Search

 27 of 132 « First Prev 25 26 27 28 29 Next Last »
The two envelopes problem
 Posted: 08 February 2012 10:04 AM [ Ignore ]   [ # 391 ]
Member
Total Posts:  114
Joined  2010-12-03
kkwan - 07 February 2012 08:34 PM
Mingy Jongo - 07 February 2012 03:23 PM

Suppose you open an envelope and it contains \$X.  While it is correct to say that the other envelope must contain either \$.5X or \$2X, it is incorrect to assume that there are even odds between them.  You must have the prior probabilities of the distribution to make that calculation.

Quite so. However, from the wiki on prior probability

The simplest and oldest rule for determining a non-informative prior is the principle of indifference, which assigns equal probabilities to all possibilities.

A shame that the principle of indifference is not epistemically justified in the slightest bit.

 Profile

 Posted: 08 February 2012 10:39 AM [ Ignore ]   [ # 392 ]
Member
Total Posts:  114
Joined  2010-12-03
StephenLawrence - 08 February 2012 01:12 AM
Mingy Jongo - 07 February 2012 03:23 PM

Suppose you open an envelope and it contains \$X.  While it is correct to say that the other envelope must contain either \$.5X or \$2X, it is incorrect to assume that there are even odds between them.  You must have the prior probabilities of the distribution to make that calculation.

Let’s say that’s right.

In that case why not just give the person making the selection the prior probabilities of the distribution and make sure the odds are even. Is that possible? If so we then get to the essence of the puzzle.

The problem is even then the equation should not work.

The reason is say you pick envelope A and conclude it’s right to switch. In that case what you should be able to do is say if I had picked the other envelope it would have been wrong to switch.

Stephen

Let’s do exactly that and observe what happens.

Given:
(1) A 50% chance that one envelope contains \$X and the other \$2X, and (2) a 50% chance that one envelope contains \$X and the other \$.5X.

If scenario 1 is the case, then there is a 50% chance you will select \$X and a 50% chance you will select \$2X.
If scenario 2 is the case, then there is a 50% chance you will select \$X and a 50% chance you will select \$.5X.

Therefore, the odds of picking \$X are 50% (.5*.5+.5*.5), the odds of picking \$2X are 25% (.5*.5+.5*0), and the odds of picking \$.5X are 25% (.5*0+.5*.5).  Therefore, your expected value without switching is \$1.125X [.5(\$X)+.25(\$2X)+.25(\$.5X)].

If you picked \$X (which you had a 50% chance of doing) and decide to switch, then there is a 50% chance you will get \$2X and a 50% chance you will get \$.5X.  Your expected value for switching is \$1.25X.
If you picked \$2X (which you had a 25% chance of doing) and decide to switch, then there is a 100% chance you will get \$X.  Your expected value for switching is \$X.
If you picked \$.5X (which you had a 25% chance of doing) and decide to switch, then there is a 100% chance you will get \$X.  Your expected value for switching is \$X.

Therefore, the overall expected value with switching is \$1.125X [.5*.5*(\$2X)+.5*.5*(\$.5X)+.25*1*(\$X)+.25*1*(\$X)], which is exactly the same as without switching.

 Profile

 Posted: 08 February 2012 11:21 AM [ Ignore ]   [ # 393 ]
Sr. Member
Total Posts:  1201
Joined  2009-05-10
kkwan - 08 February 2012 09:13 AM
GdB - 08 February 2012 05:48 AM
kkwan - 07 February 2012 07:28 PM

Your “expected value for each envelope is (A+B)/2” is the average value, not the expected value per se. The expected value is expressed in the equation above.

Yes, it is. And it is the expected value.

I think it is incorrect and misleading to consider the average value as the expected value.

The average is equivalent to the expected value if the probabilities of each possible value are equal, which they are in this case.

From the wiki on expected value

In probability theory, the expected value (or expectation, or mathematical expectation, or mean, or the first moment) of a random variable is the weighted average of all possible values that this random variable can take on.

