Yeah, in other words, it's giving a blind man a scrambled cube, having him twirl it around for a bit, and then saying, "I'm done." Sorry, getting "lucky" on a Rubik's Cube wasn't a very good way to phrase it.
Tyson Mao MSC #631 California Institute of Technology On Dec 30, 2005, at 12:26 PM, Stefan Pochmann wrote: > --- In [email protected], GameOfDeath2 > <[EMAIL PROTECTED]> wrote: >> >> P(Getting lucky on a Rubik's cube)= >> P(Getting lucky|one is on a Rubik's cube)= >> [P(Getting lucky)-P(Getting lucky|one is not on a Rubik's cube) > *P(One >> is not being on a Rubik's cube)]/P(One is on a Rubik's cube) >> >> at least assuming P(One is on a Rubik's cube)>0. > > I don't understand this, but what Tyson meant is simply: > > 1 / numberOf3x3x3CubeStates > > Cheers! > Stefan > > > > > > > > > Yahoo! Groups Links > > > > > > ------------------------ Yahoo! Groups Sponsor --------------------~--> Get fast access to your favorite Yahoo! Groups. Make Yahoo! your home page http://us.click.yahoo.com/dpRU5A/wUILAA/yQLSAA/MXMplB/TM --------------------------------------------------------------------~-> Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/speedsolvingrubikscube/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
