ike comments on Eliezer Yudkowsky Facts - Less Wrong

124 Post author: steven0461 22 March 2009 08:17PM

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Comment author: ike 01 May 2014 03:05:17PM *  1 point [-]

Technically, that would be eight-boxing. (Or 24 if you let the prize be in any box). I'll explain:

Let's say the prize is in box A. So the eight options are:

  • {EY picks box A, host opens box B, EY switches}
  • {EY picks box A, host opens box B, EY doesn't switch}
  • {EY picks box A, host opens box C, EY switches}
  • {EY picks box A, host opens box C, EY doesn't switch}
  • {EY picks box B, host opens box C, EY switches}
  • {EY picks box B, host opens box C, EY doesn't switch}
  • {EY picks box C, host opens box B, EY switches}
  • {EY picks box C, host opens box B, EY doesn't switch}

By symmetry, there are eight options for whichever box it is in, so there are 24 possibilities if you include everything.