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 $n=2,\ldots,100$:


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    $\begingroup$ This reminds me of the graph on page 31 here: amacad.org/publications/bulletin/spring2003/diaconis.pdf $\endgroup$ – James Weigandt Oct 19 '14 at 2:12
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    $\begingroup$ @JamieWeigandt: Indeed, I have added a similar image. $\endgroup$ – Joseph O'Rourke Oct 19 '14 at 11:36
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    $\begingroup$ Well, given any law on $\mathbb{N}$, you will see this kind of things happen: it is impossible to find a probability measure on $\mathbb{N}$ which gives weight $1/k$ to each of the classes modulo $k$, simultaneously for all $k$ (the weight of any number should be zero to meet that request). $\endgroup$ – Benoît Kloeckner Oct 19 '14 at 18:21

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 $s=1$.

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    $\begingroup$ Since $f(x) = (1+\log x)/x$ is convex decreasing on $x \geq e^{1/2}$, it's easy to see that the sequence $p_k = \mathbb P(\lfloor 1/(xy)\rfloor = k)$ is decreasing for $k \geq 2$. But $p_1 = \frac12(1-\log2) > \frac12(1+\log 2)-\frac13(1+\log3) = p_2$, so the distribution in question has a decreasing probability mass function supported on $\mathbb N$, making odds more prevalent than evens. $\endgroup$ – cardinal Oct 19 '14 at 4:32

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).


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