Your expected value has two variables, A and B.  It should have only one variable and as such it is only the average value. The correct expected value is expressed in the equation as:

Correct? I hope you mean correct as in that’s what’s stated in the problem, not what’s actually correct?

Signature

“What people do is they confuse cynicism with skepticism. Cynicism is ‘you can’t change anything, everything sucks, there’s no point to anything.’ Skepticism is, ‘well, I’m not so sure.’” -Bill Nye

 Profile

 Posted: 08 February 2012 06:15 PM [ Ignore ]   [ # 394 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
Mingy Jongo - 08 February 2012 10:04 AM

A shame that the principle of indifference is not epistemically justified in the slightest bit.

That is an assertion which you will have to justify.

Signature

I am, therefore I think.

 Profile

 Posted: 08 February 2012 06:43 PM [ Ignore ]   [ # 395 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
domokato - 08 February 2012 11:21 AM

The average is equivalent to the expected value if the probabilities of each possible value are equal, which they are in this case.

The probabilities are equal, but the possible values of the variable are not equal. Thus, the average value is not the expected value.

Correct? I hope you mean correct as in that’s what’s stated in the problem, not what’s actually correct?

I mean correct in the context of the TEP and also generally as stated:

From the wiki on expected value

In probability theory, the expected value (or expectation, or mathematical expectation, or mean, or the first moment) of a random variable is the weighted average of all possible values that this random variable can take on.

What do you mean by actually correct?

Signature

I am, therefore I think.

 Profile

 Posted: 08 February 2012 08:12 PM [ Ignore ]   [ # 396 ]
Member
Total Posts:  114
Joined  2010-12-03
kkwan - 08 February 2012 06:15 PM
Mingy Jongo - 08 February 2012 10:04 AM

A shame that the principle of indifference is not epistemically justified in the slightest bit.

That is an assertion which you will have to justify.

Easy.  Imagine a switch that is either on or off, but not both.  Now imagine that someone asks you what the odds of the switch being on are, given that you do not know anything about it other than it can be either on or off, but not both.  The principle of indifference states that given no other information, one should assign equal probabilities to each possible state.  However, suppose that the switch is actually in the on position 90% of the time.  Therefore, the principle of indifference leads to an incorrect conclusion in at least one case, and hence, cannot be justified for epistemic use.

 Profile

 Posted: 08 February 2012 08:50 PM [ Ignore ]   [ # 397 ]
Sr. Member
Total Posts:  7981
Joined  2009-02-26
Mingy Jongo - 08 February 2012 08:12 PM
kkwan - 08 February 2012 06:15 PM
Mingy Jongo - 08 February 2012 10:04 AM

A shame that the principle of indifference is not epistemically justified in the slightest bit.

That is an assertion which you will have to justify.

Easy.  Imagine a switch that is either on or off, but not both.  Now imagine that someone asks you what the odds of the switch being on are, given that you do not know anything about it other than it can be either on or off, but not both.  The principle of indifference states that given no other information, one should assign equal probabilities to each possible state.  However, suppose that the switch is actually in the on position 90% of the time.  Therefore, the principle of indifference leads to an incorrect conclusion in at least one case, and hence, cannot be justified for epistemic use.

That’s a straw man. Without knowing why the switch should be ON or OFF any calculation of switching does not predict an eventual outcome.
Suppose the switch is ON to prevent the launch of a missile. The 10% that the switch is placed in OFF position is used for regular testing purposes.
But this information is classified. All we know is that the switch is 90% in the ON position.
Now we are given a choice to use the switch. What possible use is the fact that we know that the switch is usually in the ON position? Personally I would say, leave the darn thing in the ON position, it seems to be working, whatever it’s function.
IMO that is not a good example.

The problem with the envelopes is that we are “given” an amount of money, which then can be switched for twice or half the amount. In either case you cannot lose. Thus there may be an incentive to gamble with the house’s money.
But now you are asked to bet out-of-pocket \$10 for a choice of envelopes which provide double or nothing. Thus if you get the empty envelope you will lose \$10, if you choose the right envelope you will double your money. Can anyone provide the “expected value” if we use this approach?

