Two Envelope Paradox

It’s been a long day for me to attempt to get this,… but it almost sounds like if you were to graph it out, it’ll look kinda like a Vertical Asymptote to infinity on the x axis…

Reading the graph left to right for every exchange…

I just have to rest a little to find the flaw…

That's true and clearly makes a nonsense of the final proposition, but there still seems to be a paradox if you're only given the choice to swap once.

Does it make any difference if you're allowed to look in the first envelope before deciding whether to swap or not?

So, let's say you can only swap once.

Looking inside the envelope wouldn't give you any useful information.

Say, you do look and there is $100 in the envelope. The other envelope contains $50 or $200. If you swap, you either lose $50 or you gain $100, both equally probable as far as I can see.

It would seem that swapping improves your expectation, even though the same logic applies to either envelope.

That's about where I had got to when I posted.

Here's a way to look at it. There are infinitely many universes with two envelopes with twice as much in one as the other, but there are only two universes where one contains $100. If we look we know we are in one of those two universes.

Given (no causality here, just starting with this assumption) that we are in one of those two: the the chances are as follows

value $ : 50 100 200

stick. - : 1/4 1/2 1/4

change : 1/4 1/2 1/4

Alternatively you could argue that the universe was chosen by the person who placed the money in the envelopes. In which case there are two possibilities:-

value $ : 50 100

stick. - : 1/2 1/2

change : 1/2 1/2

OR

value $ : 100 200

stick. - : 1/2 1/2

change : 1/2 1/2

Hmmmmm. That is what we know is really the correct "answer". But I don't think it helps a lot with the original paradox.

You've nailed it...
You can lose $50 or gain $100
So that's much better odds than lose $50 or gain $50, thus it pays to swap in the long run... mind it's highly unlikely that you'll repeatedly come across this proposition.
Del

I was just thinking… there’s Too much Las Vegas analysis going on here….

if we take greed of win/loss out of the picture.

the bottom line is, you came with empty pockets, and you walk away with money in your pocket. Which makes the odds better then Las Vegas…

give me one of them damn envelopes…

I believe Troy is correct.

Selected Envelope contains either A rupees or B rupees with equal probability:

1) A rupees = B/2 rupees (2A=B)

OR

2) B rupees = A/2 rupees (2B=A)

Conditions 1 and 2 cannot both be correct

Pemdas faux pas ....?

It's too early in the morning for that sort of thing. There is a x/2 chance of hurting people's brains with a puzzle like this.

If one envelope has 2x dollars, the other x/2, the average = (2x + x/2)/2 = 5x/4. If the experiment is done many times, sometimes swapping, sometimes not, that's the average outcome, agreeing with the puzzle.

The flaw I think is where it says "5*x/4 dollars, which is greater than x dollars. So you should swap". It doesn't follow that you should swap. 5x/4 is greater than x whether you swap or not.

Agreed. Ignore where I said "If one envelope has 2x dollars, the other x/2...." that's wrong, should be If one envelope has 2x dollars, the other x.

In your example of $50 and $100, 50/2 + 100/2 = 75, which is the average outcome if the experiment is done many times, sometimes swapping, sometimes not, as you'd expect.