The money in 2 evelopes "paradox"

[Silent A]

I don't know if this has been discussed here before.

here's the paradox ...

Suppose you're given 2 envelopes with money in them and you are told that one contains twice as much as the other (let's say they're cheques and therefore weight is not an issue). You pick one and open it up to reveal a $100 cheque. You are now offered the opportunity to switch envelopes.

Is switching +EV?

Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25.

Argument 2: It's EV neutral. If always switching was a +EV strategy then it would be more profitable to choose envelope A first and then switch to B then to just choose envelope B and not switch.

Who's right?

[Moose747]

I'm pretty sure Argument 2 is stronger, but I can't figure out why.

A very nice puzzle.

[Silent A]

[ QUOTE ]
I'm pretty sure Argument 2 is stronger, but I can't figure out why.

A very nice puzzle.

[/ QUOTE ]

Yes, I'm sure it is too. The tricky part is showing what is wrong with Argument 1's approach (rather than just showing that it must be wrong - as argument 2 does). I think I did it once, but no one ever responded with "ah, yes i see it now".

I'm hoping someone here has a good answer.

[Beermantm]

Not sure on how this qualifies as a paradox although it is intresting.

In the first equation with the envelopes you have a 100% Chance of making money so there is no gamble.

In the second equation you are asked to wager half of the first envelope contents to possibly double that envelope.

You are only +EV to switch because there was no risk to get to the first amount and now you are only risking half meaning you are a gauranteed winner regardless if you switch or not. Lets change the rules just slightly and the paradox goes away. Lets say you can only get paid out in whole dollar amounts (meaning no fractions of a dollar) and say the first envolope you pick is 1 dollar. Now you are asked to risk the full dollar amount because anything under a dollar can not get paid off. The next envolope would be (50 cents)a zero dollar value or 2 dollars. Now your back to 1:1 instead of 3:1 and probably better off just keeping the buck.

The first envelope could also be a penny and then you can not get paid because there is no coin value less than the penny. 1:1

The trick just seems to be that if the first envelope has any value at all other than 0 you can devide it by two for infinity.

The answer is you switch because you can't lose after the first pick and you want to know like wholly hell what is in the second envelope. LOL

[suited42]

you definitely switch...

[Moose747]

The logical conclusion of argument 2 (switching is EV neutral) is that when you decide to switch, there's actually a 2/3 chance you picked the envelope with more money. ie 2/3(x/2)+1/3(2x)=x

I don't see how this is possible though.

[Moose747]

Okay, there's a fairly decent explanation here: http://en.wikipedia.org/wiki/Two_envelope_paradox

I think my brain is now working again.

[NMcNasty]

[ QUOTE ]
I don't know if this has been discussed here before.

here's the paradox ...

Suppose you're given 2 envelopes with money in them and you are told that one contains twice as much as the other (let's say they're cheques and therefore weight is not an issue). You pick one and open it up to reveal a $100 cheque. You are now offered the opportunity to switch envelopes.

Is switching +EV?

Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25.

Argument 2: It's EV neutral. If always switching was a +EV strategy then it would be more profitable to choose envelope A first and then switch to B then to just choose envelope B and not switch.

Who's right?

[/ QUOTE ]

Argument 1 is flawed because you shouldn't assign the probabilities of the envelopes either being $50/$100 or $100/$200 to be 50/50. It could be that the probability the envelopes are $50/$100 is 95% and the probability they are $100/$200 is only 5%. In that case switching would be a clear mistake.

If there was some premise that says "no matter what envelope you choose the next one will either have twice as much %50 of the time and half as much %50 of the time", then argument 1 would in fact hold up.

However, simply because you cannot logically assign the probabilites of the two possibilites to be 50/50 that doesn't mean that sticking with the envelope you have is always EV neutral. There are some probabilites you should assign to each possibility P($50/$100)= x and P($100/$200) = y. What probabilties you assign have nothing to do logical factors, they are just estimated guesses based on your knowledge of the how the experiment is set up. So if you assign x to be .60 and y to be .40 you should STILL switch envelopes for a +EV.

So in conlcusion, both arguments are flawed. The decision to switch or not is relevant, but its not based on your possibilites being 50/50.

[Alan3]

I'm having trouble wrapping my head around this, but tt becomes much more clear if I change the problem slightly.

Suppose you're given 2 envelopes with money in them and you are told that one contains ONE THOUSAND TIMES as much as the other (let's say they're cheques and therefore weight is not an issue). You pick one and open it up to reveal a $100 cheque. You are now offered the opportunity to switch envelopes.

I have a belief distribution that someone is not going to just give me $100,000. If I open the $100 envelope I'm going to keep it because I expect the other envelope has $0.10.

This expectation is much smaller at the 50/100/200 range. The chance that someone is going to give me $100 is very close to the chance that someone will give me $200 and I'm not going to lament the loss of $50 so much, so maybe switching is the right answer at this level.

Increasing the "amplitude" of the problem helped me understand the "Let's make a deal" problem too.

[DrVanNostrin]

This smells like a Monty Hall varient. And it smells like a switch. If you switch when you should have stayed you'll cost yourself $99. If you stay when you should have switched you'll cost yourself $9900. Unless you somehow know that staying is correct over 99% of the time you should switch.