! (Notice that it is possible, say in the sequence $HTHTHT\dots$ to get $\infty$ for both the number of heads and the number of tails, while it's certainly not possible in the previous question to get both $N$ heads and $N$ tails.)

So you can't just write symbols like $+\infty$ in the place of ordinary finite numbers and expect to get a sensible result. In mathematics, infinity is used in altogether more delicate ways: although there are ways in which you can make it a number, or at least behave in number-like ways, it is certainly not the same sort of number as $2$ or $7$ or $\pi$ or $i$. So don't ask what $1/\infty$ is before first making sure that you know what $\infty$ is, and why you care about the result.

Furthermore, if you think that the probability of an infinite number of coin tosses all coming up heads is not zero, well, you are welcome to go away and toss an infinite number of coins, and when you're done come back and tell us what you got. The fact that you can't do this ought to make you think about what you really mean when you ask a question like "what happens when I toss infinitely many coins" and on what authority can you prononunce one or another answer to that question "right" or "wrong". The mathematical answers to this question, I would argue, are neither right nor wrong but merely convenient: they let you think about things in a simple and uniform way that is always consistent and is correct for every testable situation (and not incorrect for any other situation, since correctness makes no real sense there!)

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