Puzzle

[Ninjahedge]

It just depends on when you start measuring the chance.

If you do it from the start, you have the same chance of getting 1X or 2X no matter what you do.

The second scenario would be like this.

You pay $50 to either get $0 or get $150 on a 50/50 chance (you start with $100, you pay $100 to get $50, a loss of $50, or you pay $50 to get $200, a gain of $150). When you limit it to the actual money being risked and the time that you risk it in, it is a 3:1 net gain at 50% (I believe that swould be 1.5?)


Again, it all depends on when you start. If you have nothing, you pick one envelope and then are offered to select the other, your chances are still the same as if you picked the other from the start.

 

[Cogman]

This looks like a botched Monty Hall problem. The option to switch isn't what affects the outcome. It is the elimination of wrong choices that does.

For this to work, the host would have had to have said "The one you picked has the smaller amount of money in it". That, or there needs to be a 3rd envelope which he points out has no money in it.

Ultimately, assuming they both have the same denominations for currency, I would go with the fatter envelope.

 

[C1]

Yes, this is some weird attempt at trickery in terms of formulating a problem that looks like the Montey Hall problem.
Monty Hall was a Baysian idea (ie, conditional probability) wherein additional information was disclosed after an action/choice/event occurred.
The current marginally expressed/disclosed formulation would appear to not be the same. In any event, as always when in doubt, enumerate the sample space events and the probabilities should become evident.

 

[iCyborg]

There are links above that explain the problem, not sure why are people still guessing.

The problem is in prior probabilities:
- If there's an upper limit, say $500, then for every amount above $250 you would not switch, so it's not true that x/2 an 2x are always equally likely.
- If there's no upper limit, then by saying "for every X, there's an equal probability that it contains x/2 or 2x" one assumes a uniform distribution over all integers (or inifinitely many of them) - this cannot be done, you cannot have infinitely many numbers have some constant probability c>0 and that all the probabilities sum up to 1. Sort of the same reason why continuous distributions are always done via prob. density functions and not via assigning probabilities to individual numbers.

So no matter the case, the problem is ill posed.