Tenek

Ace of Hearts = 1/52
not(Ace of Hearts) = 51/52

There are only two possibilities, should you wish to define them as that. not(Ace of Hearts) is the sum of {(Two of Hearts)...(King of [pick whatever suit you like last])}. But they are not equally weighted. There are 51 ways to get not(Ace of Hearts) but only one way to get (Ace of Hearts) and all are equally probable, given a random deck.

The other option is that you are right, and that there is a 50% chance that in any given hour, all the molecules of air in this room are in the top cm, and a 50% chance that they are not. Since I haven't suffocated yet, there's probably something wrong with the hypothesis. :P

Jesse

OK, here's my answer to the two-envelopes problem:

Like I said before, the key is to think of the ideal probability distribution for the envelope-stuffer to use. Suppose he picks the amount to be stuffed in one envelope by picking randomly from some probability distribution defined on positive rational numbers, and then flips a fair coin to decide whether to put either half or double that amount in the other envelope. What would his initial probability distribution have to look like so that, no matter what amount I see when I open one envelope, I'll still think there's a 50% chance the other envelope will have half and a 50% chance it'll be double? The only possible answer is that he used a flat probability distribution defined from 0 to infinity, in other words, one where every rational number between 0 and infinity was equally likely. If he used some non-flat distribution, I would sometimes be able to infer whether the other envelope was more likely to contain double or half of what I found in the one I opened--for example, if most of the probability was concentrated on 0 - 100 and much less on larger numbers, then if I found 160 in the first I'd know it was more likely the other contained 80 than 320.

The key to this problem, then, is that it is actually impossible to set things up so that you will have an equal chance of picking any number from 0 to infinity. The problem of finding a flat distribution on positive rationals is equivalent to the problem of finding a flat distribution on positive integers, since it's possible to come up with a one-to-one mapping between rational numbers and integers (see here). I'll discuss the problem of a flat distribution on integers, since it's a little easier to think about. If you have something like a standard computer with a random coinflipper device built in, you can write a program to simulate almost any other probability distribution--say, one with a 1/10 probability of coming up with any number between one and 10, or one that has a 1/2 probability of picking 1, a 1/4 probability of picking 2, a 1/8 probability of picking 3, and so on for every positive integer. But there won't be any program that will give you an equal chance of picking any positive integer from 0 to infinity.

Moreover, this isn't just a limitation on what is possible if you're using "computable" probability distributions, or a limitation on what is possible in a world with our laws of physics. There is actually something inherently contradictory about a flat distribution on an infinite set of possibilities. Suppose I use this distribution to pick one number, then write it down and put it in an envelope, then use the distribution to pick another number, then write it down and put it in an envelope. You open one, and find a particular number. Now there will be a 100% chance that the number in the second envelope will be larger, since there are only a finite number of integers below the one you picked, and an infinite number above it, and each integer is equally likely. But it doesn't make sense to say that you can have two preset numbers, and no matter which you look at first, there will be a 100% chance the other is larger--this seems to show that such a flat distribution is actually an impossibility. Even God, if he obeyed the laws of logic, should not be able to randomly pick a positive integer from 0 to infinity with each integer equally likely.

This was the conclusion I came to when I first thought about this problem, but I later found some papers on the "two-envelope paradox", which seem to suggest that my conclusion is partly right but that there are some additional complications. Look at these two papers by David Chalmers, for example:

http://www.u.arizona.edu/~chalmers/misc-papers.html

The complication is that, although a flat distribution is indeed not allowed, and a flat distribution is the only one where the chances of finding double vs. half in the second envelope are precisely equal regardless of what you find in the first, there are other distributions which are technically "allowed" (although my argument about the real impossibility of such distributions might apply here too) and for which it is still true that the average expected value of taking the second envelope is higher than that of taking the first, again regardless of the amount in the first. Chalmers finds, though, that for these distributions the average expected value of either envelope will be infinite, so that this is just another example of the strange behavior you get when dealing with infinities.

SULPHUR

Why didn't you tell us all the conditions before starting

Jesse

What conditions are you referring to?

SULPHUR

I thought it was gettig two amounts of money ,one double the other and putting them into two evelopes which took no further part in the wager. Where did you describe this inthe original problem.

Jesse

I don't understand your question. It was part of the original problem that you were given two envelopes, and one contained some amount of money (you weren't told how much in advance) while the other contained double that. This was how I described it:

What do you think needed to be added to the original formulation of the problem?

SULPHUR

"flipper built in " What has this to do with two
identical envelopes. And why bring it in to the solution

SULPHUR

How are we going? Am I making any sense?

Jesse

Part of solving the problem is realizing that you have to think about different probability distributions that could be used to fill the envelopes, and how that would effect the problem. If think in these terms, you'll realize that there's only one way to guarantee that if you find X in the first envelope, the average expected return for taking the other envelope will always be 1.25X regardless of what X was: namely, to use a probability distribution where every X from 0 to infinity is equally likely. The part about the coin flipper was just to suggest that once the person stuffing the envelopes picked the X to go in the first, he could make a 50% chance the other would contain 0.5X and a 50% chance it'd contain 2X (but this isn't really necessary--he could also always put 2X in the second envelope, for example, since you don't know in advance whether you're getting the envelope he stuffed first or the one he stuffed second).

Again, realizing that you have to think about the scheme the envelope-stuffer used to decide how much money went in each envelope, and realizing that most schemes would not result in the average expected return being unchanged once you opened the first envelope and found out how much was in it, is the key to figuring out how to avoid the "paradox" of the initial statement of the problem. This is something you're supposed to figure out yourself, which is why it's not mentioned in the conditions of the problem as given.