Say you have some function $f(x)=g_1(x)\cdot\sin(g_2(x))$ and suppose that $g_1(x)$ is non-zero for all $x\in\mathbb{R}$, then $f(x)=0$ whenever $g_2(x)=r\pi$ for some $r\in\mathbb{Z}$. Let $\alpha$ and $\beta$ be such that $\beta>\alpha$, $f(\alpha)=f(\beta)=0$, and there is no $x\in(\alpha,\beta)$ such that $f(x)=0$. I'm interested in the probability that a uniformly randomly chosen $x_0\in [\alpha,\beta]$ is such that $$|f(x_0)|\geq\left|\frac{1}{\beta-\alpha}\int_{\alpha}^{\beta}f(x)\,dx\right|.$$ For example, take $f(x)=\sin x$, and let $\alpha=0$, so $\beta=\pi$. $$\frac{1}{\pi}\int_0^{\pi}\sin x\,dx=\frac{2}{\pi},$$ and a given $x_0\in [0,\pi]$ is such that $\sin x_0\geq\frac{2}{\pi}$ if and only if $\arcsin\frac{2}{\pi}\leq x_0\leq\pi-\arcsin\frac{2}{\pi}$, so the probability that a uniformly randomly chosen $x_0\in [0,\pi]$ is such that $\sin x_0\geq \frac{2}{\pi}$ is $$1-\frac{2}{\pi}\arcsin\frac{2}{\pi}.$$ Notice that, if $g_1$ is constant and $g_2$ is a constant multiple of $x$, this probability is conserved. However, I did a numerical calculation using $$f(x)=\frac{\sin(x+x^{-1})}{\sqrt{x}}$$ on two separate intervals between consecutive zeroes $\alpha$ and $\beta$ and found that the probability that an $x_0$ chosen uniformly at random in each interval is such that $$\left|\frac{\sin(x_0+x_0^{-1})}{\sqrt{x_0}}\right|\geq\left|\frac{1}{\beta-\alpha}\int_{\alpha}^{\beta}\frac{\sin(x+x^{-1})}{\sqrt{x}}\,dx\right|$$ is about $0.561$, and $$1-\frac{2}{\pi}\arcsin\frac{2}{\pi}\approx 0.561,$$ so it appears that the probability is conserved here as well. Here is why I think this has happened:
$x+x^{-1}=x(1+x^{-2})$. For all $a\in\mathbb{R}$, taking $g_2(x)=x(1+a^{-2})$ and $g_1(x)=a^{-\frac{1}{2}}$ would conserve the probability, since then $g_2$ is a constant multiple of $x$ and $g_1$ is constant. Since this is true of every $a\in[\alpha,\beta]$, for each $a\in[\alpha,\beta]$ there is a neighbourhood of $a$ where the $x^{-\frac{1}{2}}\sin(x+x^{-1})$ looks arbitrarily like $a^{-\frac{1}{2}}\sin(x(1+a^{-2}))$, so the probability is conserved for the whole function. It sort of makes intuitive sense to me, but I have no idea how to show that it is true.
Is the probability actually conserved for the example function I gave, and for what other functions is it conserved?
You wrote $\left|\frac{\sin(x_0+x_0^{-1})}{\sqrt{x}}\right|$ but I assume you mean $\left|\frac{\sin(x_0+x_0^{-1})}{\sqrt{\color{red}{x_0}}}\right|$. Aside from this, thanks for fixing the typos.
I don't think what you said can be true for general $g_1, g_2$. It is true that for very small neighborhoods $x^{-\frac{1}{2}}\sin(x+x^{-1}) \approx a^{-\frac{1}{2}}\sin(x(1+a^{-2})),$ but you are interested in the global average over the entire range $[\alpha,\beta]$, and each neighborhood approximates with a different $a,$ so the global average can be anything.
In general if you allow $g_1, g_2$ to be arbitrary then $f$ can also be (almost) any function. E.g. let $g_2(x) = x$ and $g_1(x) = \sin(x)$, then $f(x) = \sin^2(x)$ and the prob that a random point exceeds the average is $1/2$. (Indeed if you let $g_2(x) = x$ and $g_1(x) = f(x) / \sin x$, that can get you any $f(x)$ you want, except you have to specify that in a small neighborhood around each zero $g_1(x) = 0$ to avoid ${0 \over 0}$.)
I am a bit surprised at how well your conjecture works for $f(x)=\frac{\sin(x+x^{-1})}{\sqrt{x}}$ but I suspect (no proof) that it is because you tried a small range and things happened to balance out. Or maybe if you go beyond $3$ decimal places the numbers no longer match. :)