Since I wasn't satisfied with the previous response, I thought I'd give it a try, although I'm not the best at this sort of thing.
Read the two envelope paradox if you haven't already, since I'm going to refer to it a lot without repeating it here.
Okay, there's two odd conclusions here... The first is that you will always want to switch envelopes sight unseen, no matter which one you start with. That's a pretty funny conclusion, so let's have a look at the reasoning. I think the fallacy here is substituting M for "N or 2N", so let's try it with just N's.
We have four cases:
- You get N and you stay: You have N.
- You get 2N and you stay: You have 2N.
- You get N and you switch: You have 2N.
- You get 2N and you switch: You have N.
As you can see, the results are exactly the same whether you switch or not. If you always stay you always get N or 2N with .5 probability of each. If you always switch you always get 2N or N with probability .5 each. Expected value therefore being 1.5N regardless of switches.
So why is this different from the $100 scenario? In the second scenario we don't always start out with $N.
In order to make the initial value $100 for all cases, we would have to change the value of N for two of the cases, and therefore we change the probabilities. That's what happened with the "M" unknown in the above example. "M" was 1/2N or N, depending on whether the first envelope was the "high" envelope. So this (Perhaps unintentionally) tricks us into thinking the second problem is the same as the first. Effectively, we're lowering the stakes when the "high" envelope is chosen, so the actual gain when switching is raised even though the probabilities are otherwise unchanged.
As for the second part... at first glance the reasoning seems sound despite the bizarre conclusion, but in actuality there's a difference between the first switch and any subsequent switches. Obviously switching twice is the same as not switching at all, so the expected gain will be 0 when making the second switch.
Two cases for switching once:
- You had N and switched. You have 2N.
- You had 2N and switched. You have N.
Both cases are .5 probability so the expected result is 1.5N.
Two cases for switching twice:
- You had 2N and switched twice. You have 2N.
- You had N and switched twice. You have N.
So no matter what you do you still have what you started with; expected result is 1.5N.
The fallacy here is a continuation of the previous one, switching M for a complex probability and trying to compare to the initial problem.
Too bad, cause it'd be a good way to pick up some extra cash if it worked.