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The two envelopes problem
 Posted: 27 October 2012 07:34 AM [ Ignore ]   [ # 1531 ]
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GdB - 27 October 2012 06:58 AM

So according to you the total amount changes dependent on the choice of the first envelope.

The total amount does not change but it could be either 3X or 3/2X as either (X, 2X) or (X, 1/2X) are in the two envelopes (which is not dependent on the selection of any envelope).

So you say by switching envelopes the two player’s gain and loss are not the same.

The potential gain/loss is 100%/50% on switching once or any odd number of times from the respective perspective of each player:

Player A

He denotes the amount in the selected envelope as A. As A could be the smaller or the larger amount, the amounts in the two envelopes are either (A, 2A) or (A, 1/2A). On switching, the potential gain/loss is A/1/2A i.e. 100%/50%.

Player B

He denotes the amount in the selected envelope as B. As B could be the smaller or the larger amount, the amounts in the two envelopes are either (B, 2B) or (B, 1/2B). On switching, the potential gain/loss is B/1/2B i.e. 100%/50%.

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 Posted: 28 October 2012 03:49 AM [ Ignore ]   [ # 1532 ]
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kkwan - 27 October 2012 07:34 AM

Player A

He denotes the amount in the selected envelope as A. As A could be the smaller or the larger amount, the amounts in the two envelopes are either (A, 2A) or (A, 1/2A). On switching, the potential gain/loss is A/1/2A i.e. 100%/50%.

Player B

He denotes the amount in the selected envelope as B. As B could be the smaller or the larger amount, the amounts in the two envelopes are either (B, 2B) or (B, 1/2B). On switching, the potential gain/loss is B/1/2B i.e. 100%/50%.

:-)

Kkwan, it can’t be best for both player A and player B to switch because if A get’s an advantage from it then B get’s the same disadvantage.

This is one way we know the formula doesn’t work.

It seems perverse to deny it.

Stephen

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 Posted: 28 October 2012 04:00 AM [ Ignore ]   [ # 1533 ]
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kkwan - 26 October 2012 08:52 PM
StephenLawrence - 24 October 2012 11:04 PM

You’ve made this bit up. The formula doesn’t say that at all. It’s just you’ve reasoned that if you switch twice you are back to square 1.

Switching twice or any even number of times is equivalent to not switching at all.

OK, that’s right so the formula does say don’t switch twice.

But the problem is it also says switch twice.

So this is one of the contradicitions which tells us there is something wrong with the formula.

Stephen

[ Edited: 28 October 2012 04:28 AM by StephenLawrence ]
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 Posted: 28 October 2012 07:23 AM [ Ignore ]   [ # 1534 ]
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kkwan - 27 October 2012 07:34 AM
GdB - 27 October 2012 06:58 AM

So according to you the total amount changes dependent on the choice of the first envelope.

The total amount does not change but it could be either 3X or 3/2X as either (X, 2X) or (X, 1/2X) are in the two envelopes (which is not dependent on the selection of any envelope).

It is, kkwan. That’s just simple what your formula says. If X is the smallest amount then you say the total amount is 3X, if it is the biggest amount the total amount is 3X/2. That means that exchanging the envelopes determines what the total amount of money is. That is absurd.

kkwan - 27 October 2012 07:34 AM

Player A

He denotes the amount in the selected envelope as A. As A could be the smaller or the larger amount, the amounts in the two envelopes are either (A, 2A) or (A, 1/2A). On switching, the potential gain/loss is A/1/2A i.e. 100%/50%.

Player B

He denotes the amount in the selected envelope as B. As B could be the smaller or the larger amount, the amounts in the two envelopes are either (B, 2B) or (B, 1/2B). On switching, the potential gain/loss is B/1/2B i.e. 100%/50%.

Stephen is completely right.

Your trick is wiping knowledge you have under the carpet. You do this again and again. In the argument with the total amount you do not count for the fact that exchanging the envelopes has no influence on the total amount. If you think you count for that, then show me where you do.

