# All of AnotherKevin's Comments + Replies

The hundred-room problem

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 ... (read more)

All free Stanford classes, continually updated

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

1Sniffnoy10yOh, you said "and"; my apologies, I implicitly read an "or" there!
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

$N\(X\$%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

$Z\(X\$%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
$\\forall P\(\\text\{P is first a first order predicate\}\$(\exists%... (read more)

1Sniffnoy10yIf you're attempting to define N as a first order predicate, that doesn't work; you've defined it in terms of itself. You can't directly define predicates recursively; predicates must be finite. If you want to do get a "recursive" predicate you have to do quite a bit more work than that, and in particular you need tools not available in the first order theory of the reals (with addition and multiplication, as usual). Your definition of Z has additional minor problems; you mean and, not implies. (X>=0 => N(X)) is automatically satisfied for any X<0. Your last statement is correct (if a bit less general than it could be :) ), though your notation is a bit strange. (Again, assuming + and * as usual.) Might I ask what the relevance of all this is?
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.

0DanielLC11yIt's a metaphor, like the Monty Haul problem. The fact that that's not how game shows really work doesn't matter.
7jimrandomh11yThis is a very good point. Since you might be being messed with, you should run every sanity check you can think of. In increasing order of difficulty and also increasing order of value: get the room temperature from all the thermometers; take your own temperature; ask for a drink of ice water and take its temperature. You should also consider the possibility that all of the other contestants are actors with fake thermometers.