I've heard that before, it's just bad math. Why would you add the $2 to $27 when you should be subtracting it?

Oh, it seemed to me as though you were counting from the original amount as a gain, then adding instead of subtracting, and arriving at the number you did out of coincidence Well, anyways:

But that part itself is sound. The problem isn't until we begin to switch that this part becomes unsound, and it only does so because of an error the later steps failed to account for.

Perhaps ignorance is bliss, because I don't understand why symmetry affects the result when the envelopes are not symetrical... At this point it might be useful to note that I only read up to "a harder problem" on the wikipedia page.

Right but this part is wrong, AFAICT:

Just because the expected value is 1.25 does not mean that you will gain 1.25 the amount of money on a single switch. In fact, that is impossible. It means that over the course of repeated independant trials, you will arrive at an average of 1.25 times the amount of money per play.

This is why that line is wrong. Expected value does not mean average value. I means that, odds are, given n trials, you will end up with a score of n * expected value. It doesn't mean that you are certain to get that number, it's based on the probabilities involved in the original problem.

Remember, the trials need to be independent. Counting my and my cousin's winnings from the same set of envelopes is not independent.

[edit]
/me learns how to spell independent.

I found this link

Having to do with the shaking hands problem, it seems maybe matt and I are wrong.

I looked at that problem from the bottom up, using the trivial cases, making the safe assumption that there is no loss of generality. Using the same rules, if you invite no couples, the spouse shakes 0 people's hands.

If you and your spouse (Y and S) invite one couple (A and B), then some combination of A,B,S must shake 0,1, and 2 hands. Assuming A shakes only the spouse's hand, you get:

A: S
B: ?
S: A

Now, S must shake another person's hand because A shook exactly one. The only choice is:

A: S
B: S
S: A,B

But now we are stuck because A and B both only shook 1 hand. There's no way to get around it, other than for B to shake Y's hand, but that won't work because A & B will both shake 2 hands. Another contradiction happens if you say A only shakes Y hand.

A: Y
B: ?
S: ?

Now B must shake either 0 or 2 hands. If B shakes 0 hands, then S must shake two hands. But that's impossible, because S has no one else to shake. If B shakes 2 hands, then S must shake 0. But the only way for B to shake 2 hands is to shake S hand. So that case falls apart.

This only leaves the following case:

A: Y,S
S: A
B: -

Now you have a valid 2,1,0 combination. So the spouse must have shook 1 person's hand. Obviously A & B are interchangeable, but that's not relevent.

Anyway, it's reasonable to assume that there is no loss of generality here, and that as you work your way up with more couples, you maintain the same pattern with the spouse being stuck in the middle, because you are the outsider.

This was told to me by a guy from Israel in about 1990. The story concerns a particular village where there were 50 couples - 50 husbands and 50 wives. The problem was that one or more of the wives were being unfaithful. In this village, if a wife was unfaithful, she'd sleep with each of the 49 other man, without her own husband being aware of it! At the monthly village meeting (to which only men are invited), this problem came up. The town elder said, "If a man discovers that his wife has been unfaithful, he should not act straight away, but wait until midnight. On the stroke of midnight, he should shoot her through the heart."

That night, nothing happened.
The following night, nothing happened.
The next night, shots were fired.

How many wives were unfaithful?

Pete

If a man has slept with another woman, his own wife has been unfaithful.

Three were dead.

Husbands only know when other wives cheat. So if he knows two have cheated, he'll wait till third day to kill his wife.

If you didn't know anyone cheated, then you'd kill your wife on day 1 (under the riddle's assumption that it was known that at least one person cheated). If you knew one person cheated, then you'd kill your wife on day 2 if no one died previously (because if that was the only person to have cheated, then she would have been killed on day one by her husband that didn't know of anyone), etc... Anyway, it's late, and I'm going to bed. I think I described it correctly.