[ Edited: 08 February 2012 09:02 PM by Write4U ]
Signature

Art is the creation of that which evokes an emotional response, leading to thoughts of the noblest kind.
W4U

 Profile

 Posted: 08 February 2012 09:00 PM [ Ignore ]   [ # 398 ]
Member
Total Posts:  114
Joined  2010-12-03
Write4U - 08 February 2012 08:50 PM
Mingy Jongo - 08 February 2012 08:12 PM
kkwan - 08 February 2012 06:15 PM
Mingy Jongo - 08 February 2012 10:04 AM

A shame that the principle of indifference is not epistemically justified in the slightest bit.

That is an assertion which you will have to justify.

Easy.  Imagine a switch that is either on or off, but not both.  Now imagine that someone asks you what the odds of the switch being on are, given that you do not know anything about it other than it can be either on or off, but not both.  The principle of indifference states that given no other information, one should assign equal probabilities to each possible state.  However, suppose that the switch is actually in the on position 90% of the time.  Therefore, the principle of indifference leads to an incorrect conclusion in at least one case, and hence, cannot be justified for epistemic use.

That’s a straw man. Without knowing why the switch should be on or off any calculation of switching does not predict an eventual outcome.
Suppose the switch is ON to prevent the launch of a missile. The 10% that the switch is placed in OFF position is used for regular testing purposes.
But this information is classified. All we know is that the switch is 90% in the ON position.
Now we are given a choice to use the switch. What possible use is the fact that we know that the switch is usually in the ON position?

The point of the thought experiment is to show that the principle of indifference may not reflect reality in all cases—it may possibly give you one probability when another is actually correct.  I’m a bit confused at what your critique is trying to say.

 Profile

 Posted: 08 February 2012 09:01 PM [ Ignore ]   [ # 399 ]
Sr. Member
Total Posts:  1201
Joined  2009-05-10
kkwan - 08 February 2012 06:43 PM
domokato - 08 February 2012 11:21 AM

The average is equivalent to the expected value if the probabilities of each possible value are equal, which they are in this case.

The probabilities are equal, but the possible values of the variable are not equal. Thus, the average value is not the expected value.

If the possible values were equal then they wouldn’t be different values. That would just be one possible value. If two values are equal, they are the same value.

Correct? I hope you mean correct as in that’s what’s stated in the problem, not what’s actually correct?

I mean correct in the context of the TEP and also generally as stated:

From the wiki on expected value

In probability theory, the expected value (or expectation, or mathematical expectation, or mean, or the first moment) of a random variable is the weighted average of all possible values that this random variable can take on.

What do you mean by actually correct?

Let X be the smaller amount and 2X be the larger amount. The expected value is (1/2)X + (1/2)2X = (3/2)X. (I would state it in terms of A, but that is impossible - see below). That is the actual expected value. Plug in some numbers to see that it is correct and that it is also equal to the average.

I thought we had already agreed that the expected value calculation in step 7 is wrong. I thought we were just trying to figure out why it was wrong. You think it’s actually right? How can it be right if it always leads you to switch?

The expected value is the average of the amounts in both envelopes. I cannot define expected value in terms of A because the value of A changes depending on which envelope I have. This is wrong by definition since expected value (which is a weighted average, not a variable) should remain constant no matter which envelope I have.

Signature

“What people do is they confuse cynicism with skepticism. Cynicism is ‘you can’t change anything, everything sucks, there’s no point to anything.’ Skepticism is, ‘well, I’m not so sure.’” -Bill Nye

 Profile

 Posted: 08 February 2012 09:17 PM [ Ignore ]   [ # 400 ]
Sr. Member
Total Posts:  1201
Joined  2009-05-10
Write4U - 08 February 2012 08:50 PM

The problem with the envelopes is that we are “given” an amount of money, which then can be switched for twice or half the amount. In either case you cannot lose. Thus there may be an incentive to gamble with the house’s money.
But now you are asked to bet out-of-pocket \$10 for a choice of envelopes which provide double or nothing. Thus if you get the empty envelope you will lose \$10, if you choose the right envelope you will double your money. Can anyone provide the “expected value” if we use this approach?

