Infinite coins without countable additivity

Question

Suppose I flip a coin infinitely many times. What is the probability that infinitely many coins tosses will come up heads? With countable additivity, the probability is clearly 1. Can anything be said without countable additivity? That is, can the probability be anywhere between 0 and 1 when one weakens the countable additivity axiom to a finite additivity axiom?

Answer 1

We want to talk about something like a probability space, but without the countable additivity. Ordinarly we define it as a triple $(\Omega,\mathcal{F},P)$ where $\Omega$ is an arbitrary set, $\mathcal{F}$ is a $\sigma$-field of sets on $\Omega$ and $P$ is a measure on $(\Omega,\mathcal{F})$ with codomain $[0,1]$. In this example, $\Omega = \{H,T\}^\infty$ is a set of infinite coin tosses.

Well, if we don't allow countable additivity for $P$, then $P$ is technically no longer a measure, thus no longer a probability. No use of discussing it under this name and we could end here. But let's try nevertheless. For this to work, we must consider $\mathcal{F}$ as just a field of sets on $\Omega$, and not a $\sigma$-field. In other words, only finite sums of events are considered an event.

This is a problem, because the standard way to say "infinitely many coin tosses will come up heads" is as follows: suppose that $H_n$ is an event of getting $H$ in the $n^{\text{th}}$ toss and $\mathcal{F}$ be a field of sets generated by $\{H_n\}$ Then we write $$\mathcal{H}=\bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty H_n$$ It represents a following sentence: for all $n$ (the intersection), there is a $H$ after the $n^{\text{th}}$ (the sum starting from $n$) toss. As one can see, in our likeprobability space we cannot prove, that the above subset is an event, given only $H_n$. In other words, we cannot assign a probability to above event using only probabilites of finite tosses. In particular, if we only know that tossing heads once has probability $\frac12$ that is not sufficient. In order for this event to have some probability at all, we need to made it up, like define $P(\mathcal{H})=t$.

But it doesn't solve the problem! If we add $\mathcal{H}$ to the mix (i mean $\mathcal{F}$), then we also must add all sorts of sums and intersection, so that we get a new $\mathcal{F'}$ - a field of sets which contain $\mathcal{H}$. And $\mathcal{H}$ and $H_n$ are not disjont for any $n$ (the sequence $(H,H,H,\dots)$ belong to both of them) so $$P(\mathcal{H}\cup H_n) = P(\mathcal{H}) + P(H_n) + P(\mathcal{H}\cap H_n)\neq P(\mathcal{H}) + P(H_n)$$ $P(\mathcal{H}\cap H_n)$ is also undefinable at this point. However, if we define $$ P(H_i) = \frac12 \quad P(\mathcal{H}) = t \in [0,1] \quad P(\mathcal{H}\cap H_i) = \frac12 t $$ then this generates a likeprobability on $\Omega$ that answers your question.

Note: I didn't rigously prove it. I would need to prove that

• every set in the field of sets $\mathcal{F'}$ is of the form $\mathcal{H} \cap H, \mathcal{H} \cup H,\mathcal{H}' \cap H, \mathcal{H}' \cup H$ where $H \in \mathcal{F}$
• you can only define the likeprobability on intersections to uniquely determine it