Vodoo ploppies are right, we can shove our math and sims :)

[sagefr0g]

There is no relation between quantum mechanics and the envelope problem.

right i understand that, i was just wondering what the problem was with pondering over a macro world problem with qm in mind.

what ever, am i correct in understanding from all these posts, that for this particular envelope problem that it matters not if you just go with the first envelope or go with the second envelope if you want? your ev in the long run is gonna be the same?

 

[tthree]

Best game in town

right i understand that, i was just wondering what the problem was with pondering over a macro world problem with qm in mind.

what ever, am i correct in understanding from all these posts, that for this particular envelope problem that it matters not if you just go with the first envelope or go with the second envelope if you want? your ev in the long run is gonna be the same?

Unless they find a way to put negative dollars in an envelope I dont see how you can loss. RoR is 0. No investment.

 

[MangoJ]

your ev in the long run is gonna be the same?

The EV of a single game is the same whichever envelope you take.

The EV of the long run is quite different, if you're smart you can start mapping the probability distribution of the content of the envelopes.

Please don't confuse EV with long run results. The only connection between both is: The EV of the long run result is the EV of the single run, provided that all runs are independent.
All dice thrown are independent, but not each run of the envelope problem. The simple reason is, if you repeat the exact experiment, once you observe a $200 (or $50) envelope, you know the envelopes are $100/$200 (or $50/$100) and you can always make the proper decision.

 

[sagefr0g]

The EV of a single game is the same whichever envelope you take.

The EV of the long run is quite different, if you're smart you can start mapping the probability distribution of the content of the envelopes.

Please don't confuse EV with long run results. The only connection between both is: The EV of the long run result is the EV of the single run, provided that all runs are independent.
All dice thrown are independent, but not each run of the envelope problem. The simple reason is, if you repeat the exact experiment, once you observe a $200 (or $50) envelope, you know the envelopes are $100/$200 (or $50/$100) and you can always make the proper decision.

great point, imho......

so but if you don't know that some certain amount is going to be used each time in the envelopes, then the short run ev and long run ev are one and the same?

 

[MangoJ]

so but if you don't know that some certain amount is going to be used each time in the envelopes, then the short run ev and long run ev are one and the same?

How could you not know the content of the envelopes during the game, you are looking at them every time. One thing I missed stating about long run (I thought it was obvious): The long run is the concept of playing the same game, that means probabilities of all unknown elements do NOT change over time.
The dice will not morph its shape, and the payout of bets will not change with time. For the envelope game, you will learn about the probability distribution of the money in the envelope over time just by observation.

 

[assume_R]

Wizard of Odds

Not sure if anybody mentioned this yet, but the Wizard of Odds discussed his take on this problem: http://wizardofodds.com/askthewizard/258 and he has some nice links at the bottom of his post to others' solutions.

 

[tthree]

Thanks Assume_r

After looking over the various solutions I am more confused than ever. Clearly the switching logic is wrong. But does that include switching after you know the contents of one envelope? The communication of peoples ideas were not always clear. The wizard agreed with k_c. I had to read his explanation many times before I could tell where he was saying the logical error fit into the problem. Not effective communication. His explanation seems to make sense but so does the physorg links which argued my position that by breaking the equivalence of the 2 envelopes the additional information could be used to your advantage by justifying switching. They claimed to have sims to back it up. I am more confused now than before. The wizard seemed to argue both sides but I eventually understood what he was saying. I think? What I thought he was saying made sense.

I was so pleased with myself. The switching logic is flawed as I showed but maybe that means even after you open an envelope. I know given the choice I would switch because it either doesnt change your EV or increases it.

One of the links even managed to turn it into a martingale problem. Oy vey! I didnt try to follow that one. Those israelis had to make it more complicated didnt they.

I think the switch is throwing everyones logic off because it is the one you chose to begin with, so the next leap in logic the mind makes is that it is the same as simply picking the other envelope. If the problem was posed as there are two envelopes one has 1 unit and the other 2 units but the unit size is chosen at random. I show you the contents of 1 randomly selected envelope. Which do you choose that one I showed you or the other. Nobody would argue that you were wrong to pick the unopened envelope. But since the randomly selected envelope was one you initially viewed as your choice this argument seems less valid. Funny how the mind works.

 

[tthree]

ICountntrack I understand your argument. It is what the wizard argued. Like I said I am more confused than ever.

 

[tthree]

Less confused now

Ive had more time to mull the new information over. I think my observation that the randomly selected opened envelope gave you a real advantage in choosing which envelope to end up with even if it had a correlation with your initial choice. A higher expectation doesnt necessarily mean a higher return. It is like a counter sizing his bets but playing basic strategy to have an advantage over a player playing basic strategy. Just because their decisions are the same doesnt mean they have the same advantage. Or the the counter how bets the same but uses playing efficiency to have an advantage. Just because you bet the same doesnt mean you have the same advantage. The fact that you have an advantage doesnt mean you are going to have different results.

