jimmy comments on Counterfactual Mugging - Less Wrong

52 Post author: Vladimir_Nesov 19 March 2009 06:08AM

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Comment author: Vladimir_Nesov 19 March 2009 04:34:07PM *  4 points [-]

The coin toss may be known to Omega and predicted in advance, it only needs to initially have 50/50 odds to you for the expected gain calculation to hold. When Omega tells you about the coin, it communicates to you its knowledge about the toss, about an independent variable of initial 50/50 odds. For example, Omega may tell you that it hasn't tossed the coin yet, it'll do so only a thousand years from now, but it predicted that the coin will come up tails, so it asks you for your $100.

Comment author: jimmy 19 March 2009 07:48:15PM *  2 points [-]

That's just like playing "Eeny, meeny, miny, moe" to determine who's 'it'. Once you figure out if there's an even or odd number of words, you know the answer, and it isn't random to you anymore. This may be great as a kid choosing who gets a cookie (wow! I win again!), but you're no longer talking about something that can go either way.

For a random output of a known function, you still need a random input.

Comment author: fractalman 21 July 2013 04:56:18AM *  0 points [-]

The trick with eeny-meeny-miney-moe is that it's long enough for us to not consciously and quickly identify whether the saying is odd or even, gives a 0, 1, or 2 on modulo 3, etc, unless we TRY to remember what it produces, or TRY to remember if it's odd or even before pointing it out. Knowing that doing so consciously ruins its capacity, we can turn to memory decay to restore some of the pseudo-random quality. basically, by sufficiently decoupling "point at A" from "choose A" to our internal cognitive algorithms...we change the way we route visual input and spit out a "point at X".

THAT"S where the randomness of eeny-meeny-miney-moe comes in...though I've probably got only one use left of it when it comes to situations with 2 items thanks to writing this up...