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>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 f
by Mysterix 9y ago
>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
- rullelito 9y agoThis 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.
- Dylan16807 9y agoWhat 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 9y agoThe 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 9y agoYou 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"?
- Pxtl 9y agoThis assumes the client can cleanly switch between the two routers. That doesn't happen.