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The two envelopes problem
 Posted: 10 January 2012 10:32 PM [ Ignore ]   [ # 61 ]
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kkwan - 10 January 2012 09:26 PM
TimB - 10 January 2012 08:31 PM

What am I? Stephen Hawkings?  Give me an envelope.  That’s my envelope and I’m stickin to it.

Good strategy, but are you not interested to get double the amount by switching?

No, K"Monty Hall"kwan, I’m keeping envelope #1.

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 Posted: 10 January 2012 10:50 PM [ Ignore ]   [ # 62 ]
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kkwan - 10 January 2012 09:18 PM
StephenLawrence - 10 January 2012 11:44 AM

100% of £10

Yes, by fixing A= £10 as a known amount, but we don’t know the value of A…....is it the lower or higher amount and what if you switch? You will never know because you selected an envelope randomly and did not switch.

50%  of £20

Again, only if A is a fixed known value but it is not known.

So what that it is not know? What difference does that make. We can put in numbers to test what would happen.

What is this “cancelling out effect” and what/how/why does it cancel wrt the expected value? Please explain.

Double the same amount is better than half the same amount.

But factor in that we are dealing with double half the amount and half double the amount and that cancels out what otherwise would be a gain.

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 Posted: 10 January 2012 11:25 PM [ Ignore ]   [ # 63 ]
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Hey, now that I have my envelope, and know how much is in it.  Maybe I will trade.  Whadya say?

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 Posted: 10 January 2012 11:36 PM [ Ignore ]   [ # 64 ]
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kkwan - 10 January 2012 12:50 PM

a. Gain/loss = unknown
b. Gain/loss = unknown
c. Gain = A
d. Loss = A

Chance of gain A when switching is 1/4 (25%)
Chance of loss A when switching is 1/4 (25%)
Chance of unknown gain/loss by not switching is 2/4 = 1/2 (50%)

You know the correct use of statistics:

o Yes I do
x I’ll switch

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 Posted: 11 January 2012 08:26 AM [ Ignore ]   [ # 65 ]
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TimB - 10 January 2012 10:32 PM

No, K"Monty Hall"kwan, I’m keeping envelope #1.

Did you get a goat? Perhaps you should have switched to get the car?

It more fun to switch and to keep on switching.

Nothing ventured, nothing gained.

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 Posted: 11 January 2012 07:30 PM [ Ignore ]   [ # 66 ]
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kkwan - 06 January 2012 09:16 PM
domokato - 06 January 2012 01:03 PM
kkwan - 06 January 2012 08:41 AM
domokato - 05 January 2012 10:52 AM

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

Steps 4 and 5 are not conditionals.

They are only statements of what is in the other envelope if A is smaller or larger.

(emphasis mine)

Isn’t that what a conditional is?

From the wiki on indicative conditional

In natural languages, an indicative conditional is the logical operation given by statements of the form “If A then B”. Unlike the material conditional, an indicative conditional does not have a stipulated definition. The philosophical literature on this operation is broad, and no clear consensus has been reached.

Because an indicative conditional does not have a stipulated definition, one can consider steps 4 and 5 as statements of the logical operation (given A is larger or smaller).

How does that follow? And why would you think I meant indicative conditional rather than material conditional?

Yes, sorry for inaccurate terminology, but aren’t you being a bit nitpicky? I think my solution is essentially the same as the one you quoted (and Andrew’s ).

No, your solution only considered steps 4,5, and 6 which are valid whereas the proposed resolution found inconsistencies with step 7.

Step 6 is also problematic because you can’t assign probabilities to these conditionals and connect them with a logical “and”.

Hm, I don’t understand why the expectation value is infinite. Also, this solution doesn’t point out the flaw in the original argument (of the original variant) - it only posits a correct resolution. So then what’s the problem with the first proposed resolution?

...

The first proposed resolution collapses with the second variant:

The solution above doesn’t explain what’s wrong if the player is allowed to open the first envelope before being offered the option to switch. In this case, A is indeed a fixed variable. Hence, the proposed solution in the first case breaks down and another explanation is needed.

My solution doesn’t break down in the second variant, and indeed wikipedia agrees with me about step 6:

Blachman, Christensen and Utts (1996),[6] and many later authors working in probability theory, pointed out that if the amounts of money in the two envelopes have any proper probability distribution, then it is simply impossible that whatever the amount A=a in the first envelope, it is equally likely that the second contains a/2 or 2a. Thus step 6 of the argument which leads to always switching is a non-sequitur.

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 Posted: 12 January 2012 08:22 AM [ Ignore ]   [ # 67 ]
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StephenLawrence - 10 January 2012 10:50 PM

So what that it is not know? What difference does that make. We can put in numbers to test what would happen.

