Making Sense of Software EULAs

Brian E. writes "An informal Clearware.org poll indicates that 67% of the 66 respondants never or rarely completely read end-user license agreements (EULAs). Clearware.org aims to make sense of software by proposing guidelines for vendors to characterize end-user license agreements. Defined characteristics include terms and conditions found in existing EULAs that impact control over the user's experience, privacy and system security. The guideline extends on the idea of Creative Commons' commons deed and RDF/XML metadata formats. This simplifies EULA terms in a consumer friendly way similar to care labels on clothing, nutrition facts on food and warnings on hazardous materials."

Comments

Re:I thought these were unenforceable

[AhtirTano]

IANAL either, but I quote one here:

So, once again, true shrink wrap EULAs have been tested in most major jurisdictions and are valid contracts, subject to certain limitations. Terms of Service contracts, like the "EULA" found in MMOGs, are simply enforceable. There is a common perception that EULAs have not been tested in court. This is incorrect. They have been.

Don Shelkey

eula wiki

[rjnagle]

lately, I've been trying to keep track of all of the EULA's I've been agreeing to. It's overwhelming.

This wiki
http://www.gripewiki.com/index.php/EULA_Library

is trying to keep a public record of eulas (along with some analysis).

Web Sites Prevent Reading EULA

[BoRegardless]

Many download sites make you click "Accept" buttons, but if you actually try to READ the damned 15-25 pages of the EULA, you find the web site "times out" and you can't then proceed with the purchase/registration process.

No wonder people don't read them. I don't do it online anymore.

Re:Sample of 67?

[lubricated]

try +/- 12%

Re:Here's my clarification

[Pofy]

>3. If you break the license, how do i get my money back

Problem is, in most EULA, there is NO way for them to break the contract since they promise nothing and have no responsaibilites.

Re:Sample of 67?

[say]

No. The standard error (SE) is sqrt((p(1-p))/n) = sqrt((0.67*0.33)/66) = 0,057879185. If we are aiming for our standard 95 % confidence, the margin of error is 1,96*SE = 0,11344. It has a margin of error of 11 %, meaning that the true number is 67 % +- 11 % with 95 % confidence. I don't know much about the methods and people behind this test, so it's hard to say what population this statistic is correct for :-)