[Mysterix]

>Replacing your router:

>

> Vendor A: 10% broken

> Vendor B: 10% broken

> P(both A and B broken):

> 10% x 10% = 1%

>

>Replacing your router (or firmware) almost always fixes your problem.

The conclusion is false :

if router A is broken, router B still have 10% chance to be broken, the two events being independant.

P(A broken | B broken) = 10%

To get the 1% effect, advice could be :

Always buy 2 routers instead of 1

[Dylan16807]

What exactly are you saying is wrong?

If you have to replace the router, there's a 10% chance that new router is broken. But you only replace when the first router is broken, so it's 10% of 10%.

Read it as "a strategy of replacing when needed" rather than "replacing in all cases for the hell of it".

[Mysterix]

The strategy "Buy 2 routers, and if the first one fails, then use the 2nd one" is ok, and gives you the 1% result.

My (little) problem is the sentence "Replacing your router (or firmware) almost always fixes your problem.", because if the first router is broken, replacing it will only fix your problem in 10% of the cases, which is not "almost always".

[Dylan16807]

You don't actually have to buy a second router upfront, so that's not a good way to word it either.

I'm struggling to find a great way to put it. Maybe "a one-replacement backup plan gives you a 99% chance of success"? Close but not very elegant.

"Replacing your router (or firmware) fixes the problem except for 1% of all router buyers"?

[rullelito]

This is a bit like the gameshow where you select from 3 doors, and then one door is removed and you have the option to change.

[Pxtl]

This assumes the client can cleanly switch between the two routers. That doesn't happen.