Sniffnoy comments on Value Stability and Aggregation - Less Wrong

8 Post author: jimrandomh 06 February 2011 06:30PM

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Comment author: Sniffnoy 08 February 2011 02:25:10AM 1 point [-]

If I and my opponent only have A(paper), B(rock) to choose from, then always A > B. Likewise B>C, C>A.

If you and your opponent only have paper and rock to choose from, this is correct. But if that is the case, then you are not considering two options within the existing game, you are considering a different game entirely. To equate your preference for paper over rock in a game of Rock-Paper, with a preference for paper over rock in a game of Rock-Paper-Scissors, is a confusion. In that case, the scenario would read, "My opponent can throw Rock, Paper, or Scissors; if we assume I don't want to go Scissors (but my opponent does not know this), what should I do?" Within the given game, there are no intransitive preferences.