(1/2)\$0 + (1/2)\$20 = \$10
That is the expected value in an envelope. If you want your expected value from playing, you should minus your buy-in from that, which is \$10, so your expected value from playing (i.e. expected gain/loss in this case) is \$0. This makes sense. If you could lose your money or gain the same amount, with an equal chance of each, then your expected gain/loss should cancel out to \$0.

But this isn’t similar to the TEP.

Signature

“What people do is they confuse cynicism with skepticism. Cynicism is ‘you can’t change anything, everything sucks, there’s no point to anything.’ Skepticism is, ‘well, I’m not so sure.’” -Bill Nye

 Profile

 Posted: 08 February 2012 09:59 PM [ Ignore ]   [ # 401 ]
Sr. Member
Total Posts:  7981
Joined  2009-02-26
domokato - 08 February 2012 09:17 PM
Write4U - 08 February 2012 08:50 PM

The problem with the envelopes is that we are “given” an amount of money, which then can be switched for twice or half the amount. In either case you cannot lose. Thus there may be an incentive to gamble with the house’s money.
But now you are asked to bet out-of-pocket \$10 for a choice of envelopes which provide double or nothing. Thus if you get the empty envelope you will lose \$10, if you choose the right envelope you will double your money. Can anyone provide the “expected value” if we use this approach?

(1/2)\$0 + (1/2)\$20 = \$10
That is the expected value in an envelope. If you want your expected value from playing, you should minus your buy-in from that, which is \$10, so your expected value from playing (i.e. expected gain/loss in this case) is \$0. This makes sense. If you could lose your money or gain the same amount, with an equal chance of each, then your expected gain/loss should cancel out to \$0.

But this isn’t similar to the TEP.

But is that not the essential problem. If your first envelope contains \$10, it is now your money. Then you are asked if you want to invest that money to buy an envelope which has either \$5 or \$20.  But the actual investment is only \$5, because if the other envelope contains lesser amount you only lose \$5, where if the envelope contains \$20 you double your money. Thus switching has an equal chance of gain vs loss. The difference is that the gain is more than you can lose.
In effect you are betting only \$5 of your original \$10 on the chance of winning \$10 more.

Signature

Art is the creation of that which evokes an emotional response, leading to thoughts of the noblest kind.
W4U

 Profile

 Posted: 08 February 2012 11:53 PM [ Ignore ]   [ # 402 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
Mingy Jongo - 08 February 2012 08:12 PM
kkwan - 08 February 2012 06:15 PM
Mingy Jongo - 08 February 2012 10:04 AM

A shame that the principle of indifference is not epistemically justified in the slightest bit.

That is an assertion which you will have to justify.

Easy.  Imagine a switch that is either on or off, but not both.  Now imagine that someone asks you what the odds of the switch being on are, given that you do not know anything about it other than it can be either on or off, but not both.  The principle of indifference states that given no other information, one should assign equal probabilities to each possible state.  However, suppose that the switch is actually in the on position 90% of the time.  Therefore, the principle of indifference leads to an incorrect conclusion in at least one case, and hence, cannot be justified for epistemic use.

I don’t see how this can be right.

Firstly we should assume determinism. That is because the switch is on 90% of the time, that has nothing to do with indeterminism, that will be because of what is done in the room and at what times and how much natural light there is and so on.

Assuming determinism the objective probability of the switch being on is either 1 or 0. Any probability other than 0 or 1 comes from wider circumstances than the precise circumtances.

In this case the appropriate circumstances are wider circumstances than the circumstances in which the light switch is on 90% of the time, because the subject doesn’t know that. They include the light switch being off 90% of the time and that is equally probable.

