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The two envelopes problem
Posted: 20 January 2012 12:26 AM   [ Ignore ]   [ # 136 ]
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I keep monitoring this thread.  I don’t know why.  It’s like I’m in some sort of twilight zone two envelope purgatory.

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Posted: 20 January 2012 12:49 AM   [ Ignore ]   [ # 137 ]
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TimB - 20 January 2012 12:26 AM

I keep monitoring this thread.  I don’t know why.  It’s like I’m in some sort of twilight zone two envelope purgatory.

1
I know!  why on earth has this thread gone on so long??

This ISN’T like the ‘Monty Hall problem’ (in which you’re given some information when Monty reveals one of the doors), we’re given NO information ~ and when you’re given an either/or choice between 2 unknown things, there isn’t any way to devise a ‘strategy’ to pick the one with the most $ in it.

1

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Posted: 20 January 2012 06:48 AM   [ Ignore ]   [ # 138 ]
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Axegrrl - 19 January 2012 02:28 PM

Now, if this thread were ABOUT the ‘Monty Hall’ problem, it would have been as interesting, AND we all could have agreed and understood that you should always switch for that last door!

Great little exercise in probabilities-that-aren’t-what-you-expect smile

In the Monty Hall problem, because of the higher probability of picking a goat and once a door is opened revealing a goat, it means there is a higher probability of winning a car and thus one should switch.

In the two envelopes problem (TEP), there is no such guide to switch at all because the probabilities of selecting the envelope with the higher or lower amount are the same. Also, we don’t know whether the envelope we have selected has the higher or the lower amount.  Hence, we are in a quandary.

If we do not switch, we will not know whether we have gained or lost anything at all.

OTOH, if we switch, we could gain 100% or lose 50% of what we have.

In the argument to switch, the expected value is 5/4 A, (where A is the amount in the envelope we have selected and it is greater than A), so we should switch. But, if we switched, the same reasoning recommends us to switch again ad infinitum, which is absurd. Expected value is the only guide we have. Generally, it is considered as the fundamental principle in decision theory and it is reliable. However, in the TEP it leads us to perpetual circularity and absurdity. Why is it so?

These aspects of the TEP differentiates it from the Monty Hall problem (which is more clearcut wrt switching) and as it is, there is no agreed consensus wrt switch or not to switch.

This makes the TEP much more interesting, teasing and taxing on our intellects. The solution could be simple, but where is it?

grin

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Posted: 20 January 2012 06:59 AM   [ Ignore ]   [ # 139 ]
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TimB - 20 January 2012 12:26 AM

I keep monitoring this thread.  I don’t know why.  It’s like I’m in some sort of twilight zone two envelope purgatory.

It is mesmerizing, is’nt it.  cheese

Mathematicians, physicists, philosophers, logicians, probability and decision theorists etc. have been cracking their minds/brains over the simple TEP for more than 50 years with no agreed solution.

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Posted: 20 January 2012 07:02 AM   [ Ignore ]   [ # 140 ]
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domokato - 19 January 2012 03:15 PM

Yes, I know what the wiki says, and I disagree with it.

Why do you disagree?

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Posted: 20 January 2012 07:08 AM   [ Ignore ]   [ # 141 ]
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domokato - 19 January 2012 03:20 PM

The expected value may turn out to be an impossible amount, but it still has to be calculated from possible values.

Quite so. However, the expected value is a fundamental principle in decision theory and it is known to be reliable.

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Posted: 20 January 2012 07:27 AM   [ Ignore ]   [ # 142 ]
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Write4U - 19 January 2012 07:08 PM

Regardless of what amount is in the first envelope, you still do not know the EXPECTED value of the second envelope. It is either larger or smaller. 50/50 either way.

That is why it is known as the expected value.

From the wiki

The expected value may be intuitively understood by the law of large numbers: the expected value, when it exists, is almost surely the limit of the sample mean as sample size grows to infinity. More informally, it can be interpreted as the long-run average of the results of many independent repetitions of an experiment (e.g. a dice roll). The value may not be expected in the ordinary sense—the “expected value” itself may be unlikely or even impossible (such as having 2.5 children), just like the sample mean.

Generally, the expected value is known to be reliable if the sample size is large enough.

What is obvious is that the amount the second envelops could be either larger or smaller, but we don’t know which is which even though it is 50/50.

Hence, we have to rely on the calculated expected value as a guide.

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Posted: 20 January 2012 08:25 AM   [ Ignore ]   [ # 143 ]
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domokato - 19 January 2012 06:45 PM

That’s irrelevant. For the first proposed solution you don’t have that information either, so 4 and 5 would be no more right than 4a and 5a by your critique.

4 and 5 must be interpreted together with 1,2, and 3:

  1. I denote by A the amount in my selected envelope.
  2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
  3. The other envelope may contain either 2A or A/2.
  4. If A is the smaller amount the other envelope contains 2A.
  5. If A is the larger amount the other envelope contains A/2.

