In the other hand, problem 1 is poorly defined and contains bad assumptions. Considering the condition “gain A or lose A/2″ that will not only leave us with $100 (initial), $200 or $50 envelopes (look,2 possibilities already), this will imply going down to $25, $12.5, $6.25… and upb to $200, $400, $800… The possibilities wining/losing-50% are then split by the fact you can chose between 2 different envelopes apart from the one you have in your hand (your baseline on each time you select to stay or switch) that’s what 1.25 stands for.

The two envelopes contain random values. Assigning them values of $50, $100, and $200 as in the examples is done without loss of generality, and is a standard method of making a proof or explanation more readable.

Absolutely, except in this case assigning the envelopes values comes WITH loss of generality. So, you can’t do it. It would be like if I said, “Prove that all even numbers are divisible by two,” and you started your proof with, “Okay, assume the even number is 12…”

Philip says: I submit to you that there is no difference between the following two answers:

(A) There is not enough information to solve the problem.

(B) It doesn’t matter if you switch envelopes.

Zorel says: I submit to you that this is wrong. These two are mutually exclusive.

If you were God, you would know which envelope contained more money and know whether to switch. By the problem’s very statement there is “not enough information” to make the right choice, so the question is how to make the best GUESS. If there is “not enough information” to make a good guess, then the answer to the question obviously must be that it doesn’t matter what your guess is.

Thus, “two envelopes filled with a random amount of money” is meaningless unless you either set the bounds or specify the distribution you’re talking about.

It’s not meaningless, the envelopes were obviously filled somehow, but since you DON’T KNOW how, you can’t use it to try to solve the problem.

There’s no such thing as a counterproof.

Of course there is. You can nitpick the finer points of grammar if you want to waste your time, but if somebody shows me a proof that is wrong, what am I supposed to do, NOT point out the inconsistency?

What you wrote isn’t the flaw in Solution 1. It’s… speculation, and it’s incorrect at that. It still doesn’t answer the question “What’s wrong with Solution 1?”

So what’s the flaw in Solution 1? As it turns out, absolutely nothing.

Solution 1 is completely correct. There does indeed exist a rigorous logical proof of “If the amount of money in both envelopes is uniformly distributed and one contains twice the amount the other contains, then it is advantageous to switch.”

It seems you’ve committed an egregious logical error: claiming that something is wrong WITHOUT offering a counter-proof, or indeed any reason at all.

The problem with solution 1 is simple: once you switch envelopes, your envelope STILL contains either double or half the other envelope, which means that if it was advantageous to switch before, it is now advantageous to switch back, which is a contradiction. If you want to claim that solution 1 is correct, you MUST address that contradiction. Otherwise, your argument carries no weight.

Also, you mentioned the “uniform distribution” of the money in the envelopes, which is an assumption that was NEVER stated, and therefore you are using information that you don’t have, which is another logical error.

That’s why you have to be careful with proofs of contradiction. All a proof by contradiction will tell you is that one of your assumptions is incorrect. It won’t tell you which.

Proofs by contradiction are often used to prove things wrong (for obvious reasons, ha ha). If you want to prove that 1 is the largest positive integer, I can simply say the word “2″ and I have successfully contradicted your assertion. Likewise, as mentioned above, there is a very simple contradiction in solution 1. If you want to claim that solution 1 is the correct solution, you’re going to HAVE to address that contradiction and get around it somehow.

The Wikipedia article you link to on the two envelopes paradox goes into more detail, including the correct answer, which has not been discussed yet.

First, I’d like to respond to a few statements here:

Sigh says: These are different problems, there is nothing to see here.

The two envelopes contain random values. Assigning them values of $50, $100, and $200 as in the examples is done without loss of generality, and is a standard method of making a proof or explanation more readable.

Philip says: I submit to you that there is no difference between the following two answers:

(A) There is not enough information to solve the problem.

(B) It doesn’t matter if you switch envelopes.

I submit to you that this is wrong. These two are mutually exclusive. The question of which is correct isn’t clear, though, because your problem is phrased in English and English can be ambiguously interpreted. If I had to make a call, the way I would interpret the question would make ‘A’ the correct answer.

