Puzzle

[Nathelion]

I'm having a bit of trouble with this puzzle a friend sent me, so I'm hoping the HT community can shed some light on the problem.

Bob is playing a game. There are two envelopes in front of him. There is money in both. One has twice as much money as the other.
Bob must choose one of the envelopes and open it. He is then given the option to switch to the other envelope. Should he switch?

Here is the mystifying part: Assume that there is x money in the first envelope. If he doesn't switch, he will get x money. If he switches, he will get either 2x or 0.5x money with equal probability. The expectation value will be 2x*.5 + .5x*.5 = 1.25x money. So it is always better to switch.
But then he doesn't even need to open the first envelope, he could just switch right away.

??????

 

[CycloWizard]

The correct response is to grab both envelopes and run. Other than that, I'm not sure where you're running into trouble.

 

[Nathelion]

My problem is that I don't see how opening the envelope gives you any new information. OK, so there's x money in there. Let's go back to before we opened any envelopes. Let's assume that there's x money in the left envelope. Grabbing the right envelope now has the expectation value of giving you 1.25x money, whereas the left only gives you x. But you could just as easily have defined x to be the contents of the right envelope, in which case the left one would give you more.

 

[uclabachelor]

I'm having a bit of trouble with this puzzle a friend sent me, so I'm hoping the HT community can shed some light on the problem.

Bob is playing a game. There are two envelopes in front of him. There is money in both. One has twice as much money as the other.
Bob must choose one of the envelopes and open it. He is then given the option to switch to the other envelope. Should he switch?

Here is the mystifying part: Assume that there is x money in the first envelope. If he doesn't switch, he will get x money. If he switches, he will get either 2x or 0.5x money with equal probability. The expectation value will be 2x*.5 + .5x*.5 = 1.25x money. So it is always better to switch.
But then he doesn't even need to open the first envelope, he could just switch right away.

??????

He should keep the original and bet his friend $100 that if his friend switches, the two will end up with the same earnings over the long run.

 

[borisvodofsky]

My problem is that I don't see how opening the envelope gives you any new information. OK, so there's x money in there. Let's go back to before we opened any envelopes. Let's assume that there's x money in the left envelope. Grabbing the right envelope now has the expectation value of giving you 1.25x money, whereas the left only gives you x. But you could just as easily have defined x to be the contents of the right envelope, in which case the left one would give you more.

OMG n00b, the tricky is in the causal order. He must've already CHOSEN an envelope if he's to SWITCH to another one.

 

[Nathelion]

OMG n00b, the tricky is in the causal order. He must've already CHOSEN an envelope if he's to SWITCH to another one.

But in the hypothetical case where we haven't even opened the envelopes yet, we have literally not gained any new information. How can we draw the conclusion that one envelope is better than another if we have no new information?

 

[Nathelion]

It seems that just the act of assigning a variable to represent the amount of money in an envelope causes trouble. If we let x represent the amount of money in one envelope, then the other envelope becomes more attractive. But the same holds true if we let x represent the money in the other envelope.

 

[Howard]

I don't think it matters if switching happens.

 

[Nathelion]

I don't think it matters if switching happens.

What do you mean? You lost me, sorry

 

[Howard]

What do you mean? You lost me, sorry

Utilizing the option to switch envelopes, IMO, should not result in any statistical change.

 

[DominionSeraph]

never mind

 

[edcarman]

This is a similar probability problem the Monty Hall problem involving a game show, three doors, a car and some goats.

Wikipedia has a comprehensive write-up:
http://en.wikipedia.org/wiki/Monty_Hall_problem
http://en.wikipedia.org/wiki/Two_envelopes_problem

 

[Nathelion]

This is a similar probability problem the Monty Hall problem involving a game show, three doors, a car and some goats.

Wikipedia has a comprehensive write-up:
http://en.wikipedia.org/wiki/Monty_Hall_problem
http://en.wikipedia.org/wiki/Two_envelopes_problem

I understand the Monty Hall problem. This is similar, but not the same. Thank you for the second wikipedia link though.

 

[Nathelion]

Yes, the trick is that the (envelope 1(1) + envelope 2(1) + envelope 1(2) + envelope 2(2))/2 does not equal 2x. It just looks that way because you've redefined the variable.

I'm not sure I understand entirely what you mean. Could you explain more verbosely?

 

[humanure]

http://www.maa.org/devlin/devlin_0708_04.html

 

[crashtestdummy]

I'm not sure I understand entirely what you mean. Could you explain more verbosely?

You calculate the expectation of earnings as if there were actually three envelopes, not two. You're assuming that your envelope has a value x, and that the other envelope has either a value of 0.5x or 2x. This is not the case. It would imply that there is a factor of 4 in the range of possible earnings. It's the equivalent of giving me $10, and the choice of taking one of two envelopes, one with $5 and one with $20.

In reality, since the value of your envelope is unknown, your envelope has either the value 1x or 2x, and the other envelope is valued at 1x or 2x, meaning that your expected earnings in BOTH CASES is 1.5x.

 

[firewolfsm]

I see it this way, rather than thinking that there are two envelopes, think that by switching you accept a 50 percent chance of either halving or doubling, this chance is intrinsically good because there is more to gain than to lose. Basically, ending up with the smaller envelope 50% of the time is fine. The bet is not in choosing an envelope because there is no game to that, merely a random choice, but in the switch.

 

[ModestGamer]

odds of money 100%

odds of more then the smallest amount 50%

There is no difference.

