thomblake comments on Solving the two envelopes problem - Less Wrong

32 Post author: rstarkov 09 August 2012 01:42PM

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Comment author: thomblake 09 August 2012 03:05:08PM *  6 points [-]

Generally, people do take R and B to be fixed amounts of money - then, it's worth pointing out that the expected value of the blue envelope is not .5*.5R+.5*2R, because R there stands for 2 different amounts of money.

If the amounts of money are $5 and $10, then R in the first case stood for $10 and R in the second case stood for $5, so it should really be .5*.5($10)+.5*2($5), and the expected value is just $7.50.

Comment author: mwengler 30 August 2012 02:31:16PM *  0 points [-]

Yes, I like your simplification.

To amplify it and make the results explicit, Suppose my envelope strategy is to flip a fair coin and
heads: R=$5 B=$10
tails: R=$10 B=$5.

Then
E(R) = E(B) = $7.50
E(B)/E(R) = E(B)/E(R) =1
E(R/B) = E(B/R) = 1.25

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