If they opened the other envelope every time you would be at an advantage not switching with this information even though you never changed your decision. Advantage and results are not the same. That is the paradox. The mind equates the two because of the 100% correlation to your random selection either positive when never switching or negative when always switching. Like the lotto ticket example you both have a 1/12,000,000 chance that you picked the correct numbers and you know the added information tells you that you absolutely have a 1/2 chance that you hold the winning ticket. Your results are always the same but your odds that you hold the winning ticket changed drastically given the new information. I bet you would have no trouble selling that lotto ticket for lots of money even though your result is still the same and the odds you picked the correct numbers have not changed. Only the odds you hold the winning ticket has changed.

Translating this to the problem says independent of any correlation to your initial pick you improve your expectation by picking the unopened envelope. But that says nothing about improving your result as can be seen when the other envelope is opened every time causing you to never switch. Opening the envelope breaks the equivalence between envelopes and can be used to increase your EXPECTATION.

 

[MangoJ]

Without knowing the prize distribution to begin with, you can't assign possible amounts to the second envelope.
http://wizardofodds.com/askthewizard/258

Thanks Wizard, exactly my guess on this paradox.

 

[tthree]

Thanks Wizard, exactly my guess on this paradox.

Mangoj, I understood what the wizard was saying. Would you say the game is different if a random envelope is selected and opened and then you pick the unopened because it either has twice or half the known amount with equal likelihood? I was having trouble processing this.

My previous post stated were I was left. Let me quantify.
The only problem I have with what the wizard said is if you know you have $100 in one envelope you cannt say you have either $50 or $200.

Given you are shown one of the two envelopes at random, it contains $100.
There is a 50% chance you chose the larger envelope. If that is true then the other envelope contains $50.
The other 50% of the time you chose the smaller envelope. If that is true the other envelope contains $200.
Therefore choosing the other envelope rather than $100 risks $50 to gain $100.

I just cannt see a flaw in that logic. Not one part of it can even be argued to be wrong. I totally understand everything else he said. I question the logic of the last way of stating no advantage in switching. It just seemed forced in comparison. It may not have a flaw but it certainly not as straight forward.

Can anyone show me the flaw in the above logic. It covers 100% of the remaining possibilities. Why is it wrong?

 

[assume_R]

My previous post stated were I was left. Let me quantify.
The only problem I have with what the wizard said is if you know you have $100 in one envelope you cannt say you have either $50 or $200.

Given you are shown one of the two envelopes at random, it contains $100.
There is a 50% chance you chose the larger envelope. If that is true then the other envelope contains $50.
The other 50% of the time you chose the smaller envelope. If that is true the other envelope contains $200.
Therefore choosing the other envelope rather than $100 risks $50 to gain $100.

I just cannt see a flaw in that logic. Not one part of it can even be argued to be wrong. I totally understand everything else he said. I question the logic of the last way of stating no advantage in switching. It just seemed forced in comparison. It may not have a flaw but it certainly not as straight forward.

Can anyone show me the flaw in the above logic. It covers 100% of the remaining possibilities. Why is it wrong?

Your logic is incorrect that there is a 50% chance in both your stated scenarios. Consider the following:

Okay so I am the maker of the envelopes. I put $100 and $200 in each envelope. Now I know there is a 50% chance that you'll choose one or the other. That's a 50% chance for sure. You choose the $100 envelope half the time. Your EV is +$100 by switching. You choose the $200 envelope half the time. Your EV is -$100 by switching.

This time I am the maker of the envelopes. I put $50 and $100 in each envelope. Now I know there is a 50% chance that you'll choose one or the other. That's a 50% chance for sure. You choose the $100 envelope half the time. Your EV is -$50 by switching. You choose the $50 envelope half the time. Your EV is +$50 by switching.

Those are the only ways you can use the 50% number.

 

[revrac]

Mangoj, I understood what the wizard was saying. Would you say the game is different if a random envelope is selected and opened and then you pick the unopened because it either has twice or half the known amount with equal likelihood? I was having trouble processing this.

My previous post stated were I was left. Let me quantify.
The only problem I have with what the wizard said is if you know you have $100 in one envelope you cannt say you have either $50 or $200.

Given you are shown one of the two envelopes at random, it contains $100.
There is a 50% chance you chose the larger envelope. If that is true then the other envelope contains $50.
The other 50% of the time you chose the smaller envelope. If that is true the other envelope contains $200.
Therefore choosing the other envelope rather than $100 risks $50 to gain $100.

