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The two envelopes problem
 Posted: 20 January 2012 02:04 PM [ Ignore ]   [ # 151 ]
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kkwan - 20 January 2012 06:48 AM

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

`
kkwan, when there’s no information about either envelope and we only have one choice, there ISN’T any “expected value” because we have no information from which to glean any ‘expectation’.

When you have 2 envelopes in front of you, know nothing about their contents, and have only one choice to make, there ISN’T any justification for the assertion “so we should switch”.

`

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 Posted: 20 January 2012 02:08 PM [ Ignore ]   [ # 152 ]
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domokato - 20 January 2012 10:44 AM

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.

1

No information with which to glean probability means that any suggestion of “expected value” or “should switch” are groundless.

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 Posted: 20 January 2012 07:51 PM [ Ignore ]   [ # 153 ]
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domokato - 20 January 2012 10:46 AM

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 Posted: 20 January 2012 08:01 PM [ Ignore ]   [ # 154 ]
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domokato - 20 January 2012 10:51 AM

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.

The expected value was not calculated using impossible values. In fact, it was calculated using possible values as:

Thus, step 7 is impeccably valid and your objection is not justified.

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 Posted: 20 January 2012 09:07 PM [ Ignore ]   [ # 155 ]
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Axegrrl - 20 January 2012 02:00 PM

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 \$.

What I wrote must be taken in the context of post 130 where I was comparing the TEP to the Monty Hall problem i.e in comparing initial selection probabilities. It should not be taken to mean that the solution is simple.

The probabilities are the same in the initial selection process where we select one envelope randomly as the two envelopes are identical.

However, we donâ€™t know whether the envelope we have selected has the higher or the lower amount.  Hence, we are in a quandary. This is intriguing because we have selected an envelope with an unknown amount and comparative value wrt what is in the other envelope.

After the selection process comes the switching process (whereby the offer to switch is considered) and there is no such guide to switch or not to switch as in the Monty Hall problem.

This is the second quandary. It is also intriguing because the expected value (as a guide) leads to the logical absurdity to switch ad infinitum.

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 Posted: 20 January 2012 09:23 PM [ Ignore ]   [ # 156 ]
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Axegrrl - 20 January 2012 02:04 PM

kkwan, when there’s no information about either envelope and we only have one choice, there ISN’T any “expected value” because we have no information from which to glean any ‘expectation’.

When you have 2 envelopes in front of you, know nothing about their contents, and have only one choice to make, there ISN’T any justification for the assertion “so we should switch”.

There is the information that in one of the envelopes there is twice the amount of the other.

With that information, only two identical envelopes and A as the amount in the selected envelope, we can calculate the expected value as in step 7:

And:

8. This is greater than A, so I gain on average by swapping.

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 Posted: 20 January 2012 09:55 PM [ Ignore ]   [ # 157 ]
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Buit you overlook that the selected envelope could contain 2A, in which case your expected value is reversed and yields less on average by swapping.

Assuming that the first envelope contains A is a fallacy of assigned value.

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 Posted: 20 January 2012 11:02 PM [ Ignore ]   [ # 158 ]
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kkwan - 20 January 2012 09:20 AM
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

The first answer is that what you can gain or lose is the difference between the two envelopes with equal chance of gaining or losing.

The next step is to explain what’s wrong with relating that to what’s in your envelope.

My solution is that really the position is equivelent to the starting postiion.

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?

By asking If I were to pick A would that be the best selection. The answer is no because there is double or half A in the other envelope with a 50/50 probability, so B is best.

It works just as well from the start.

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 Posted: 21 January 2012 12:50 AM [ Ignore ]   [ # 159 ]
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kkwan - 20 January 2012 09:07 PM

Hence, we are in a quandary. This is intriguing because…...

`
Yet again, you lose me at the ‘intrigue’ part ;)

What’s intriguing about recognizing that we have know way of knowing what the ‘expected value’ is given the amount of (or lack of) information provided?

What’s intriguing about recognizing that 50/50 chances means that’s there’s no way of discerning whether or not we should switch?

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 Posted: 21 January 2012 12:53 AM [ Ignore ]   [ # 160 ]
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`
Someone on another board looked at this and said:

this is a misapplication of expected value i think. it’s not a continuous function in this case.

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 Posted: 21 January 2012 01:01 AM [ Ignore ]   [ # 161 ]
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`
Someone else commented:

the problem is in the incorrect use of “expected value.”  Another way of looking at is that the expected value is not the value that one would expect is in the other envelope, it’s simply a fancy statistical term for the average of the amounts in the two envelopes.

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 Posted: 21 January 2012 03:34 AM [ Ignore ]   [ # 162 ]
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Another try. The text as kkwan quoted (bold by me):

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.

Assume one envelope has A dollars in it, the other 2A.

Now let’s change the procedure a little: you pick an envelope, and you show the contents to me: it is A dollars. Now I take the other envelope, retract me from your view, and toss a coin. Dependent on the result I put A/2 dollars into it, or 2A dollars. Now you can loose A/2 to take the other envelope, or gain A. Now your formula applies. But the difference with the original formulation is clear: the total amount is fixed, it is 3A. In the second it is variable: it is minimal 1.5A (you took the envelope with A, and I tossed for 0.5A), maximal 6A (you took the envelope with 2A and then I tossed for doubling it, so 6A).

It shows the same fact again and again: in your example the A’s are not the same, and you cannot combine them in the way you do.

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 Posted: 21 January 2012 11:34 AM [ Ignore ]   [ # 163 ]
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Write4U - 20 January 2012 09:55 PM

Buit you overlook that the selected envelope could contain 2A, in which case your expected value is reversed and yields less on average by swapping.

Assuming that the first envelope contains A is a fallacy of assigned value.

In algebra, an unknown quantity is always denoted as A or X, not 2A or 2X, which presupposes a numerical value of the unknown quantity, a contradiction.

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 Posted: 21 January 2012 11:48 AM [ Ignore ]   [ # 164 ]
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StephenLawrence - 20 January 2012 11:02 PM

The first answer is that what you can gain or lose is the difference between the two envelopes with equal chance of gaining or losing.

The next step is to explain what’s wrong with relating that to what’s in your envelope.

My solution is that really the position is equivelent to the starting postiion.

Gain or loss must be computed in the context of each situation.

The starting position is not equivalent to the switching position wrt selection and circumstance.

By asking If I were to pick A would that be the best selection. The answer is no because there is double or half A in the other envelope with a 50/50 probability, so B is best.

It works just as well from the start.

The same argument will apply vice versa, so A is best.

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 Posted: 21 January 2012 12:00 PM [ Ignore ]   [ # 165 ]
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Axegrrl - 21 January 2012 12:50 AM

What’s intriguing about recognizing that we have know way of knowing what the ‘expected value’ is given the amount of (or lack of) information provided?

What’s intriguing about recognizing that 50/50 chances means that’s there’s no way of discerning whether or not we should switch?

The expected value can be calculated with the information provided. It is intriguing because the expected value (as a guide) leads to the logical absurdity to switch ad infinitum.

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