TITLE: Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$
QUESTION [5 upvotes]: Given for example $\omega_1$ coin tosses (i.e. a mapping from the elements of $\omega_1$ to $\{H,T\}$ with independent probabilities half), what is the probability that there is an infinite subsequence subinterval [corrected following comments] consisting only of heads?
Is this question even well defined? Does it depend on the set theory axiomatisation?
For a countable sequence of tosses (i.e. $<\omega_1$) the answer is presumably zero, as an infinite subinterval of heads is as likely as any of the uncountably many other possible infinite subintervals.

REPLY [4 votes]: Taking "subsequence" to mean "subinterval", the set you are describing is not measurable, and in fact has outer measure $1$ and inner measure $0$.  Indeed, let $S\subset\{0,1\}^{\omega_1}$ be the set of sequences which are constant with value $1$ on some infinite interval in $\omega_1$.  Suppose $B\subset S$ is a Borel set.  Then there is some $\alpha<\omega_1$ and some $A\subseteq \{0,1\}^\alpha$ such that $B=A\times\{0,1\}^{[\alpha,\omega_1)}$.  It is then clear that every element of $A$ must be constant with value $1$ on some infinite interval in $\alpha$. There are only countably many such infinite intervals, and the set of sequences that are constant on each one has measure zero.  So $A$ must have measure zero, and hence $B$ has measure zero.  This shows $\omega_1$ has inner measure zero.
Now suppose $B\supset S$ is Borel, and let $B=A\times\{0,1\}^{[\alpha,\omega_1)}$ as before.  An element of $S$ can have any restriction to $\alpha$ (since its infinite interval of $1$s could come after $\alpha$), so $A$ must be all of $\{0,1\}^\alpha$.  Thus $B=\{0,1\}^{\omega_1}$ and has measure 1.  Hence $S$ has outer measure $1$.  
Since the inner measure and outer measure of $S$ do not agree, $S$ is not measurable.  That is, the probability of the event you are describing happening is undefined, at least for the standard definition of "probability".