Envelope Dilemma?

Interesting take Rudy. However I think we can assume that we don't NEED the money and are simply trying to play the optimal game. I think when we get into utility principle then we are missing out on the rest of the problem.


Excellent answers. The answer is obvious to me that switching and not switching have the same EV no matter what the disparity factor of x is between envelopes. I just can't prove it.

The people that say you had a 50% chance of having the bigger one to begin with and switching increases expectation because of the disproportional payoff seem correct also.

Clearly there is no solution to this problem. I haven't looked at the Wizards answer and I am not really interested in looking it up.

Unimpressed with the Wizards answer. He used the fact that the gameshow prize of 50k is a big prize and that knowing that you already have a big prize then that offsets the EV of switching.

Kinda a stupid answer to a hypothetical question is (to use a real life example of 50k is a lot of money!), like that isn't avoiding the problem.

Anyway apparently this debate will go on forever so I will resign myself to say that it doesn't matter at all considering you could choose either envelope in the first place and be right 50% of the time and there is an obvious paradox because the other answer of, switching regardless, seems to make sense also when you consider that the second envelope has a 50% chance of containing twice as much compared to a 50% chance of containing half as much.

I am not impressed with the Wizard's answer. He dodged the question and never admitted that it was simply a paradox. Couched the situation in a real life context of 50k being alot of money to offset the chances. Not good. Ok back to sports for this amigo.

Perhaps I'm missing something, but if we exclude 1) marginal utility theory and 2) assume a 0.5 probability regardless of the amount initially found, the EV's for the two choices is the same; hence, there is no difference in either keeping the first envelope or swapping. The EV is the same for both.

Let's assume $50,000 and $100,000

Choice A (never swap, keep the first envelope)
====> $50,000 (stop, win $50,000)
====> $100,000 (stop, win $100,000)
EV (Choice A) = 0.5*50,000 + 0.5*100,000 = You'd win around $75,000 on average over a long series of trials

Choice B (swap always, take the second envelope)
====> $50,000 (swap) ======> $100,000 (stop, win)
====> $100,000 (swap) =====> $50,000 (stop, win)
EV (Choice B) = 0.5*$100,000 + 0.5*$50,000 = You'd keep $75,000 on average over a long series of trials

Algebra:
EV (A) = 0.5*X + 0.5*2X = 1.5X
EV (B) = 0.5*2X + 0.5*X = 1.5X

Is it more complicated?

No more dilemma. The obvious answer is also the correct answer.

Jeez I almost thought that there was a paradox there.

You're over-thinking it.

You are offered a bet with a 50% chance of winning with a 2 to 1 payoff.

Betting to lose 25 vs. winning 50

Swap.

It's an ill-posed problem.

You do not have enough information to make the proper decision.

This problem is a bit more complicated than is being discussed here.

The paradox lies in the fact that you should always switch envelopes after opening whatever envelope you chose. This would imply that it would be beneficial to switch envelopes prior to opening, which isn't the case, hence the paradox.

Regardless of the amount of money in the first envelope you should always switch to the other envelope without hesitation immediately after learning of the amount in the first envelope. All learning the amount in the first envelope does for you is provide a basis for you to calculate the probabilieis of how much you will end up with.

Matador made the assumption that the envelope would contain either 50000 or 100000 but there are actually three possibilities, if envelope A contained 50000 (call this amount X) envelope B could contain either 25000 (X/2) or 100000 (2X). Since there is a 50% chance that envelope B contains each of the other two possible sums then since X isn't greater than or equal to .5*(X/2)+.5(2X) then the correct action is to change to envelope B.

To create an analogy that any gambler should be able to understand imagine that someone gave you 50000. Immeditaely after giving you the 50000 they gave you an opportunity to wager on the flip of a fair coin. Heads they would double your money (you end with 100000), tails they would take half your money (you end with 25000). Since the fair odds on the flip of a coin would be double or nothing it would obviously be in your best interest to wager on the flip of the coin. This is the exact situation created with the envelope paradox.

FrankB, Can you give us the correct answer?

This is not a conditional probablilty question. Nothing has been exposed as a result of opening one envelope. The crux of the dilemma is that after opening an envelope, you can calculate an EV based on partial knowledge. The people that say "switch" are implicitly assuming that they can produce a +EV by switching, but the probability of having the envelope with the bigger amount is still 50/50. This is different than the proposition Baker describes, where you are getting better odds on one end of a coin flip than the other. In that case, you would always go for the gold because you're getting 2/1 odds. In the envelope case, it may LOOK to you like it's a coin flip based on your partial knowledge, but the outcome is predetermined when the envelopes were sealed. One has more, one has less, and you learned nothing from opening one envelope that gives you any clue about the contents of the other.

"Ask Marilyn" in the Sunday Parade Mag solved this problem some time last year - the bad news is I can't remember her math. - sorry.