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The two envelopes problem
 Posted: 13 August 2012 12:41 PM [ Ignore ]   [ # 1441 ]
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kkwan - 13 August 2012 08:31 AM
GdB - 13 August 2012 07:35 AM

Now following Write4U, there are 15 and 30 dollar cents in the envelopes. What is (X,2X) and (X,X/2)?

(X, 2X) is (15, 30) and (X, 1/2X) is (15, 15/2) in dollars/cents.

But that only works in England where they do have half pence…...

Actually you have answered your own question by saying “there are 15 and 30 dollar cents in the envelopes”. If that is the case then where does the envelope with 7 1/2 cents come from?
Just because I don’t “know” does not change the actual amounts in the envelopes, i.e. X, 2X (where X is the smaller amount), or X, 1/2 X (where X is the larger amount). In either case the amounts remain at 15, 30.

[ Edited: 13 August 2012 03:41 PM by Write4U ]
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 Posted: 13 August 2012 10:13 PM [ Ignore ]   [ # 1442 ]
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kkwan - 13 August 2012 08:11 AM

It is true if and only if you know X and 2X are actually in the two envelopes.

Right.

Just for completeness, I repeat the statements here:

I have two envelopes with amounts X and 2X:

If I have the envelope with X, then the other one contains 2X.
If I have the envelope with 2X, then the other one contains X.

Now is the following statement true about these envelopes?

One envelope contains twice as much as the other.

Yes or no?

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 Posted: 13 August 2012 10:20 PM [ Ignore ]   [ # 1443 ]
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Write4U - 13 August 2012 12:41 PM

But that only works in England where they do have half pence…...

kkwan will probably introduce quarter pence…

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 Posted: 20 August 2012 03:30 AM [ Ignore ]   [ # 1444 ]
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Write4U - 13 August 2012 12:41 PM
kkwan - 13 August 2012 08:31 AM
GdB - 13 August 2012 07:35 AM

Now following Write4U, there are 15 and 30 dollar cents in the envelopes. What is (X,2X) and (X,X/2)?

(X, 2X) is (15, 30) and (X, 1/2X) is (15, 15/2) in dollars/cents.

But that only works in England where they do have half pence…...

If the amounts are expressed in dollars and cents, then 15/2 is \$7.50.

Actually you have answered your own question by saying “there are 15 and 30 dollar cents in the envelopes”. If that is the case then where does the envelope with 7 1/2 cents come from?
Just because I don’t “know” does not change the actual amounts in the envelopes, i.e. X, 2X (where X is the smaller amount), or X, 1/2 X (where X is the larger amount). In either case the amounts remain at 15, 30.

It is ridiculous to insist on cents only wrt to 15 and 30 (with one cent is the minimum denomination) as we know currency is discrete. There is no issue wrt the amounts in dollars and cents as stated above.

Because you don’t know, you cannot specifically say whether (X, 2X) or (X, 1/2X) are in the two envelopes i.e. either (15, 30) or (15, 15/2) which are not equivalent in dollars and cents.

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 Posted: 20 August 2012 04:03 AM [ Ignore ]   [ # 1445 ]
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GdB - 13 August 2012 10:13 PM

I have two envelopes with amounts X and 2X:

If I have the envelope with X, then the other one contains 2X.
If I have the envelope with 2X, then the other one contains X.

Now is the following statement true about these envelopes?

One envelope contains twice as much as the other.

Yes or no?

Yes, if and only if, you know those amounts are actually in the two envelopes.

However, in the TEP, you don’t know as it is also possible:

There are two envelopes with (X, 1/2X) which also satisfy “one envelope contains twice as much as the other”

Hence, if you select an envelope and denote the amount in it as A:

1. If A is the smaller amount, the other envelope contains 2A.

2. If A is the larger amount, the other envelope contains 1/2A.

Agreed?