With your player A and B here you do the same. You do not account for the fact that the loss of player A is the gain of B, and the other way round: if A has the biggest amount, then B has the smallest, or the other way round. You treat the views of A and B as independent, but they are not. You do not count for the fact that once the amounts are in the envelopes, they do not change.

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 Posted: 29 October 2012 12:45 AM [ Ignore ]   [ # 1535 ]
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I think the situation is this:

If we do the puzzle for any two numbers we get the correct answer that there is nothing to be gained or lost by switching, so don’t switch.

But we can also compare two different situations we might be in, as Kkwan does and the question remains, if we do so why do we seem to get the answer that we should switch?

One might answer “that just isn’t how to do it” but that’s not obviously so, it looks like this other way should work as well and this illusion is particularly strong when we open our envelope.

The point of opening the envelope is it seems it makes no difference.

So I think the solution to the puzzle should:

A) Show that opening the envelope does make a difference.

Or

B) Be a formula that gets us exactly the same result as the result we get when we don’t look.

Both look tricky.

B) looks tricky because it’s hard to escape that if I open the envelope and discover \$20 that I can gain \$20 or lose \$10. So the only thing seems to be to deny that the probability is 50/50. But that brings it’s own problems as it appears that the only reason not to switch would be that the probability was lower than 50/50 and yet how could opening the envelope make the probability of having the smaller amount lower than 50/50?

One might try answering we just don’t know what the probability is and so we don’t have the information to make a judgement. But that doesn’t seem to work because we did have before we opened the envelope, so why not after? :)

It doesn’t seem good enough to say “we dunno” when we open the envelope, so don’t switch, whilst before we opened it we had a formula that showed us that rather than “we dunno” we did know that switching made no different.

Stephen

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 Posted: 29 October 2012 01:54 AM [ Ignore ]   [ # 1536 ]
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StephenLawrence - 29 October 2012 12:45 AM

I think the situation is this:

If we do the puzzle for any two numbers we get the correct answer that there is nothing to be gained or lost by switching, so don’t switch.

But we can also compare two different situations we might be in, as Kkwan does and the question remains, if we do so why do we seem to get the answer that we should switch?

One might answer “that just isn’t how to do it” but that’s not obviously so, it looks like this other way should work as well and this illusion is particularly strong when we open our envelope.

The point of opening the envelope is it seems it makes no difference.

So I think the solution to the puzzle should:

A) Show that opening the envelope does make a difference.

Or

B) Be a formula that gets us exactly the same result as the result we get when we don’t look.

Both look tricky.

B) looks tricky because it’s hard to escape that if I open the envelope and discover \$20 that I can gain \$20 or lose \$10. So the only thing seems to be to deny that the probability is 50/50. But that brings it’s own problems as it appears that the only reason not to switch would be that the probability was lower than 50/50 and yet how could opening the envelope make the probability of having the smaller amount lower than 50/50?

One might try answering we just don’t know what the probability is and so we don’t have the information to make a judgement. But that doesn’t seem to work because we did have before we opened the envelope, so why not after? :)

It doesn’t seem good enough to say “we dunno” when we open the envelope, so don’t switch, whilst before we opened it we had a formula that showed us that rather than “we dunno” we did know that switching made no different.

Stephen

It is a mental Mathematical “illusion”..... :cheese:

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 Posted: 30 October 2012 09:20 PM [ Ignore ]   [ # 1537 ]
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StephenLawrence - 28 October 2012 03:49 AM

Kkwan, it can’t be best for both player A and player B to switch because if A get’s an advantage from it then B get’s the same disadvantage.

This is one way we know the formula doesn’t work.

It seems perverse to deny it.

Not necessarily so.

Because of the symmetry of the TEP with both players not knowing whether they have the smaller or the larger amount, both should switch for the potential gain/loss of 100%/50%.

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 Posted: 30 October 2012 09:29 PM [ Ignore ]   [ # 1538 ]
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StephenLawrence - 28 October 2012 04:00 AM

OK, that’s right so the formula does say don’t switch twice.

But the problem is it also says switch twice.

So this is one of the contradicitions which tells us there is something wrong with the formula.

