Comments for Gowers's Weblog

Comment on When are two proofs essentially the same? by Paul

Paul — Mon, 12 May 2008 23:48:52 +0000

A good problem to think of in this context would be the classic theorem
“Whenever a rectangle is tiled by rectangles each of which has at least one integer side, then the tiled rectangle has at least one integer side.”

Wagon offers 14 different proofs in “Fourteen proofs of a result about tiling of a rectangle” and he even classifies them according to how they generalize…

http://www.jstor.org/stable/2322213?seq=1

Comment on When are two proofs essentially the same? by Konstantin Ziegler’s Weblog

Konstantin Ziegler’s Weblog — Wed, 07 May 2008 11:17:43 +0000

[...] Gowers asks When are two proofs essentially the same? For example, it is often possible to convert a standard inductive proof into a proof by [...]

Comment on Probability paradox II by Ehud Friedgut

Ehud Friedgut — Tue, 06 May 2008 12:13:27 +0000

If we formalize everything we have no problem with the fact that there are two random variables X and Y and two ways to partition our probability space such that in one way the conditional expectation
of X is always larger, and in the other way Y always wins.

What bothers us is the implications of this on our actions in “real life”
There are two operations we want to do:
1) Choose between the envelopes in a manner that maximizes our expectation.
2) Look at the content of the envelope we’re holding, in order to decide what to do.

Obviously these two operations do not “commute” since before looking at the content the expectation is infinite, so action number 1 is meaningless.
The whole “paradox” comes from our intuition that if whenever we do action 2 first we do the same thing in action 1 then the operations do commute. i.e. it is meaningful to do action 1 first.

This is just a fallacy of our limited intuition that usually
deals with random variables with finite expectation.
I like to think of this as a version of Schroedinger’s cat:
by looking at the content of the envelope we cause a collapse of the wave function.

Comment on Probability paradox II by Gil

Gil — Tue, 06 May 2008 11:03:11 +0000

Another variant on how to how to tell this paradox would be that after you get the envalopes you have to pay 1 dollar for switching the envalopes and 1 dollar to look at the content of the envalope you have.

Comment on Probability paradox II by Gil Kalai

Gil Kalai — Tue, 06 May 2008 10:58:57 +0000

There is some resemblence between this paradox and the Newcomb paradox. Also there there are reasons that the dominant strategy be rejected but for the Newcomb’s paradox this is not because of unrealistic probabilistic distribution but because of strange logical connections between your action and a prediction that took place before the actions.

As for this paradox, I do not see how the argument that the situation is untrue because the probability distribution is supposed to have infinite expectation can be rejected by the counter argument with the heaven story (either Tim’s or Terry’s version).

Comment on Probability paradox II by Ben

Ben — Mon, 28 Apr 2008 16:02:27 +0000

I find it interesting that no one is talking about the initial probabilities of n. If you open an envelope with $1000 in it, it either means n=3 and you picked the small envelope, or n=2 and you picked the big envelope. But the problem states that n=2 is twice as likely as n=3. I’m not saying this has any relevance to the decision to switch. In fact, I don’t think it does. However, the reason why it’s not relevant is important.

Comment on Probability paradox II by Jack

Jack — Fri, 25 Apr 2008 17:01:03 +0000

In fact, in the long run, it doesn’t matter if you switch envelopes or stick with your first choice. If the 2 envelopes have X and 2X in them, a person who always sticks will end up with an average of 1.5X and so will a person who always switches.

Comment on Probability paradox II by nicolas

nicolas — Wed, 23 Apr 2008 13:01:55 +0000

Anonymous, you’re hired.

Comment on A paradox in probability by nicolas

nicolas — Wed, 23 Apr 2008 12:55:42 +0000

Ok so I know the point of this was not the paradox itself but to hell with it.

The crux of the paradox lies for me in that x itself is a random variable, and we dont recognize it as such when supposedly taking expectancy to compare outcomes : after taking unconditional expectancy, which is supposed to be a number, we end up with x, which is a random variable. This is not possible, and we have been fooled.
The way it is presented, we are considering the case where $x=2a$, and say the other envelope has a 1/2 proba of having $2x=4a$. this case do not exist and we know so, hence it should not appear in our computation…
So we take an envelope, call its value $x$. the other envelope contains either $2x$ if x=a, or $x/2$ if $x=2a$
the expectancy of it value is $3/2a$. the expectancy of the other envelope is $3/2a$, there is no incentive to switch.

Comment on Open problems concerning card games by harrison

harrison — Mon, 21 Apr 2008 15:20:23 +0000

Hmm. I’ve seen a few places (Wikipedia, Guy) that mention that Conway listed this as an “anti-Hilbert problem,” but I can’t find an actual reference to Conway’s original article/statement. Anyone know one?