Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ 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.
 A: 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".
A: In reference to the countable sequence of tosses, the probability of any individual subsequence might be zero, but that is not the same as the probability of there existing an infinite subsequence of heads.  The only way to not have an infinite subsequence of heads is if, after some finite point, you only get tails from then on.  Now, the probability of getting only tails from then on would be 0, and so the probability of its complement, equivalent to the probability of there being a subsequence of heads, would be $1-0=1$.
Considering that same logic, I would imagine the answer would be the same for $\omega_1$ tosses...although I can't quite wrap my head around uncountably many discrete events like coin tosses.