I just cannt see a flaw in that logic. Not one part of it can even be argued to be wrong. I totally understand everything else he said. I question the logic of the last way of stating no advantage in switching. It just seemed forced in comparison. It may not have a flaw but it certainly not as straight forward.

Can anyone show me the flaw in the above logic. It covers 100% of the remaining possibilities. Why is it wrong?

Its not that complicated but the false information is confusing it. Basically the issue is probabilities are being assigned without any more information on whats actually in the envelopes..you have 2 envelopes, x and 2x. Look at icountntrack's post, number 54, its a PERFECT mathmatical explanation, could not of put it any better.

When you open one and there is 100 in it, you actually don't have any more information than when you started with. You know you have 2 choices x and 2x and you have 100. You can NOT now assume 50 and 200 are the choices, thats a false assumption. All you know is one has x and one has 2x. You have 50% chance of gaining X and 50% of losing X = 0.

 

[assume_R]

If you want to take it further, you can assume each of my above paragraphs happen 50% of the time (assume the envelope maker is equally likely to start with any amount, and given you know that $100 was chosen by you). So 25% of the time you win $50, 25% of the time you lose $50, 25% of the time you win $100, 25% of the time you lose $100.

But that 50% wasn't stated in the original formulation.

Even though it also yields an EV of 0

 

[assume_R]

You can NOT now assume 50 and 200 are the choices, thats a false assumption.

But you can assume that it'll either be a choice between $50 and $100, or between $100 and $200. You don't know the probability of either of those cases are. But in either case (as I posted), your EV is 0 from switching.

 

[revrac]

But you can assume that it'll either be a choice between $50 and $100, or between $100 and $200. You don't know the probability of either of those cases are. But in either case (as I posted), your EV is 0 from switching.

Without a doubt. I agree that you would then know you have one of those situations. knowing two situations are possible is not assuming false information but still gives you no more "decisional" information about which envelope to choose. All you know is in one situation X is 50 and in one X is 100 with a 50% probability of each situation. You still don't know if you have x or 2x meaning 50% chance of gaining x and 50% chance of losing x.

 

[tthree]

Thanks asume_r I understand

Your logic is incorrect that there is a 50% chance in both your stated scenarios. Consider the following:

Okay so I am the maker of the envelopes. I put $100 and $200 in each envelope. Now I know there is a 50% chance that you'll choose one or the other. That's a 50% chance for sure. You choose the $100 envelope half the time. Your EV is +$100 by switching. You choose the $200 envelope half the time. Your EV is -$100 by switching.

This time I am the maker of the envelopes. I put $50 and $100 in each envelope. Now I know there is a 50% chance that you'll choose one or the other. That's a 50% chance for sure. You choose the $100 envelope half the time. Your EV is -$50 by switching. You choose the $50 envelope half the time. Your EV is +$50 by switching.

Those are the only ways you can use the 50% number.

Thanks Assume. The bold print and the for sure helped drive it home. You cannt add the if then and still have the 50% unless you are sure all possible subsets are represented equally. The $100 samples may only exist as the larger or smaller envelope or any percentage in between. I knew the wizard was saying that but it still didnt make sense. Sometimes I feel foolish. That should have been more obvious especially with how many times it was explained to me. I always understood it partially but now I see the big picture. Thank you everyone.

 

[assume_R]

It's not foolish. If the envelope maker came up to you beforehand and whispered, "I flipped a coin before I filled these envelopes to decide if I should put $50 & $100 or if I should put $100 & $200," then you would know that half the time the other envelope contained $50 and half the time the other envelope contained $200. In that case it would be correct to switch! But the envelope maker never said that. That is the flawed assumption.

 

[Gamblor]

Not sure if anybody mentioned this yet, but the Wizard of Odds discussed his take on this problem: http://wizardofodds.com/askthewizard/258 and he has some nice links at the bottom of his post to others' solutions.

Ah thanks aR, wasn't aware the wizard had a take on this

 

[Gamblor]

Wizard might be wrong. It seems like he's slipping in extra assumptions. He is stating that 50% of the time its either $50 and $100 or $100 and $200. Which is derived from the situation when we find $100 in the envelope. OK, lets assume this is the case, what about the time we chose $50 in the envelope (or the $200)? What do we that say, there's a 50% chance there was $25 and $50 and $50 and $100 all along? And so forth?

We do not have this god-like overview of whats in the envelope. All we know is that we find $100, and one envelope contains twice the other, so we should switch because the EV is $125. Now, I don't necessarily disagree that we might be incorrectly applying EV in this case, but seems sound to me.