Parity of $\lfloor 1/(x y) \rfloor$ not equally distributed A curious puzzle for which I would appreciate an explanation.
For $x$ and $y$ both uniformly and independently distributed in $[0,1]$, 
the value of $\lfloor 1/(x y) \rfloor$ has a bias toward odd numbers.
Here are $10$ random trials:
$$51, 34, 1, 239, 9, 4, 2, 1, 1, 1 $$
with $7$ odd numbers.
Here are $10^6$ trials, placed into even and odd bins:

 
 
 


About 53% of the reciprocals are odd.
If I use the ceiling function instead of the floor, the bias reverses, with
approximately 47% odd. And finally, if 
I round to the nearest integer instead,
then about 48% are odd.

None of these biases appear to be statistical or numerical artifacts
(in particular, it seems that the 47% and 48% are numerically distinguishable),
although I encourage you to check me on this.
Update.
To supplement Noam Elkies' answer,
a plot of $x y = 1/n$ for $n=2,\ldots,100$:

 
 
 
 
 

 A: You're dividing the square $S = \{ (x,y) \colon 0 < x < 1, 0 < y < 1\}$
into two regions according to the parity of $\lfloor 1/(xy) \rfloor$,
separated by the segments of the hyperbolas $xy = 1/n$ ($n=2,3,4,\ldots$)
contained in $S$.  There's no reason to expect that the two regions
have the same area.  If I did this right, the area between the $n$-th
hyperbola and the top right corner of the square is
$$
A(n) := 1 - \frac{1 + \log n}{n}
$$
so the discrepancy between odd and even values of $\lfloor 1/(xy) \rfloor$ is
$$
(A(2)-A(1)) - (A(3)-A(2)) + (A(4)-A(3)) - + \cdots
$$
which is numerically $0.066556553635\ldots$ according to the gp calculation
A(n) = 1 - (log(n)+1)/n
sumalt(n=1, (-1)^n*(A(n)-A(n+1)))

So we expect about 53.33% odd and 46.67% even values,
which seems consistent with your experiment.
P.S. Using a formula I found in
MO Question 140547,
I gather that this number $0.066556553635\ldots$ has the closed form
$$
(\log 2)^2 + \bigl(2 (1 - \gamma) \log 2\bigr) - 1,
$$
where $\gamma$ is Euler's constant $0.5772156649\ldots$.
P.P.S. I see that I didn't address the end of the original question:
"If I use the ceiling function instead of the floor, the bias reverses, [...]
if I round to the nearest integer instead, then about 48% are odd."
The first part is clear because changing
$\lfloor 1/(xy) \rfloor$ to $\lceil 1/(xy) \rceil$
switches even and odd values (except in the negligible case that
$1/(xy)$ is an exact integer).  For the nearest-integer function,
the discrepancy between odd and even values is
$$\bigl(A(3/2)-A(1)\bigr) 
- \bigl(A(5/2)-A(3/2)\bigr) 
+ \bigl(A(7/2)-A(5/2)\bigr) 
- + \cdots
$$
which evaluates numerically to $-0.03500998166\ldots$
(using sumalt in gp as before), which again
is consistent with observation (48.25% odd, 51.75% even).
There's still a "closed form" for this discrepancy, but
more complicated:
$$
-3 + 4 \log(2)
 + \pi \bigl(1 + \log(\pi/2) - \gamma - 4 \log\Gamma(3/4) \bigr).
$$
This requires evaluation of 
$\log(1)/1 - \log(3)/3 + \log(5)/5 - \log(7)/7 + - \cdots$,
which can be achieved by differentiating the functional equation for 
the Dirichlet L-function
$L(s,\chi_4) = 1 - 3^{-s} + 5^{-s} - 7^{-s} + - \cdots$
and evaluating at $s=1$.
A: I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers.  Here are some more details.  Let $Z$ be the random variable $xy$.  Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if 
$$
Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots
$$
Note that this set has measure more than $1/2$.  However, the distribution of $Z$ is of course not uniform on $[0,1]$ (it is actually skewed towards $0$ instead of $1$).  I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectly split between even and odd for say $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z \geq 1/5$ (where one easily sees that odd wins out).    