OTOH, 4a and 5a stand by themselves as:

4a. If you have the envelope with 2A, then the other envelope contains A
5a. If you have the envelope with A, then the other envelope contains 2A

A is defined in 1 as “the amount in my selected envelope”, but in 4a and 5a, the terms 2A and A are considered as fixed amounts whereas in 3 the other envelope may contain either 2A or A/2.

There is a crucial difference in meaning because we don’t know which envelope has what and this is expressed in 3 as may whereas in 4a and 5a it is stated that “if…......“which implies it can be known.

4 and 5 refer to A which is defined, but 4a and 5a refer to different numerical values of A but A is not defined at all. Why not 2 and 1 instead since A is not defined?

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Posted: 20 January 2012 09:00 AM   [ Ignore ]   [ # 144 ]
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GdB - 20 January 2012 12:16 AM

Then tell me which information you have in my formulation that you do not have in the original formulation. I tell you, I have 2 envelopes, one with an amount of A and the other with 2A. I do not even tell you what A is. Now pick an envelope. Do you have the smaller or the bigger amount?

If A is undefined, you might as well just say 1 and 2 instead.

A is defined in 1 as “the amount in the envelope I selected”. That is all there is to it.

Whereas A as defined in 1 has no numerical value per se, your A has.

Because of the numerical value of A in your formulation (which we don’t know at all), the information is already considered as known whereas A as defined in 1 has no such information bias.

Hence, your A is not the same as the A in 1.

LOL

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Posted: 20 January 2012 09:20 AM   [ Ignore ]   [ # 145 ]
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StephenLawrence - 20 January 2012 12:12 AM

We know that it is because we should not relate the gain or loss to what is in your envelope. We just need a bit more explanation.

Why not and what more do you need to determine gain or loss?

No, the question is to switch or not to switch, so two possible outcomes. It’s just the same as the starting position.

Try the puzzle from the starting position. Call the envelopes A and B. I can work out that it’s best to select B:

If I were to select A what should I do if offered the chance to switch? Answer switch to B, therefore B is best.

What’s wrong with that?

Of course, to switch or not to switch has two possible outcomes, but what I wrote was “OTOH, in switching, there is only one outcome”.

OTOH, the starting position is different because it is a selection and not a switching process.

How do you work out that it is best to select B?

And how do you know B is best in switching?

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Posted: 20 January 2012 09:32 AM   [ Ignore ]   [ # 146 ]
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Axegrrl - 20 January 2012 12:49 AM

This ISN’T like the ‘Monty Hall problem’ (in which you’re given some information when Monty reveals one of the doors), we’re given NO information ~ and when you’re given an either/or choice between 2 unknown things, there isn’t any way to devise a ‘strategy’ to pick the one with the most $ in it.

Exactly. We are in a quandary. All we have is the expected value as a guide and it indicates switch and switch perpetually. This is hopeless and futile, like a supertask.

Of course, we could switch X times and call it quits.

LOL

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Posted: 20 January 2012 10:44 AM   [ Ignore ]   [ # 147 ]
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Axegrrl - 20 January 2012 12:49 AM
TimB - 20 January 2012 12:26 AM

I keep monitoring this thread.  I don’t know why.  It’s like I’m in some sort of twilight zone two envelope purgatory.

1
I know!  why on earth has this thread gone on so long??

This ISN’T like the ‘Monty Hall problem’ (in which you’re given some information when Monty reveals one of the doors), we’re given NO information ~ and when you’re given an either/or choice between 2 unknown things, there isn’t any way to devise a ‘strategy’ to pick the one with the most $ in it.

1

The problem isn’t to devise a strategy to pick the best envelope; the problem is figuring out what is wrong with the proposed strategy.

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Posted: 20 January 2012 10:46 AM   [ Ignore ]   [ # 148 ]
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kkwan - 20 January 2012 07:02 AM
domokato - 19 January 2012 03:15 PM

Yes, I know what the wiki says, and I disagree with it.

Why do you disagree?

I’ve already explained that. Read pages 7-8 or so

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Posted: 20 January 2012 10:51 AM   [ Ignore ]   [ # 149 ]
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kkwan - 20 January 2012 07:08 AM
domokato - 19 January 2012 03:20 PM

The expected value may turn out to be an impossible amount, but it still has to be calculated from possible values.

Quite so. However, the expected value is a fundamental principle in decision theory and it is known to be reliable.

kkwan…what are you talking about? My point was that step 7 was calculating expected value using impossible values, not that expected value is not a reliable principle in decision theory. And since it is using impossible values, it is an invalid expected value calculation. Why was it posited using impossible values in the first place? Because of steps 4-6, and I’ve already explained what’s wrong with them.

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Posted: 20 January 2012 02:00 PM   [ Ignore ]   [ # 150 ]
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kkwan - 20 January 2012 06:48 AM

The solution could be simple, but where is it?

`
You offered it yourself:

“In the two envelopes problem (TEP), there is no such guide to switch at all because the probabilities of selecting the envelope with the higher or lower amount are the same”

If one is given one choice, there’s no way of discerning or ‘strategizing’ how to pick the one with the most $.

Beyond that, i’m not really getting what’s so ‘intriguing’ about this ‘quandary’.

`

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