Veggie says: I did the experiment in 10000 trials (read: rows) in Excel. There did not appear to be any advantage between the strategy of switching or staying.

No, you didn’t. The problem as stated is “two envelopes filled with money”. What you probably simulated was “two envelopes filled with a random integer amount of money between 1 and 4,294,967,296″ or something like that.

And now, for our answer.

Your problem statement is: “two envelopes filled with money”, but you do not specify how much money.

There are several different interpretations of this statement, but given the context it is unlikely that this is a question about our socioeconomic condition and how much a game show is likely to afford to put in an envelope, the assumption is that the envelopes are filled with a random amount of money.

Now, at this point, astute readers should be asking themselves “Is that a rigorous statement?”

Well, usually, it is. If you say “random”, that means a uniformly random distribution. “Randomly pick an integer from 1 to 10″ means that there’s a 10% probability each for 1, 2, etc.

Uniformly random distributions even exist for uncountably infinite sets. You can choose a random real number between 0 and 1. It’ll have a 90% chance of being greater than 0.1, etc.

However: You cannot choose a random number greater than 0, since continuous uniform distributions exist only for bounded sets.

Thus, “two envelopes filled with a random amount of money” is meaningless unless you either set the bounds or specify the distribution you’re talking about.

Philip says: The question is now to understand why solution 1 is incorrect. Solution 1 casually slips in a third envelope under the radar. It tricks you into thinking that there are three possible values: x, 2x, and 0.5x, when in reality there are only two values. Although you don’t know what the values are, one of two scenarios must be the case: (A) the two envelopes are x and 2x, or (B) the two envelopes are x and 0.5x. There is no third case where all three values are in play at once. It is then easy to see that both cases A and B degenerate into the second solution.

This is… wrong.

A mentor once told me: “There’s no such thing as a counterproof.” By this, he meant that if someone can prove that a “proof” is wrong, that “proof” was never a real proof in the first place – it was wrong.

What you wrote isn’t the flaw in Solution 1. It’s… speculation, and it’s incorrect at that. It still doesn’t answer the question “What’s wrong with Solution 1?”

So what’s the flaw in Solution 1? As it turns out, absolutely nothing.

Solution 1 is completely correct. There does indeed exist a rigorous logical proof of “If the amount of money in both envelopes is uniformly distributed and one contains twice the amount the other contains, then it is advantageous to switch.”

So how can that be possible? The answer is by ex falso quodlibet, also known as the principle of explosion.

That’s why you have to be careful with proofs of contradiction. All a proof by contradiction will tell you is that one of your assumptions is incorrect. It won’t tell you which.

Here’s a classic example: The proof that 1 is the largest positive integer:

Let n be the largest positive integer. By contradiction, assume n != 1. n^2 is larger than n. But n is the largest positive integer. Contradiction! Therefore, n = 1.

There’s nothing wrong with the proof itself. But it turns out, the contradicted assumption isn’t that n != 1 (which is correct), it’s the assumption that the largest positive integer even exists.

The same thing is going on here: The only thing wrong about Solution #1 is the assumption that the uniform distribution exists for unbounded sets.

Pick ½ 1 2
Switch #1 1 2 ½
Switch #2 2 ½ 1

as there appears to be equal chance, it does not matter.

If the two alternatives presented in the above sentence are equally likely, then Solution 1 applies.

On the other hand, if you _assume_ the envelopes were first filled with x and 2x dollars and then randomly assigned to two people, then yes, that’s very symmetrical, and Solution 2 applies. But such a process is nowhere mentioned in the problem statement.

(Well, I am overstating my position a bit. Actually I think it’s an ambiguous problem statement. But if I had to choose, I would say Interpretation/Solution 1 is better supported by the literal meaning of the words in the problem statement.)

Concerning the two children riddle, yes that is the one I am talking about. I’ve seen many people who need to formulate this problem by introcucing a meeting with one of the childs. This can be done, but only with a good reason to assume that given the circumstances of the meeting, the child you met could only be a boy. Indeed, this is more impressive because it really seems to be equivalent to saying that the first child is a boy and asking the sex of the second one, but it is slightly different and it still works. You can even say which child is a boy, because it’s the one you met.