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The two envelopes problem
 Posted: 08 January 2012 09:37 AM [ Ignore ]   [ # 16 ]
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Write4U - 07 January 2012 09:30 PM

As long as I don’t have to buy the envelopes and have no investment, I gain either way. If A has \$1, I gain 1\$. If B has \$2, I gain \$2. I can’t lose anything either way, there is no penalty involved. Can’t look a gift horse in the mouth….

Exactly.

So, why not take a calculated gamble, switch to double the money or lose half of it? It is not double or nothing. In any case, because you did not have to pay anything to choose the envelopes, you lose nothing.

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 Posted: 08 January 2012 09:47 AM [ Ignore ]   [ # 17 ]
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kkwan - 08 January 2012 09:27 AM

It is double or half, not double or nothing.

There is no trick.

So, why should one not switch?

There is a trick Kkwan because it’s irrelevent if it’s double or half. What’s relevent is that the amount of money you could gain is the same amount as the amount of money you could lose.

So if you have an envelope with £200 in it and that’s the smallest amount you would gain £200 by switching.

If you have an envelope with £400 in it and that’s the largest amount you would lose £200 by switching.

It’s as simple as that.

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 Posted: 08 January 2012 10:32 AM [ Ignore ]   [ # 18 ]
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StephenLawrence - 08 January 2012 09:47 AM

If you have an envelope with £400 in it and that’s the largest amount you would lose £200 by switching.

It’s as simple as that.

Yep. Expressed more generally: if the one envelope has A dollars, and the other 2A dollars, you gain or you lose A dollars. I don’t understand why people are really discussing this.

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 Posted: 08 January 2012 10:37 AM [ Ignore ]   [ # 19 ]
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GdB - 08 January 2012 10:32 AM

I don’t understand why people are really discussing this.

I’m really discussing it to point out we shouldn’t really be discussing it.

And it’s always good to try and learn the best way to explain why.

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 Posted: 08 January 2012 10:48 AM [ Ignore ]   [ # 20 ]
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GdB - 08 January 2012 10:32 AM
StephenLawrence - 08 January 2012 09:47 AM

If you have an envelope with £400 in it and that’s the largest amount you would lose £200 by switching.

It’s as simple as that.

Yep. Expressed more generally: if the one envelope has A dollars, and the other 2A dollars, you gain or you lose A dollars. I don’t understand why people are really discussing this.

Almost right, you are never given both envelopes i.e. 3A.  You start with 0A and can only choose one envelope which is then given to you.
Regardless if the envelope contains 1A or 2 A, you always gain.
This is not a mathematical question. It is a psychological question.

[ Edited: 08 January 2012 11:00 AM by Write4U ]
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 Posted: 08 January 2012 11:23 AM [ Ignore ]   [ # 21 ]
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StephenLawrence - 08 January 2012 10:37 AM

I’m really discussing it to point out we shouldn’t really be discussing it.

And it’s always good to try and learn the best way to explain why.

That’s perfectly fine. I just waited till the correct answer came along…

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 Posted: 08 January 2012 09:03 PM [ Ignore ]   [ # 22 ]
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StephenLawrence - 08 January 2012 09:47 AM

There is a trick Kkwan because it’s irrelevent if it’s double or half. What’s relevent is that the amount of money you could gain is the same amount as the amount of money you could lose.

So if you have an envelope with £200 in it and that’s the smallest amount you would gain £200 by switching.

If you have an envelope with £400 in it and that’s the largest amount you would lose £200 by switching.

It’s as simple as that.

It is, so to switch or not to switch?

By not switching, one is following the proverb “a bird in the hand is worth more than two in a bush” i.e. it is preferable to have a small but certain advantage than a mere potential of a greater one.

OTOH, if one could gain £200 by switching or only lose £200 leaving one with £200, why not switch? The potential gain is worth the gamble as the odds are in your favor on average.

This is a problem in decision theory.

From this paper
HERE

From the abstract:

The two-envelope problem (or exchange problem) is one of maximizing the payoff in choosing between two values, given an observation of only one. This paradigm is of interest in a range of fields from engineering to mathematical finance, as it is now known that the payoff can be increased by exploiting a form of information asymmetry. Here, we consider a version of the ‘two-envelope game’ where the envelopes’ contents are governed by a continuous positive random variable. While the optimal switching strategy is known and deterministic once an envelope has been opened, it is not necessarily optimal when the content’s distribution is unknown. A useful alternative in this case may be to use a switching strategy that depends randomly on the observed value in the opened envelope. This approach can lead to a gain when compared with never switching. Here, we quantify the gain owing to such conditional randomized switching when the random variable has a generalized negative exponential distribution, and compare this to the optimal switching strategy. We also show that a randomized strategy may be advantageous when the distribution of the envelope’s contents is unknown, since it can always lead to a gain.

As a game:

The two-envelope problem can be thought of as a game where there is a ‘house’ and a ‘player.’ The objective for the player is to gain money from the house. The version of the two-envelope problem that we consider is one in which the house selects a value x and seals the amounts \$x and \$(2x) into two identical envelopes. We assume here that x is an outcome of a continuously valued nonnegative random variable, X. The player chooses one envelope at random, opens it, observes the value, y, then decides whether to keep or switch envelopes in order to maximize the payoff.

