Randomly Unlucky

For a non-leap year, since $365 = 7 \times 52 + 1$ , so the calendar moves forward 1 day. If the 1st of January is a Sunday this year, then it would be a Monday next year.
For a leap year, since $366 = 7 \times 52 + 2$ , so the calendar moves forward 2 days (after Feb 28th). Since we simply cycle through the days, it seems very equally likely that the 13th day of the month is any of the days of the week.

Let's investigate further to see why this intuition works. Recall that every 4 years, we get a leap year. This suggests that after $28 = 4 \times 7$ years, we would have done a complete cycle and are back to the original day of the week. For example, if we record down when the first of January for each year from 2001 to 2028 (inclusive), we obtain:

However, because every 100 years is not a leap year, as such this would mess up the pattern. In particular, from 2001 to 2028, 2056 or 2084, each weekday is equally likely. However, from 2000 to 2112, since 2100 is not a leap year, the pattern no longer holds . Our intuition is thrown off because we forgot to account for the other edge cases.

Further complicating this, is that every 400 years is not a leap year! Let's consider the number of days in these 400 years. There are 97 leap years in this period, or a total of $400 \times 365 + 97 = 146097$ days. Since this is a multiple of 7, if there is any discrepancy during this period, it would persist throughout time, instead of being evened out in 2800 years. It remains to check what happens to the 13th of the month over a period of 400 years.

As it turns out, the frequency of the 13th day of the month is:

As such, Friday is the most likely day of the week!

Using a computer and Zeller's Congruence , we can directly count the number of times the 13th falls on the days of the week. Since the Gregorian calendar cycles every $400$ years, let parameter $n$ be the number of the month from Year $1$ to Year $400$ , or $4800$ months in all. Then

$m=$ Mod $\left[n+9,12\right]+3$

$y=\left\lceil\dfrac{n}{12}\right\rceil$

$k=$ Mod $[y,100]$

$j=\left\lfloor\dfrac{y}{100}\right\rfloor$

$q=13$ for 13th day off the month

$h=$ Mod $\left[ \left( q + \left\lfloor \dfrac{13(m+1)}{5} \right\rfloor + k + \left\lfloor \dfrac{k}{4} \right\rfloor + \left\lfloor \dfrac{j}{4} \right\rfloor + 5j \right) ,7 \right]$

where $h$ is the day of the week, $\left(2,3,4,5,6,0,1\right)$ for (Sun, Mon, Tues, Wed, Thurs, Fri, Sat). $\;$ When tabulated, the results are

$\left(687, 685, 685, 687, 684, 688, 684\right)$

and so the 13th falls on a Friday more than any other day of the week.

Edit: If you're curious about any of the other days of the month, particularly the 29th, 30th, and the 31st, it turns out that the distribution is always the same, except for rotation. For example, the dates 6th, 13th, 20th, 27th have the same distribution

$\left(687, 685, 685, 687, 684, 688, 684\right)$

while dates 3rd, 10th, 17th, 24th, 31st have the following distribution, the most frequent day of the week being Tuesday

$\left(687, 684, 688, 684, 687, 685, 685\right)$

Or, you could just guess Friday by the common thing - Friday the 13th.