Mozilla would go with another partner and/or change their structure to accept donations for the browser.

If money wasn't a consideration perhaps ddg become a good partner. That combo could shift as much as 5% of google traffic away from search.

Because another partner would pay to be the default search engine of a browser with declining market share. Would you want to count on donations to fund a company the size of Mozilla?

Google pays Apple a reported $8 billion a year to be the default search engine for iOS. It pays Mozilla less than $600 million (https://www.cnet.com/news/google-firefox-search-deal-gives-m...).

Doesn’t that tell you how little Firefox traffic is worth?

How is $600mil/yr "little"? That's a metric f-ton of money by any standards except "companies that are larger than countries."

Googling around, a few sources tell me there are about 250 million Firefox users; Mozilla currently seems to say 210 million: https://data.firefox.com/dashboard/user-activity

As for iOS, this source says there were 900 million iPhone users (doesn't include iPads) as of Jan 2019: https://www.idownloadblog.com/2019/01/30/900-million-iphones... , and this one says 1.5 billion Apple devices (probably includes Macs) as of Q1 2020: https://9to5mac.com/2020/01/28/apple-hits-1-5-billion-active...

I'll arbitrarily guess 1.2 billion iOS users (the rest being Macs). In that case, there are maybe 6x as many iOS users as Firefox users. Therefore, if Google paid the same per Firefox user as they did per iOS user, they would have paid $1.3 billion for Firefox users. That's a difference of about 2x per user from what they actually paid.

iOS users have demonstrably paid for (or been gifted by people who paid for) premium electronic devices, which I would expect makes them more valuable advertising targets. I'm mildly surprised that the difference isn't larger, but I don't have a good feel for advertising expenditures.

There's also the part where iOS Safari is pretty much the only browser that people will want to use on the platform.