What percent of primitive Pythagorean triples have an even number as their smallest leg?

This turns out to be a reasonably complicated question. To answer a question of the form "what proportion of an infinite set", one first has to decide on an ordering of that infinite set.

The most convenient ordering on Pythagorean triples $(a,b,c)$ comes from the classical parametrization $$ a = k(m^2-n^2), \quad b=k(2mn), \quad c=k(m^2+n^2), $$ where $m>n>0$ are relatively prime integers, not both odd, and $k$ is a positive integer. One can then count approximately how many Pythagorean triples there are with $1\le k,m,n\le x$, and how many of them have either $k$ even or $b$ as the smaller side. Those for which $b$ is the smaller side—that is, for which $2mn < m^2-n^2$, or $(\frac mn)^2 - 2\frac mn-1 > 0$—correspond to numbers $m,n$ with $m>(1+\sqrt2)n$. Out of all pairs with $m>n>0$, this corresponds to a proportion of $\frac1{1+\sqrt2} = \sqrt2-1$. Of course the even $k$ correspond to a proportion of $\frac12$. So the triples $(k,m,n)$ yielding an odd shorter side comprise a proportion $\big(1-(\sqrt2-1)\big)(1-\frac12) = 1-\frac1{\sqrt2}$, meaning that those yielding an even shorter side comprise a proportion $\frac1{\sqrt2}$.

There are some assumptions being swept under the rug—for example, that $k$ being even and $2mn$ being less than $m^2-n^2$ are asymptotically independent; and also that these proportions don't change when we restrict to relatively prime pairs $(m,n)$ that are not both odd. But I believe these assumptions can be verified with a lengthier argument.

So in conclusion: under this ordering, the percentage of Pythagorean triples with the shorter leg even seems to be $\frac1{\sqrt2} \approx 70.71\%$. (And if we restrict to primitive Pythagorean triples—those for which the three sides are relatively prime—then the $k$ variable disappears, and the percentage then becomes $\sqrt2-1 \approx 41.42\%$.)

The most natural ordering probably comes not from saying that $k,m,n\le x$, but rather that all three sides of the triangle are less than $y$, so that $k(m^2+n^2)\le y$. In this case, instead of the proportion of the triangle with vertices $(0,0)$, $(x,0)$, and $(x,x)$ that lies under the line $m=(\sqrt2+1)n$, I believe we should take the proportion of the circular wedge $\{m^2+n^2\le y,\, m>n\}$ that lies under that line—and that proportion turns out to be exactly $\frac12$! So under this ordering, the percentage of Pythagorean triples with the shorter leg even seems to be $\frac34$, and the percentage of primitive Pythagorean triples with the shorter leg even seems to be $\frac12$.