AnotherKevin

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The ML class has definitely been dumbed down. The on campus class' notes and assignments are publicly viewable here.

Gödel and Bayes: quick question

Z is defined correctly. When X >= 0 the formula becomes N(X) AND TRUE when X < 0 the formula becomes TRUE AND N(0-X).

Otherwise I was confused. I was trying to define N implicitly which I should have recognized as invalid. Explaining what I was trying to say at the end would be pointless given that I didn't say it and it's also wrong =P. Mea culpa

Gödel and Bayes: quick question

Either I am confused or this discussion is confused.

N(X) iff (X=0) || ((X > 0) && N(X-1)) iff X is natural or 0

Z(X) iff ( (X >= 0) -> N(X) ) && ( (X < 0) -> N(0 - X) ) iff X is an integer

equivalently

%20\iff%20(X=0)%20\vee%20((X%20%3E%200)%20\wedge%20N(X-1))%20\iff%20) X is a natural number

%20\iff%20(%20(X%20%3E=%200)%20\implies%20N(X)%20)%20\wedge%20(%20(X%20%3C%200)%20\implies%20N(0%20-%20X)%20)%20\iff%20) X is an integer

I'm also under the impression that the algebraic numbers are countable, dense in R, and that

(\exists%20x((x%20\in%20\mathbb%20R%20)%20\vee%20P(x))%20\iff%20\exists%20y%20((\text{y%20is%20algebraic})%20\vee%20P(y))%0A)

Edit: note to all, mixing latex and plain text on a line looks messy. Further edited for formatting due to lack of preview.

An Anchoring Experiment

I suppose the correct value is probably 10 million

Navigating disagreement: How to keep your eye on the evidence

The thermometer answer is wrong, you're ignoring that you're on a game show On a game show the producers try to organize things such that few people (or only one person) wins a challenge. As such I would expect all but one thermometer to be in error. Furthermore by watching old episodes of the show I could tell if only one thermometer will be right or if several contestants also succeed at each challenge and therefore either pick the small clump or the lone outlier.

Attention Lurkers: Please say hi

Since you asked, hi.

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 .99. If we're both blindfolded and copied and both show up in a 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?