getting lucky on a rubiks cube? sounds uncomfortable, and potentially dangerous
----- Original Message ----
From: Stefan Pochmann <[EMAIL PROTECTED]>
To: [email protected]
Sent: Friday, December 30, 2005 12:26:23
Subject: [Speed cubing group] Re: (Off topic)3 doors...
--- 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
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