The money in 2 evelopes \"paradox\"
 

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?

Re: The money in 2 evelopes \"paradox\"
 

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

A very nice puzzle.

Re: The money in 2 evelopes \"paradox\"
 

[ 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.

Re: The money in 2 evelopes \"paradox\"
 

you definitely switch...

Re: The money in 2 evelopes \"paradox\"
 

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.

Re: The money in 2 evelopes \"paradox\"
 

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

I think my brain is now working again.

Re: The money in 2 evelopes \"paradox\"
 

[ 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.

Re: The money in 2 evelopes \"paradox\"
 

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.

Re: The money in 2 evelopes \"paradox\"
 

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.

Re: The money in 2 evelopes \"paradox\"
 

This has been discussed before.

Always switching and never switching from the first pick are
EV neutral (you may simply use a symmetry argument).

The more interesting result is that there are INFINITELY
MANY strategies of switching that are +EV, assuming that you
can generate a uniform random variable over the unit
interval (0,1).