Unlike the roulette, there is no house edge or expected value in its favor.

From the wiki I cited:

Game theory

This problem can be considered from the point of view of game theory, where we make the game a two-person zero-sum game with outcomes win or lose, depending on whether the player ends up with the lower or higher amount of money. One player chooses the joint distribution of the amounts of money in both envelopes, and the other player chooses the distribution of Z. The game does not have a “solution” (or saddle point) in the sense of game theory. This is an infinite game and von Neumann’s minimax theorem does not apply.

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 Posted: 08 January 2012 09:32 PM [ Ignore ]   [ # 23 ]
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Write4U - 08 January 2012 10:48 AM

Almost right, you are never given both envelopes i.e. 3A.  You start with 0A and can only choose one envelope which is then given to you.
Regardless if the envelope contains 1A or 2 A, you always gain.
This is not a mathematical question. It is a psychological question.

That’s certainly true, so I don’t understand why Stephen and GdB don’t look at it from this simple accounting perspective.

OTOH, one can consider it as a problem in deciding whether to switch or not to switch.

There is the potential advantage in switching verses the potential loss (win more/less), so why not switch?

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 Posted: 08 January 2012 10:02 PM [ Ignore ]   [ # 24 ]
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kkwan - 08 January 2012 09:32 PM
Write4U - 08 January 2012 10:48 AM

Almost right, you are never given both envelopes i.e. 3A.  You start with 0A and can only choose one envelope which is then given to you.
Regardless if the envelope contains 1A or 2 A, you always gain.
This is not a mathematical question. It is a psychological question.

That’s certainly true, so I don’t understand why Stephen and GdB don’t look at it from this simple accounting perspective.

OTOH, one can consider it as a problem in deciding whether to switch or not to switch.

There is the potential advantage in switching verses the potential loss (win more/less), so why not switch?

If both envelopes are unopened, the odds are 50/50. I see no advantage in switching. You might be switching the envelope containing 2A for the one containing 1A. Without a third “known” probability, there is no advantage either way. As the conclusion above shows it is an infinite problem. The actual amounts in the envelopes are irrelevant. The probability factor of “picking” 2A remains @ 50%.

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 Posted: 09 January 2012 12:45 AM [ Ignore ]   [ # 25 ]
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Here I have a better one (those who already know it, do not give the solution immediately):

The quiz master shows you 3 boxes. One contains a price, the other boxes are empty. You choose a box, say box B. Now the quiz master opens one of the other boxes which he knows is empty (with 3 boxes with one price there always is another empty box). Now he asks you: do you stick to box B, or do you switch?

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 Posted: 09 January 2012 01:29 AM [ Ignore ]   [ # 26 ]
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kkwan - 08 January 2012 09:03 PM

OTOH, if one could gain £200 by switching or only lose £200 leaving one with £200, why not switch?

bold by me.

Stephen

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 Posted: 09 January 2012 02:24 AM [ Ignore ]   [ # 27 ]
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GdB - 09 January 2012 12:45 AM

Here I have a better one (those who already know it, do not give the solution immediately):

The quiz master shows you 3 boxes. One contains a price, the other boxes are empty. You choose a box, say box B. Now the quiz master opens one of the other boxes which he knows is empty (with 3 boxes with one price there always is another empty box). Now he asks you: do you stick to box B, or do you switch?

Ok, gut reaction is switching makes no difference, so assuming that’s wrong it’s a cool puzzle. But how could it be wrong? I’m trying to move towards an answer and think I might vaguely have it, but still, when I think it through It doesn’t convince me.

So here is the idea, in white.  The clue is in “(with 3 boxes with one price there always is another empty box).” what this means is I’m getting no extra information about the probability of the money being in my box by the quiz maker opening an empty box. Assuming the quiz master opens box C what I am getting is information that changes the probability of the prize being in box A only. As the probability of the prize being in my box remains the same and the probability of the money being in box A increases, I should switch.

[ Edited: 09 January 2012 03:12 AM by StephenLawrence ]
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 Posted: 09 January 2012 03:39 AM [ Ignore ]   [ # 28 ]
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So: We start with a 1 in 3 chance of the prize being in box B and a 2 in 3 chance of the prize being either in box A or C. Once box C is revealed to be empty there is now a 1 in 3 chance of the prize being in Box B as before and a 2 in three chance of the prize being in box A

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 Posted: 09 January 2012 04:29 AM [ Ignore ]   [ # 29 ]
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Well,
———->i reread the original problem as posed and information given. Nowhere did I read 3 envelopes. The flaw was that it was a false problem, a trick question.

[ Edited: 09 January 2012 04:32 AM by Write4U ]
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 Posted: 09 January 2012 06:23 AM [ Ignore ]   [ # 30 ]
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StephenLawrence - 09 January 2012 03:39 AM

So: We start with a 1 in 3 chance of the prize being in box B and a 2 in 3 chance of the prize being either in box A or C. Once box C is revealed to be empty there is now a 1 in 3 chance of the prize being in Box B as before and a 2 in three chance of the prize being in box A

Yep.

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