Is $\frac{1}{\infty}$ equal zero?

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.

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.

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).

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 $n \rightarrow \infty$.