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 Posted: 20 August 2012 04:09 AM [ Ignore ]   [ # 1446 ]
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kkwan - 20 August 2012 04:03 AM
GdB - 13 August 2012 10:13 PM

I have two envelopes with amounts X and 2X:

If I have the envelope with X, then the other one contains 2X.
If I have the envelope with 2X, then the other one contains X.

Now is the following statement true about these envelopes?

One envelope contains twice as much as the other.

Yes or no?

Yes, if and only if, you know those amounts are actually in the two envelopes.

One envelope contains twice as much as the other.

Fits the situation:
I have two envelopes with amounts X and 2X.

The rest is fantasy you add to it.

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 Posted: 25 August 2012 12:30 AM [ Ignore ]   [ # 1447 ]
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GdB - 20 August 2012 04:09 AM

One envelope contains twice as much as the other.

Fits the situation:
I have two envelopes with amounts X and 2X.

The rest is fantasy you add to it.

But, X and 1/2X also fit the description:

One envelope contains twice as much as the other.

So, it is not uniquely X and 2X.

There are two possible distinct events (X, 2X) or (X, 1/2X), not one, for any finite value of X.

Are you denying (X, 1/2X) as a possible distinct event?

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 Posted: 25 August 2012 02:37 AM [ Ignore ]   [ # 1448 ]
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kkwan - 25 August 2012 12:30 AM

Are you denying (X, 1/2X) as a possible distinct event?

Yes. It is just another description, given that you describe the 2 fixed amounts in the envelopes.

There are only 2 amounts, and by describing them as (X,2X) or (X,X/2) makes no difference. Of course not for the same X. If you take the same X you are describing 3 amounts, and there are just 2.

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 Posted: 27 August 2012 07:35 AM [ Ignore ]   [ # 1449 ]
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GdB - 25 August 2012 02:37 AM

Yes. It is just another description, given that you describe the 2 fixed amounts in the envelopes.

There are only 2 amounts, and by describing them as (X,2X) or (X,X/2) makes no difference. Of course not for the same X. If you take the same X you are describing 3 amounts, and there are just 2.

It is not just another description as (X, 2X) and (X, 1/2X) are not equivalent.

(X, 2X) or (X, 1/2X) are two distinct possible events for any finite value of X.

This is because you have no definite information wrt whether (X, 2X) or (X, 1/2X) are in the two envelopes.

Hence, either (X, 2X) or (X, 1/2X) are in the two envelopes as both events satisfy the requirement “one envelope contains twice the amount in the other”.

The issue of 3 amounts is not relevant as there are only 2 amounts in each possible mutually exclusive event.

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 Posted: 27 August 2012 10:38 PM [ Ignore ]   [ # 1450 ]
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2 Amounts, one relationship.

(X,2X) and (X,X/2) represent the same relationship.

And as X is an unknown (you have no idea which amounts really are in the envelopes ((1,2) or (1234,2468))), (X,2X) and (X,X/2) are both correct descriptions of the relationship between the same 2 amounts.

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 Posted: 01 September 2012 07:14 AM [ Ignore ]   [ # 1451 ]
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GdB - 27 August 2012 10:38 PM

2 Amounts, one relationship.

(X,2X) and (X,X/2) represent the same relationship.

And as X is an unknown (you have no idea which amounts really are in the envelopes ((1,2) or (1234,2468))), (X,2X) and (X,X/2) are both correct descriptions of the relationship between the same 2 amounts.

Not quite so as you expressed it wrt the “same 2 amounts”.

(X, 2X) and (X, 1/2X) relate to 2 different distinct possible events for any finite value of X.

For instance, if X is 10, then either (10, 20) or (10, 5) are in the two envelopes.

Generally, for any finite value of X, there are 2 distinct possible events (X, 2X) or (X, 1/2X), not one.

If you only consider one possibility, then your description of the TEP is incomplete and misleading.

You can only consider one possibility if and only if you know what are actually in the two envelopes by acquaintance but, in the context of the TEP, you don’t.

Therefore, both possibilities must be considered as either one is a possible event.

