What is the statistical likelihood of a person being sick exactly in 14 day intervals? I have a co-worker who calls out exactly every other Saturday in 14 day intervals like clockwork. It is always that exact day in the 14 period cycle. My question, as a not so statistician-savvy individual, is what are the (astronomical) odds that an individual could be sick during this interval for a period lasting over six months?
 A: The context of your question seems to imply that you want to use this probability to strengthen your believes about your co-worker cheating. If this is the case then note, with a big exclamation mark, that this is not a correct statistical practice. Sadly this type of prosecutors fallacy is made by professionals as well. 

1) We need to first get a correct view of your data(collection) (before doing any analysis). It is risky to already start spitting out numbers before we actually know what went on in the data. It might make us less critical when we obtain some satisfying number.
The anecdote of being sick every second Saturday for half a year is much too vague. The calculations from Santy.8128 should not be used. It is an extreme number that can be reduced by several orders.  


*

*One thing is that it only considered the exact 14 day interval on a Saturday. (but there are 14 different way to have a 14 day interval)

*But more importantly it assumes that your co-worker is working 7 days per week. (which I assume is not the case)

*It doesn't incorporate errors or possible imperfections in your data.
e.g. (this is just one example) do you know, for sure, she is sick, not just calling out, exactly every second Saturday? Already the probability to be sick is not evenly distributed throughout the week (you may find certain lists for this, but note that those are averages and on an individual/group level it might be even more non-homogeneous). But also the calling out may be even more non-equal distributed. The person might be sick and still go to work (and this process could be different on certain days in comparison to others). So an even pattern of calling out might not mean that the underlying reason to call out, being sick, was uneven.
..
So what we need to know for a good statistical data analysis, is 1) more precise information on how you obtained your data (more than an anecdote, also more context like the fraction of workdays, for this person, which are a Saturday), 2) more information on the distribution of sick leave among others in a similar situation (we should not assume homogeneity, e.g. longer working days or different moments of the day/night shifts might be called out more likely).

2) We need to use a Bayesian analysis when we wish to use the 'calculated probability' to say something about 'the probability that somebody is cheating' (I am not sure this is the implied underlying question in of your original question, but it seems like that and is at least noteworthy to mention for by-passers with similar questions).
So the question, when you observe some data $x$ is not only 'how likely is this data $x$ to come from a person', but instead it is better to ask 'how likely is it that, conditional on data $x$, that the person is a cheating person'. 
Some very simple expression the probability of being a cheater after observing $X$, using say $P(\text{cheating}) = x$, $P(\text{not cheating}) = y$, $P(X|\text{cheating}) = a$, $P(X|\text{not cheating}) = b$ Then:
$$P(\text{cheating}|X)  = P(\text{cheating}) \frac{P(X|\text{cheating})}{P(X)} = \frac{1}{1 + \frac{yb}{xa}}$$
which if you put in the number $b = 10^{-20}$ is still very close to one (unless you set $a$ to something equally low, or the factor x/y is low), but when you set more realistic values for $a$, $b$ include $y/x$ which I believe should be much larger than one, then this type of equation would be more 'fair' to use.
The Bayesian analysis may come across as including subjective guessed numbers for prior probabilities (although they can always be obtained from good secondary analyses if you like), but it provides a much better representation of the probability when one is actually looking for P(a|x) instead of P(x|a). It shows how we should connect multiple pieces of data, we can not just use the probability for the data alone if we wish to express the probability for something else.

In practice, these type of statistical computations are based on too simplistic models and the term 'probability' becomes very confusing (it is the probability as a term in a calculation, the probability for error in the calculation is not included).
It is much better to get additional more trustworthy data (the 14-day pattern can be at best an indication). For instance get the person to a doctor. Who knows they have something like hay-fever or another sickness that can be triggered by external factors that possibly occur in a repetitive way.
A: This would be a very simple heuristic approach to it (under many assumptions). 
Your company allows $1$ sick day every $14 \mbox{ days}$. From here let us assume that the company considers the baseline probability of a person being sick on a day as $p=1/14$. Let us also assume that the events of getting sick on a day is independent of other days (which is generally not true)
Now, in 6 months (180 days) lets us assume that the person has taken 12 sick days (2 each month, only on the second Saturdays). Thus, the chances of that would be the probability that a person gets sick on a fixed set of $12$ days out of $180$ days and not being sick for the remaining $168$ days. [Note that this is NOT a binomial random variable since the variable is not just the number of sick days but the exact set of days on which the person gets sick]
$\mbox{Probability of getting sick every alternate Saturday for 6 months}=\left(\frac{1}{14}\right)^{12} \times \left(\frac{13}{14}\right)^{168} \approx 10^{-20} = \mbox{one in a hundred trillion }\ $
