Will a given pattern ever show up in an infinite random sequence of 0s and 1s? Here the pattern is a finite or infinite sequence of 0s and 1s, not necessarily consecutive, for example, $\lbrace 1, *, 1, *, 1 \rbrace$ and $\lbrace 0, *,  0, *, 0, *, \ldots \rbrace$ ($ * $, hole position not cared), which is to be moved along a given infinite binary sequence for a match. 
If the pattern is finite, the probability of that pattern ever showing up in a randomly selected infinite binary sequence is $1$ (random in the sense of flipping a fair coin). 
If the pattern is infinite but has only a finite number of holes, then the set of binary sequences containing that pattern is dense but countable in $[0, 1)$ (viewing a binary sequence as a binary fraction). Therefore, the probability is $0$. 
However, if the pattern contains an infinite number of holes, then the set of binary sequences that contain that pattern is uncountable since the holes are essentially an infinite sequence of 0s and 1s without constraint. 
What is the probability then, that is, for a pattern with infinite many holes?
 A: There are a couple of ways to interpret the non-consecutive issue in your question. 
One way is that you are fixing a template with infinitely many positions, and in some of those positions, a definite digit value appears, and in others, there is a hole. Now, a given binary sequence conforms with the template if there is a way to place it over the sequence such that at the determined places, the digit values agree. Thus, the digits of the template are non-consecutive, but fixed distance apart. In this case, the measure of the collection of sequences conforming with the template will be zero.  This is because for any starting position, the template divides the measure in half for each additional digit that is specified at the corresponding position. So if you do this infinitely many times, the measure must be zero. Meanwhile, the set of conforming sequences will be uncountable, if there are infinitely many holes in your template. 
Another way to interpret your question is that you allow the non-consecutive bits to expand to any finite size.  Thus, a template 01101110110etc. is followed by any sequence that can be obtained by inserting any finite number of digits between any two of the digits in the template, or at the front. In this case, the measure will be full measure 1, because any sequence containing infinitely many 0s and infinitely many 1s will conform with any template, and this is a measure one set, because the remaining sequences are eventually constant, and there are only countably many of those.  
