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The two envelopes problem
Posted: 09 January 2012 09:06 AM   [ Ignore ]   [ # 31 ]
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Write4U - 08 January 2012 10:02 PM

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%.

From the wiki on odds

In probability theory and statistics, where the variable p is the probability in favor of the event, and the probability against the event is therefore 1-p, “the odds” of the event are the quotient of the two, or

4f17f0e6a236a293c7b31b98265ee875.png

And:

While the mathematical probability of an event has a value in the range from zero to one, “the odds” in favor of that same event lie between zero and infinity.

So, for the 2 envelopes, the probabilities are 1/2 and 1-1/2=1/2, and the odds in favor or not in favor of an event are both:

1/2 divided by 1-1/2, which is 1.

Knowing the probabilities (1/2) and the odds (1) gives one no guide to switch or not to switch because the odds for or not for an event are both 1.

What is more useful is the expected value and that is, in step 7:

c0f75c1a69af64e06f77ce0ec051c958.png

Since A is the amount in my selected envelope and the expected value is greater than A, I gain on average by switching.

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Posted: 09 January 2012 09:11 AM   [ Ignore ]   [ # 32 ]
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kkwan - 09 January 2012 09:06 AM

Since A is the amount in my selected envelope and the expected value is greater than A, I gain on average by switching.

Kkwan, what we know is you have a 50/ 50 chance of gaining or losing.

And that what you can gain or lose is the same amount.

So what we know is the sums being used are the wrong sums.

Stephen

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Posted: 09 January 2012 08:11 PM   [ Ignore ]   [ # 33 ]
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StephenLawrence - 09 January 2012 09:11 AM

Kkwan, what we know is you have a 50/ 50 chance of gaining or losing.

This is the same as saying that the odds (in favor or not in favor are both 1), i.e. the odds are even.

And that what you can gain or lose is the same amount.

Yes, but wrt two different situations.

1. Select A=1, switch and gain A (100%)

2. Select A=2, switch and lose A/2 (50%)

It is not the same amount, but the comparison of the percentage gain/loss in each situation (the relative gain verses the relative loss) which matters.

Thus, in situations 1/2, by switching gain/loss is 100%/50%.

OTOH, if the gain/loss is 100%/100%, then one should not switch.

So what we know is the sums being used are the wrong sums.

Is it so? Analyze steps 1 to 7 carefully.

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Posted: 09 January 2012 09:35 PM   [ Ignore ]   [ # 34 ]
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I still believe that the sums contained in the envelopes are irrelevant

It is a Schrodinger’s cat problem.

Unopened, A = 1 and 2
unopened, B = 1 and 2

If A = 1 then B = 2
If A = 2 then B = 1

Only on opening A or B will the actual amount become materialized. The odds for the actual sum in either envelope is 50/50
The actual amounts contained in either envelope can not be determined, until they are opened. Take your pick and hope for a little luck.

Einstein was stumped by this very problem in regard to entanglement. He came up with a solution of a matched pair of gloves. He posited that if box 1 had a left hand glove, box 2 at a great distance had to have the right hand glove. But he was proven wrong.

Come to think of it, maybe I’m wrong with my equations… cheese

[ Edited: 09 January 2012 09:44 PM by Write4U ]
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Posted: 09 January 2012 11:04 PM   [ Ignore ]   [ # 35 ]
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kkwan - 09 January 2012 08:11 PM
StephenLawrence - 09 January 2012 09:11 AM

Kkwan, what we know is you have a 50/ 50 chance of gaining or losing.

This is the same as saying that the odds (in favor or not in favor are both 1), i.e. the odds are even.

And that what you can gain or lose is the same amount.

Yes, but wrt two different situations.

It’s irrational to take that into account.

That’s the solution.

Stephen

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Posted: 09 January 2012 11:22 PM   [ Ignore ]   [ # 36 ]
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Write4U - 09 January 2012 09:35 PM

It is a Schrodinger’s cat problem.

It isn’t. The money is already in both envelopes.

Write4U - 09 January 2012 09:35 PM

Einstein was stumped by this very problem in regard to entanglement. He came up with a solution of a matched pair of gloves. He posited that if box 1 had a left hand glove, box 2 at a great distance had to have the right hand glove. But he was proven wrong.

Was he? Check your sources…

Don’t state anything you are not sure of. Know the limits of your knowledge.

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Posted: 09 January 2012 11:28 PM   [ Ignore ]   [ # 37 ]
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KKwan,

I think I see the problem?

You’re viewing it as similar to this:

I have £10

On the toss of a coin I’m offered double or half so I should do it.

In the this case double or half is two different amounts £5 or £ 10 so I can gain more than I can lose.

In your example they are not different amounts so I cannot gain more than I can lose.

You’re applying the logic of double or half one figure, to double or half two different figures.

As the two different figures are themselves double or half compared to each other, this cancels out.