In the problem, we do not know the amounts in the two envelopes except that one has twice the other, thus A and 2A. Putting in numbers makes no difference to the expected value.

If one selects an envelope at random and do not switch, the gain/loss is unknown because one does not know whether the other envelope has double or half the amount.

OTOH, when one switches envelopes the expected value can be calculated and it is always more than A, therefore it is advantageous to switch.

But factor in that we are dealing with double half the amount and half double the amount and that cancels out what otherwise would be a gain.

What do you mean by that statement?

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 Posted: 12 January 2012 09:40 AM [ Ignore ]   [ # 68 ]
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domokato - 11 January 2012 07:30 PM

How does that follow? And why would you think I meant indicative conditional rather than material conditional?

From the same wiki:

The material conditional does not always function in accordance with everyday if-then reasoning. Therefore there are drawbacks with using the material conditional to represent if-then statements.

You wrote:

Step 6 is also problematic because you can’t assign probabilities to these conditionals and connect them with a logical “and”.

Why not?

My solution doesn’t break down in the second variant, and indeed wikipedia agrees with me about step 6:

Blachman, Christensen and Utts (1996),[6] and many later authors working in probability theory, pointed out that if the amounts of money in the two envelopes have any proper probability distribution, then it is simply impossible that whatever the amount A=a in the first envelope, it is equally likely that the second contains a/2 or 2a. Thus step 6 of the argument which leads to always switching is a non-sequitur.

In the argument:

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.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

In step 1, A is defined as the “amount in my selected envelope”.

In step 3, the amount in the other envelope is either 2A or A/2.

Steps 4 and 5 follows logically from steps 1 and 3. There is no inconsistency.

Step 6 combines the information provided in steps 1 to 5.

In the second variant, A is now a fixed variable. The proposed resolution was about the inconsistency of treating A as a random variable. This is no longer the case.

Your solution states that:

Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

It is immaterial to resolve whether A is the larger or smaller amount though the situations of A being smaller or larger are considered in steps 4 and 5.

Your quote from the wiki is about probability distributions and the impossibility of equal probabilities of 2a and a/2 which is not related your solution at all.

[ Edited: 12 January 2012 09:53 AM by kkwan ]
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 Posted: 12 January 2012 10:48 AM [ Ignore ]   [ # 69 ]
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kkwan - 12 January 2012 09:40 AM
domokato - 11 January 2012 07:30 PM

How does that follow? And why would you think I meant indicative conditional rather than material conditional?

From the same wiki:

The material conditional does not always function in accordance with everyday if-then reasoning. Therefore there are drawbacks with using the material conditional to represent if-then statements.

But those drawbacks don’t apply here.

You wrote:

Step 6 is also problematic because you can’t assign probabilities to these conditionals and connect them with a logical “and”.

Why not?

From the proposed resolution on wikipedia:

This follows from the recognition that the symbol A in step 7 is effectively used to denote two different quantities, committing the fallacy of equivocation.

This fallacy applies to step 6 for the same reason. And since 7 derives from 6, it follows that 4-6 is sufficient to point out the problem.

My solution doesn’t break down in the second variant, and indeed wikipedia agrees with me about step 6:

Blachman, Christensen and Utts (1996),[6] and many later authors working in probability theory, pointed out that if the amounts of money in the two envelopes have any proper probability distribution, then it is simply impossible that whatever the amount A=a in the first envelope, it is equally likely that the second contains a/2 or 2a. Thus step 6 of the argument which leads to always switching is a non-sequitur.

In the argument:

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.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

In step 1, A is defined as the “amount in my selected envelope”.

In step 3, the amount in the other envelope is either 2A or A/2.

Steps 4 and 5 follows logically from steps 1 and 3. There is no inconsistency.

Step 6 combines the information provided in steps 1 to 5.

In the second variant, A is now a fixed variable. The proposed resolution was about the inconsistency of treating A as a random variable. This is no longer the case.

I don’t see how the second variant makes A a fixed variable. It is still the outcome of a random process - I don’t see how knowing its value makes it “fixed”.

Your solution states that:

Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

It is immaterial to resolve whether A is the larger or smaller amount though the situations of A being smaller or larger are considered in steps 4 and 5.

Your quote from the wiki is about probability distributions and the impossibility of equal probabilities of 2a and a/2 which is not related your solution at all.

Hm, I think you’re right

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 Posted: 12 January 2012 12:10 PM [ Ignore ]   [ # 70 ]
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kkwan - 12 January 2012 08:22 AM
StephenLawrence - 10 January 2012 10:50 PM

But factor in that we are dealing with double half the amount and half double the amount and that cancels out what otherwise would be a gain.