Not necessarily so, as 10 only recommends switching and we know switching twice is equivalent to not switching.

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 Posted: 30 October 2012 10:01 PM [ Ignore ]   [ # 1539 ]
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GdB - 28 October 2012 07:23 AM

It is, kkwan. That’s just simple what your formula says. If X is the smallest amount then you say the total amount is 3X, if it is the biggest amount the total amount is 3X/2. That means that exchanging the envelopes determines what the total amount of money is. That is absurd.

It is not absurd as either (X, 2X) or (X, 1/2X) are in the two envelopes. This is not dependent on whether any envelope is selected.

However, if an envelope is selected, the amount in it is denoted as A and as A could be either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes which is consistent with the above.

Your trick is wiping knowledge you have under the carpet. You do this again and again. In the argument with the total amount you do not count for the fact that exchanging the envelopes has no influence on the total amount. If you think you count for that, then show me where you do.

Of course, the total amount does not change but we don’t know what is the total amount. It could be either (X, 2X) or (X, 1/2X) are in the two envelopes. Thus, the total amounts could be either 3X or 3/2X.

With your player A and B here you do the same. You do not account for the fact that the loss of player A is the gain of B, and the other way round: if A has the biggest amount, then B has the smallest, or the other way round. You treat the views of A and B as independent, but they are not. You do not count for the fact that once the amounts are in the envelopes, they do not change.

Neither player knows whether their selected envelope has the smaller or the larger amount and as such, the potential gain/loss for both players on switching is 100%/50%.

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 Posted: 30 October 2012 10:44 PM [ Ignore ]   [ # 1540 ]
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StephenLawrence - 29 October 2012 12:45 AM

I think the situation is this:
If we do the puzzle for any two numbers we get the correct answer that there is nothing to be gained or lost by switching, so don’t switch.

Of course, if we assume (X, 2X) are in the two envelopes (which is not justified because unless we know by looking into the two envelopes which we are not allowed to in the TEP), it is obvious that if an envelope is selected, the amount in it is either X or 2X.

So, on switching, gain or loss is X.

That is tantamount to reducing/simplifying the TEP to non TEP.  :cheese:

But we can also compare two different situations we might be in, as Kkwan does and the question remains, if we do so why do we seem to get the answer that we should switch?

However, if either (X, 2X) or (X, 1/2X) are in the two envelopes, the TEP is more complex.

Now, if an envelope is selected and the amount in it is denoted as A and as A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

So, on switching, gain is A or loss is 1/2A.

This is the TEP.

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 Posted: 31 October 2012 12:04 AM [ Ignore ]   [ # 1541 ]
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kkwan - 30 October 2012 10:01 PM
GdB - 28 October 2012 07:23 AM

It is, kkwan. That’s just simple what your formula says. If X is the smallest amount then you say the total amount is 3X, if it is the biggest amount the total amount is 3X/2. That means that exchanging the envelopes determines what the total amount of money is. That is absurd.

It is not absurd as either (X, 2X) or (X, 1/2X) are in the two envelopes. This is not dependent on whether any envelope is selected.

However, if an envelope is selected, the amount in it is denoted as A and as A could be either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes which is consistent with the above.

Sorry kkwan, that doesn’t do. You have not said one word against my argument, and are just parroting your self.

kkwan - 30 October 2012 10:01 PM

Your trick is wiping knowledge you have under the carpet. You do this again and again. In the argument with the total amount you do not count for the fact that exchanging the envelopes has no influence on the total amount. If you think you count for that, then show me where you do.

Of course, the total amount does not change but we don’t know what is the total amount. It could be either (X, 2X) or (X, 1/2X) are in the two envelopes. Thus, the total amounts could be either 3X or 3/2X.

Again, you do not show where you use the knowledge that the total amount stays the same independent. It is the essential difference between TEP and the situation where you choose an envelope, and then afterwards somebody (knowing which amount you have choosen!) gives you an envelope that has twice or half the amount of your envelope in it. Show me where you use the knowledge that the total amount is the same independent of player’s choice.

kkwan - 30 October 2012 10:01 PM

With your player A and B here you do the same. You do not account for the fact that the loss of player A is the gain of B, and the other way round: if A has the biggest amount, then B has the smallest, or the other way round. You treat the views of A and B as independent, but they are not. You do not count for the fact that once the amounts are in the envelopes, they do not change.

