AnotherKevin comments on The hundred-room problem - Less Wrong

0 Post author: APMason 21 January 2012 06:12PM

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Comment author: AnotherKevin 22 January 2012 01:15:07AM -2 points [-]

Your confusion confuses me.

If after removing the blindfold in the blue room you offer every copy of me the same pair of bets of "pay me X dollars now and I'll give you 1 dollar if the coin was tails or pay me 1-X dollars and get 1 dollar if was heads" I bet tails when X < .99 and take heads when when .99<X<=1 to make the most profit over the collective. If you don't know the result of the coin flip nor the colour of my room I then also use the same strategy when you only offer the bet to one of me; I expect to profit when X <> .99. If we're both blindfolded and copied and both show up in a <colour> room I will offer you the bet "pay me $.49 and get $1 if the coin flip resulted in (tails when in a red room, heads when in a blue room) and the 100 copies of me will have $48 more than the copies of you (which we will split if we meet up later). If I clone you, someone else flips the coin and paints the rooms, then I open one room I offer you the same bet and make an expected (.99 x $0.49 - .01 x $0.51) = $0.48.

If all 100 copies of me have our sight networked and our perception is messed up such that we can't see colours (but we can still tell at least one of us is in a blue room (say this problem is triggered by anyone on the network seeing the colour blue)); then we agree with the 50-50 odds on heads-tails. If all of my copies have bug-free networked sight I I'll take bets whenever (X <> 1 given 99 blue rooms AND X <> 0 given 99 red rooms).

Is there another meaning of probability that I'm missing? Does this clarify the informational value of learning the colour of the room you wake up in?