Stephen

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Posted: 09 January 2012 11:31 PM   [ Ignore ]   [ # 38 ]
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Write4U - 09 January 2012 09:35 PM

Only on opening A or B will the actual amount become materialized.

Are you being serious?

Stephen

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Posted: 09 January 2012 11:40 PM   [ Ignore ]   [ # 39 ]
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StephenLawrence - 09 January 2012 11:31 PM
Write4U - 09 January 2012 09:35 PM

Only on opening A or B will the actual amount become materialized.

Are you being serious?

Stephen

Schrodinger was when saying the cat was both dead and alive in the closed box. Does it matter how many boxes with cats there are?
I see little difference when I don’t know which envelope contains 10 or 20. Before opening one envelope is both A and B.

[ Edited: 09 January 2012 11:44 PM by Write4U ]
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Posted: 10 January 2012 12:36 AM   [ Ignore ]   [ # 40 ]
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GdB - 09 January 2012 11:22 PM
Write4U - 09 January 2012 09:35 PM

It is a Schrodinger’s cat problem.

It isn’t. The money is already in both envelopes.

Write4U - 09 January 2012 09:35 PM

Einstein was stumped by this very problem in regard to entanglement. He came up with a solution of a matched pair of gloves. He posited that if box 1 had a left hand glove, box 2 at a great distance had to have the right hand glove. But he was proven wrong.

Was he? Check your sources…

Don’t state anything you are not sure of. Know the limits of your knowledge.

Sorry, I did not state it correctly.
I cited this only in context of entanglement v quantum mechanics. But I see underlying similarities vis a vis the envelope problem and schrodinger’s cat.

Wiki,

The Einstein Podolsky Rosen paradox

Here is the paradox summed up:
It is one thing to say that physical measurement of the first particle’s momentum affects uncertainty in its own position, but to say that measuring the first particle’s momentum affects the uncertainty in the position of the other is another thing altogether. Einstein, Podolsky and Rosen asked how can the second particle “know” to have precisely defined momentum but uncertain position? Since this implies that one particle is communicating with the other instantaneously across space, i.e. faster than light, this is the “paradox”.

edited from original quote.

[ Edited: 10 January 2012 01:01 AM by Write4U ]
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Posted: 10 January 2012 01:38 AM   [ Ignore ]   [ # 41 ]
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Write4U - 10 January 2012 12:36 AM

But I see underlying similarities vis a vis the envelope problem and schrodinger’s cat.

The only similarity that you do not know what is in the box or envelope. But in the topic discussed here there is no quantum process involved, which is the essence of Schrödinger’s cat.

Einstein was not wrong about the consequence of QM: that when there is a left hand glove in box A there will be a right hand glove in Box B. This conclusion was absurd for him. Only in the last statement he was wrong. Quantum reality is absurd, experimentally proven.

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Posted: 10 January 2012 02:15 AM   [ Ignore ]   [ # 42 ]
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GdB - 10 January 2012 01:38 AM
Write4U - 10 January 2012 12:36 AM

But I see underlying similarities vis a vis the envelope problem and schrodinger’s cat.

The only similarity that you do not know what is in the box or envelope. But in the topic discussed here there is no quantum process involved, which is the essence of Schrödinger’s cat.

Einstein was not wrong about the consequence of QM: that when there is a left hand glove in box A there will be a right hand glove in Box B. This conclusion was absurd for him. Only in the last statement he was wrong. Quantum reality is absurd, experimentally proven.

ok, but still, in the envelope problem and the mathematical equations cited.
When there is an unopened envelope which might contain either 10 or 20, is that not the same as Schrodingers unopened box in which the cat may or may not be dead.  The theory states that the cat may be considered both dead and alive. Why not an unopened envelope theoretically containing both 10 and 20, of which only one emerges in reality?

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Posted: 10 January 2012 02:18 AM   [ Ignore ]   [ # 43 ]
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Write, maybe you should stop writing…

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Posted: 10 January 2012 02:33 AM   [ Ignore ]   [ # 44 ]
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GdB - 10 January 2012 02:18 AM

Write, maybe you should stop writing…

Maybe you should stop treating this ridiculous envelope paradox so seriously.
I was having a little fun with it.

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Posted: 10 January 2012 07:07 AM   [ Ignore ]   [ # 45 ]
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Write4U - 09 January 2012 09:35 PM

I still believe that the sums contained in the envelopes are irrelevant

A can be any amount in the envelope that you selected.

It is a Schrodinger’s cat problem.

Not quite so, though from the wiki:

History of the paradox

It is also mentioned in a 1953 book on elementary mathematics and mathematical puzzles by the mathematician John Edensor Littlewood, who credited it to the physicist Erwin Schroedinger.

smile

Come to think of it, maybe I’m wrong with my equations… cheese

With A and B it becomes more convoluted and you get more confused.  cheese

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