Neither player knows whether their selected envelope has the smaller or the larger amount and as such, the potential gain/loss for both players on switching is 100%/50%.

And what is 100%/50% in money? If there is A has X and B 2X, then 100% of X equals X, and 50% of 2X equals X, so their gain and loss are the same. Are you saying that in terms of amounts (not percentages) loss and gain of A and B are different? (Just for your information: this is a yes/no question.)

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 Posted: 31 October 2012 12:58 AM [ Ignore ]   [ # 1542 ]
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kkwan - 30 October 2012 09:20 PM
StephenLawrence - 28 October 2012 03:49 AM

Kkwan, it can’t be best for both player A and player B to switch because if A get’s an advantage from it then B get’s the same disadvantage.

This is one way we know the formula doesn’t work.

It seems perverse to deny it.

Not necessarily so.

Because of the symmetry of the TEP with both players not knowing whether they have the smaller or the larger amount, both should switch for the potential gain/loss of 100%/50%.

This is impossible. One person’s gain is the others loss in this set up.

Stephen

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 Posted: 31 October 2012 01:02 AM [ Ignore ]   [ # 1543 ]
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kkwan - 30 October 2012 09:29 PM
StephenLawrence - 28 October 2012 04:00 AM

OK, that’s right so the formula does say don’t switch twice.

But the problem is it also says switch twice.

So this is one of the contradicitions which tells us there is something wrong with the formula.

Not necessarily so, as 10 only recommends switching and we know switching twice is equivalent to not switching.

The formula does recommend switching a second time, if you get the chance.

So the problem is it does and it doesn’t.

Stephen

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 Posted: 31 October 2012 01:10 AM [ Ignore ]   [ # 1544 ]
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kkwan - 30 October 2012 10:44 PM
StephenLawrence - 29 October 2012 12:45 AM

I think the situation is this:
If we do the puzzle for any two numbers we get the correct answer that there is nothing to be gained or lost by switching, so don’t switch.

Of course, if we assume (X, 2X) are in the two envelopes (which is not justified because unless we know by looking into the two envelopes which we are not allowed to in the TEP), it is obvious that if an envelope is selected, the amount in it is either X or 2X.

What we can do is stick with a pair of numbers.

We don’t have to consider more than 1 pair of numbers.

You have no reasonable objection to doing that.

Stephen

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 Posted: 04 November 2012 09:57 PM [ Ignore ]   [ # 1545 ]
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GdB - 31 October 2012 12:04 AM

Sorry kkwan, that doesn’t do. You have not said one word against my argument, and are just parroting your self.

Your argument is flawed because it is based on the premise that the total amount has changed which is not so.

Again, you do not show where you use the knowledge that the total amount stays the same independent. It is the essential difference between TEP and the situation where you choose an envelope, and then afterwards somebody (knowing which amount you have choosen!) gives you an envelope that has twice or half the amount of your envelope in it. Show me where you use the knowledge that the total amount is the same independent of player’s choice.

Is it not true that either (X, 2X) or (X, 1/2X) are in the two envelopes irrespective of choice as per the TEP?

Hence, either 3X or 3/X are the total amounts in the two envelopes.

And what is 100%/50% in money? If there is A has X and B 2X, then 100% of X equals X, and 50% of 2X equals X, so their gain and loss are the same. Are you saying that in terms of amounts (not percentages) loss and gain of A and B are different? (Just for your information: this is a yes/no question.)

You cannot determine that (X, 2X) are in the two envelopes because you don’t know. It is either (X, 2X) or (X, 1/2X) are in the two envelopes.

As such, if the selected envelope contains an amount denoted as A and as A could be the smaller or the larger amount, which means either (A, 2A) or (A, 1/2A) are in the two envelopes, on switching, potential gain/loss is A /1/2A.

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