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.

Szekely G.J. Pcaradoxes in probability theory and mathematical statistics,

it is a large listing of paradoxes with references to literature.

Another thing, this time regarding the paradox proposed in the reply above by mmm: It is assumed that, by ’symmetry’, both participants have the same chance to win, thus the paradox. The assumption is, of course, wrong. You would only have true symmetry if you both had the same amount of money in your respective wallets at the start.

The relative amounts of money each player has is critical to deciding who wins, and it is implicit in the problem statement that, lacking knowledge of how much more (or less) money i have than the other person, i can assume that i occupy a central (’symmetric’ position in the overall distribution of money contained in people’s wallets in general.

Now, that assumption is obviously at fault. If i have no money at all, i’m guaranteed never to lose. And if only few people carry more than 100 dollars at once in their wallets but i happen to have 150 right now, i’m more than likely to lose.

So, your expected gain depends on how much money you’re carrying in your wallet right now, and you are certainly aware of that amount. And even if you had no way of knowing the general distribution of money inside people’s wallets, you couldn’t claim an expected win based on ’symmetry’. You’d have to say that your expected gain is indeterminate.

Reworking the example to have amounts $x$ and $kx$ in the envelopes, with $k$ any constant other than 2, works as well.

Now, why does the analysis in the problem statement appear paradoxical? The analysis is certainly correct (refer to the calculation in the 3rd paragraph of the post); it’s just that it’s an analysis for a different problem. Consider:

Suppose i have 10 dollars. A friend offers me the following proposition: I give him my 10 dollars, and in return he will flip a coin. If it comes out heads, he’ll give me 5 dollars; tails, and he’ll give me 20 dollars (the $x/2$ and $2x$ envelopes, respectively). Now, should i accept his offer? Of course! As the calculation in the post shows, my expected gain is 5/4*10 dollars, minus my original 10 dollars, so i come out winning 2.5 dollars on average.

In conclusion, the paradox stems from applying the incorrect analysis to the problem. As we mostly try to find problems in the analysis itself (that is, we neglect to see if the analysis is really meant to solve THIS problem), but the development of the analysis is indeed correct, we get the “What the hell?” feeling.

I do not know a lot about probability, but I think this paradox can be resolved very simply. It does not matter how (by which distribution) the sums in the envelopes have been determined. All we know is that we are given the information “there are two envelopes, one contains amount A and the other contains amount 2A”. This is something which sounds quite similar to, but is very different from the information “you may open one envelope. If you find amount A in it, you ‘ll be given the possibility to open a second one which will contain amount A/2 or 2A, each with probability 1/2″. Let us call these situations (1) and (2). In situation (1) (the one we are really considering) you open an envelope. It contains amount X and we don’t know if X=A or X=2A. Both are possible with equal probability 1/2. If we change, with probability 1/2 we go from A to 2A, or from 2A to A. No gain or less is to be expected and obviously, the same holds if we do not change.

On the other hand if we are in situation (2), where the amount in the next envelope actually depends on the amount in the first one (that’s the whole point: it does not in situation (1)!!), then naturally one has to go for the second one. If initially one has found amount x, then by iterating this procedure, our obtained amount will do a random walk on x.2^Z (Z={integers}).

Q1) What happens at 12? bag is empty
Q2) If the ball taken out is selected randomly? prob 1 the bag will be empty at 12.

These are more curious(to me at least) than important but helped me keep some students awake.

http://ideas.repec.org/p/wpa/wuwpga/0310004.html

what seems to be an earlier version of it.

And I had simplified the conclusion of their work: there is no assumption that the ranking gives probability 1/2 to the two possibilities, only that the ranking is no constant (i.e. that neither X> Y nor X< Y holds all the time).

BTW, is a preprint of Samet/Samet paper available online?

In both cases, his expected gain is the half of the sum of the amounts in the two envelopes.