I'm having a bit of trouble with this puzzle a friend sent me, so I'm hoping the HT community can shed some light on the problem.

Bob is playing a game. There are two envelopes in front of him. There is money in both. One has twice as much money as the other.
Bob must choose one of the envelopes and open it. He is then given the option to switch to the other envelope. Should he switch?

Here is the mystifying part: Assume that there is x money in the first envelope. If he doesn't switch, he will get x money. If he switches, he will get either 2x or 0.5x money with equal probability. The expectation value will be 2x*.5 + .5x*.5 = 1.25x money. So it is always better to switch.
But then he doesn't even need to open the first envelope, he could just switch right away.

??????

 

[Paperdoc]

Here's the way I figure it:
There are TWO scenarios with known probabilities. In each, the first choice made completely determines the result of the second, even though it is not known in advance.

Scenario #1: Pick One of the envelopes at random and DO NOT change. We know one has $x, one has $2x. If the event (choosing ONE envelope) is repeated many times, the probability of picking the one with $x is 0.5, while the probability of picking the one with $2x also is 0.5.The average amount found in the envelope is $1.5x.

Scenario #2: Pick one envelope at random, then CHANGE. Same contents. Same probabilities! The probability of picking the one with $x FIRST is still 0.5, but when you switch, that makes the probability of ending up with the one containing $2x 100% of the first probability, also 0.5 again! This probability is COMPLETELY determined by the result of the first choice (even though at that point we do not know whether the first one had $x or $2x, because we have not been told "x".) That is, the probability for the contents of the second envelope is NOT independent of the first probability. Likewise, the probability of ending up (after the switch) with the one containing $x is 0.5. SAME RESULT AS BEFORE, average amount "found inside" is still $1.5x.

 

[Nathelion]

http://www.maa.org/devlin/devlin_0708_04.html

Thank you. That explains it.

 

[ModestGamer]

Here's the way I figure it:
There are TWO scenarios with known probabilities. In each, the first choice made completely determines the result of the second, even though it is not known in advance.

Scenario #1: Pick One of the envelopes at random and DO NOT change. We know one has $x, one has $2x. If the event (choosing ONE envelope) is repeated many times, the probability of picking the one with $x is 0.5, while the probability of picking the one with $2x also is 0.5.The average amount found in the envelope is $1.5x.

Scenario #2: Pick one envelope at random, then CHANGE. Same contents. Same probabilities! The probability of picking the one with $x FIRST is still 0.5, but when you switch, that makes the probability of ending up with the one containing $2x 100% of the first probability, also 0.5 again! This probability is COMPLETELY determined by the result of the first choice (even though at that point we do not know whether the first one had $x or $2x, because we have not been told "x".) That is, the probability for the contents of the second envelope is NOT independent of the first probability. Likewise, the probability of ending up (after the switch) with the one containing $x is 0.5. SAME RESULT AS BEFORE, average amount "found inside" is still $1.5x.

NO.

the odds are 50/50 on the larger amount no matter how many times you swap envelopes.

 

[Paperdoc]

ModestGamer, we are agreeing. If the odds are 50% on the larger amount, then they are 50% on the smaller amount, and the average is still 0.5 * $x + 0.5 * $2x = 1.5 $x.

 

[epidemis]

But that is just the average amount you will get for choosing any envelope. It doesn't tell you anything about whether it's good to switch or not.

 

[CSMR]

Bob is playing a game. There are two envelopes in front of him. There is money in both. One has twice as much money as the other.
Bob must choose one of the envelopes and open it. He is then given the option to switch to the other envelope. Should he switch?

Here is the mystifying part: Assume that there is x money in the first envelope. If he doesn't switch, he will get x money. If he switches, he will get either 2x or 0.5x money with equal probability. The expectation value will be 2x*.5 + .5x*.5 = 1.25x money. So it is always better to switch.
But then he doesn't even need to open the first envelope, he could just switch right away.

??????

The argument you are making is like this: if I see open the first envelope and find an amount x in the envelope, it could have come from (x,2x) or (x,1/2x) with equal probability. This is not true.

If we fix a concrete prior belief, we can see where you are going wrong:
Imagine this problem: Bob believes the envelopes contain ($1,$2) or ($2,$1) with equal probability. This belief satisfies your assumption that one envelope always contains twice as much as the other, and your implicit assumption of symmetric probabilities (x,y) and (y,x) equally likely.
Suppose he finds an envelope with $2. Your argument is he should switch because it could be ($2,$1) and ($2,$4) with equal proability. But in fact his updated belief is ($2,$1) with probability 1 and he shouldn't switch.

In general the amount of money seen will convey information about whether the envelope chosen is the better or worse of the two. (If it were not so, the paradox you describe would hold, therefore it must be true for all priors.)

Possibly you are also assuming a uniform prior belief on the amount of money in each envelope, which doesn't exist. Trying to do something impossible can easily generate paradoxes.

 

[Unmoosical]

If he switches, he will get either 2x or 0.5x money with equal probability.

The problem you are having is because of your explaination. You give two values for x.


When you state 2x, x is equal to the money in the envelope with the smaller amount.
When you state .5x, x is equal to the money in the envelope with the larger amount.

You've given 2 values to x.


In terms of Envelop containing the lesser amount.
Scenario: Open envelope 1: You should get x or 2x.
Scenario: Open envelope 2: You should get 2x or x.

In terms of Envelope containing the larger amount.
Scenario: Open envelope 1: You get .5x or x
Scenario: Open envelope 2: You get x or .5x


As others have said it's a 50/50 chance.