What do you mean by that statement?

I mean the same thing as I’ve already said.

Double or half is an advantage if dealing with the same amount.

But in this case if you get double it’s because you have half of what is in the other envelope and visa versa.

This does cancel out.

Stephen

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 Posted: 12 January 2012 01:46 PM [ Ignore ]   [ # 71 ]
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StephenLawrence - 12 January 2012 12:10 PM
kkwan - 12 January 2012 08:22 AM
StephenLawrence - 10 January 2012 10:50 PM

But factor in that we are dealing with double half the amount and half double the amount and that cancels out what otherwise would be a gain.

What do you mean by that statement?

I mean the same thing as I’ve already said.

Double or half is an advantage if dealing with the same amount.

But in this case if you get double it’s because you have half of what is in the other envelope and visa versa.

This does cancel out.

Stephen

What if you have double the amount in the envelope in hand? Then by switching you lose half. The amounts are irrelevant as you do not “know” which is in either envelope. You only “know” that you will end up with at least half.  But switching does not increase your chances at all.

[ Edited: 12 January 2012 01:49 PM by Write4U ]
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 Posted: 12 January 2012 02:01 PM [ Ignore ]   [ # 72 ]
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The fallacy lies in assuming you have half in hand and by switching you can only gain. But you have an equal chance of having double in hand and thus you can only lose by switching.
The reality is that you never switch.  Having the envelope in hand is not any different than having both envelopes on the table. All you can do is make a choice. The assumption of switching is an illusion. You do the same thing mentally when both envelopes are on the table, “shall I pick this one or the other one”?  50/50 either way.

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 Posted: 13 January 2012 08:08 AM [ Ignore ]   [ # 73 ]
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domokato - 12 January 2012 10:48 AM

But those drawbacks don’t apply here.

From the wiki on conditional sentence

....the intuitive interpretation of conditional statements in natural language does not always correspond to the definition of this mathematical relation. Modelling the meaning of real conditional statements requires the definition of an indicative conditional….

You wrote:

From the proposed resolution on wikipedia:

This follows from the recognition that the symbol A in step 7 is effectively used to denote two different quantities, committing the fallacy of equivocation.

This fallacy applies to step 6 for the same reason. And since 7 derives from 6, it follows that 4-6 is sufficient to point out the problem.

As A has a known and fixed value in the second variant, there is no fallacy of equivocation

I don’t see how the second variant makes A a fixed variable. It is still the outcome of a random process - I don’t see how knowing its value makes it “fixed”.

In the second variant, the player is allowed to open the first envelope before being offered the option to switch. So, A is known, fixed and neither is it the outcome of a random process.

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 Posted: 13 January 2012 10:41 AM [ Ignore ]   [ # 74 ]
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StephenLawrence - 12 January 2012 12:10 PM

Double or half is an advantage if dealing with the same amount.

Whether double or half is an advantage depends on the situation.

1. If you selected A and switched, you gain A (100%)

2. If you selected 2A and switched you lose A (50%)

In 1, you gain 100% whereas in 2, you only lose 50% of the amount you selected.

In comparing 1 with 2, the potential gain in 1 is greater than the potential loss in 2.

And in both situations you do not end up with nothing.

However, in both situations, you don’t know what is in the other envelope and whether you will win or lose. But you can anticipate the above and with the expected value as 5/4A, it is an advantage to switch as you gain on average.

But in this case if you get double it’s because you have half of what is in the other envelope and visa versa.

This does cancel out.

You should only compare gain/loss from what you started with.

Half or double of the other envelope is equivalent to double or half of the envelope you selected.

Both describe the same situations. It is tautological.

Thus, there is nothing to cancel out.

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 Posted: 13 January 2012 11:11 AM [ Ignore ]   [ # 75 ]
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Let’s say in the beginning there are two envelopes offered to you: one with \$5 and one with \$10 inside. Since your choice NOT to take an envelope is not an option then your chances are 1in2 that you will pick the higher prize, \$10.

Now, you pick an envelope. Let’s say it’s the envelope with \$10. You now have \$10 in your possession. Given another offer and KNOWING that you have \$10 makes the unknown envelope a chance of either containing a \$5 OR A \$20 BILL. Therefore your next choice would be a 1in3 chance that you will pick the highest prize (\$5 or \$20 if you make the switch or \$10 if you do not make the switch).

You are better off NOT switching because your odds would then be worse (1in3) (\$5, \$10 or \$20) instead of the first pick (1in2) (\$5 or \$10) of picking the envelope with the best prize.

[ Edited: 13 January 2012 11:29 AM by LuckyBezel ]
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