How do you exclude a page from showing a widget?

Better way to exclude a page and sub-pages of certain page is to use Widget Content,
which you can list certain page and sub-pages of certain page, where you want to exclude widget.
If you set this way, certain widget doesn’t exist is blog page and sub-pages of blog pages:
Target by URL:
https:…/blog/
https:…/blog/*

Alternative you can create your own function as use it as a condition. Easiest this can be done by using Code Snippet plugin.

function showInBlogPage(){
$blog=stristr('https://' . $_SERVER['HTTP_HOST']. $_SERVER['REQUEST_URI'],'/your-site-name/blog'); 
return $blog;
}

use this as condition:
!showInBlogPage()

So If I don’t want the widget to show up on the page… https://www.domain.com/blog………..or any of the blog posts…..e.g. https://www.domain.com/blog/bestbluewidget

what would the conditional tag look like?

is_page(‘https:…./blog/*’)

No, is_page doesn’t support direct web-address.

In my proposal !showInBlogPage() is the condition.

My proposal !showInBlogPage() works, if blog page is not blog article page but other page. Blog articles has their own logic, where main page – sub-page system doesn’t exist.