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 Posted: 01 September 2012 07:59 AM [ Ignore ]   [ # 1452 ]
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kkwan - 01 September 2012 07:14 AM

(X, 2X) and (X, 1/2X) relate to 2 different distinct possible events for any finite value of X.

For instance, if X is 10, then either (10, 20) or (10, 5) are in the two envelopes.

Generally, for any finite value of X, there are 2 distinct possible events (X, 2X) or (X, 1/2X), not one.

If you only consider one possibility, then your description of the TEP is incomplete and misleading.

No. There are only 2 amounts, so you only must consider one possibility.

kkwan - 01 September 2012 07:14 AM

You can only consider one possibility if and only if you know what are actually in the two envelopes by acquaintance but, in the context of the TEP, you don’t.

You know that one of the amounts is twice the other one amount. You already said that (X,2X) describes that information. That is all you need, that covers all possible TEP pairs: every TEP pair can be described by (X,2X).

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 Posted: 01 September 2012 08:38 AM [ Ignore ]   [ # 1453 ]
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GdB - 01 September 2012 07:59 AM

No. There are only 2 amounts, so you only must consider one possibility.

There are 2 amounts in each possible event, but there are two distinct possible events. Therefore, you must consider both possibilities as you have no definite knowledge wrt which event is actual.

You know that one of the amounts is twice the other one amount. You already said that (X,2X) describes that information. That is all you need, that covers all possible TEP pairs: every TEP pair can be described by (X,2X).

Not quite so as the information is ambiguous in the context of the TEP.

Both (X, 2X) or (X, 1/2X) fit the information that “one envelope contains twice the amount in the other” and they are not equivalent for any finite value of X.

As such, both possible events must be considered for a complete description of the TEP.

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 Posted: 01 September 2012 09:20 AM [ Ignore ]   [ # 1454 ]
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kkwan - 01 September 2012 08:38 AM
GdB - 01 September 2012 07:59 AM

No. There are only 2 amounts, so you only must consider one possibility.

There are 2 amounts in each possible event, but there are two distinct possible events. Therefore, you must consider both possibilities as you have no definite knowledge wrt which event is actual..

How can 2 amounts, one being twice the other, be 2 events? If you think about which amounts could be in the envelopes, i.e. fit the description then the answer is endless many. (X,2X) and (X,X/2) describe for any of these amounts the same relationship “being twice the other”. Your idee fixé is that X should be the same. If (5,10) is in the envelopes it is perfectly described by both, just not for the same X. Why must it be for the same X?

It seems you think you have different amounts when you describe them as “one amount is twice the other” or as “one amount is half of the other”. You think both descriptions are not equivalent?

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 Posted: 05 September 2012 07:01 PM [ Ignore ]   [ # 1455 ]
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GdB - 01 September 2012 09:20 AM

How can 2 amounts, one being twice the other, be 2 events? If you think about which amounts could be in the envelopes, i.e. fit the description then the answer is endless many. (X,2X) and (X,X/2) describe for any of these amounts the same relationship “being twice the other”. Your idee fixé is that X should be the same. If (5,10) is in the envelopes it is perfectly described by both, just not for the same X. Why must it be for the same X?

There are 2 amounts in each of the 2 possible events. You are conflating amounts with events.

It is not so as for any finite value of X (irrespective of the value of X), there are two distinct possible events (X, 2X) or (X, 1/2X) which satisfy the relationship “being twice the other”.

It seems you think you have different amounts when you describe them as “one amount is twice the other” or as “one amount is half of the other”. You think both descriptions are not equivalent?

Notwithstanding that both (X, 2X) and (X, 1/2X) can be described as such does not mean they are equivalent for any finite value of X. They are distinct possible events.

Equivalent descriptions does not imply equivalent values.

For instance, if X is 100, then (100, 200) is not equivalent to (100, 50).

In general, for any finite value of X, (X, 2X) and (X, 1/2X) are not equivalent i.e. they are distinct possible events.

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