Edit: Also what else can you do than assign a probability of 1/2 to on and 1/2 to off? You could shrug your shoulders and say “I dunno, I can’t assign a probability”, but I think that just means the probability is 50% . I imagine a beliefometer between 0 and 100. Dunno being 50%. The beliefometer measures what you should believe given what you know.

Stephen

[ Edited: 09 February 2012 01:25 AM by StephenLawrence ]
 Profile

 Posted: 09 February 2012 12:02 AM [ Ignore ]   [ # 403 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
Mingy Jongo - 08 February 2012 10:39 AM
StephenLawrence - 08 February 2012 01:12 AM
Mingy Jongo - 07 February 2012 03:23 PM

Suppose you open an envelope and it contains \$X.  While it is correct to say that the other envelope must contain either \$.5X or \$2X, it is incorrect to assume that there are even odds between them.  You must have the prior probabilities of the distribution to make that calculation.

Let’s say that’s right.

In that case why not just give the person making the selection the prior probabilities of the distribution and make sure the odds are even. Is that possible? If so we then get to the essence of the puzzle.

The problem is even then the equation should not work.

The reason is say you pick envelope A and conclude it’s right to switch. In that case what you should be able to do is say if I had picked the other envelope it would have been wrong to switch.

Stephen

Let’s do exactly that and observe what happens.

Given:
(1) A 50% chance that one envelope contains \$X and the other \$2X, and (2) a 50% chance that one envelope contains \$X and the other \$.5X.

If scenario 1 is the case, then there is a 50% chance you will select \$X and a 50% chance you will select \$2X.
If scenario 2 is the case, then there is a 50% chance you will select \$X and a 50% chance you will select \$.5X.

Therefore, the odds of picking \$X are 50% (.5*.5+.5*.5), the odds of picking \$2X are 25% (.5*.5+.5*0), and the odds of picking \$.5X are 25% (.5*0+.5*.5).  Therefore, your expected value without switching is \$1.125X [.5(\$X)+.25(\$2X)+.25(\$.5X)].

If you picked \$X (which you had a 50% chance of doing) and decide to switch, then there is a 50% chance you will get \$2X and a 50% chance you will get \$.5X.  Your expected value for switching is \$1.25X.
If you picked \$2X (which you had a 25% chance of doing) and decide to switch, then there is a 100% chance you will get \$X.  Your expected value for switching is \$X.
If you picked \$.5X (which you had a 25% chance of doing) and decide to switch, then there is a 100% chance you will get \$X.  Your expected value for switching is \$X.

Therefore, the overall expected value with switching is \$1.125X [.5*.5*(\$2X)+.5*.5*(\$.5X)+.25*1*(\$X)+.25*1*(\$X)], which is exactly the same as without switching.

What all the examples do that are like this, is call X something other than what is in your envelope.

In kkwan’s example X is what is in your envelope.

Although other equations are right I’m sure, that doesn’t tell us what is wrong with kkwan’s equation.

If you were to say X shouldn’t be viewed as what is in kkwan’s envelope because…................... that might solve it. (Edit: I see domokato has.)

Stephen

[ Edited: 09 February 2012 12:11 AM by StephenLawrence ]
 Profile

 Posted: 09 February 2012 12:09 AM [ Ignore ]   [ # 404 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
domokato - 08 February 2012 09:01 PM

I cannot define expected value in terms of A because the value of A changes depending on which envelope I have. This is wrong by definition since expected value (which is a weighted average, not a variable) should remain constant no matter which envelope I have.

This is interesting, is that right? I dunno.

Stephen

 Profile

 Posted: 09 February 2012 12:18 AM [ Ignore ]   [ # 405 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
Write4U - 08 February 2012 09:59 PM

But is that not the essential problem. If your first envelope contains \$10, it is now your money. Then you are asked if you want to invest that money to buy an envelope which has either \$5 or \$20.  But the actual investment is only \$5, because if the other envelope contains lesser amount you only lose \$5, where if the envelope contains \$20 you double your money. Thus switching has an equal chance of gain vs loss. The difference is that the gain is more than you can lose.
In effect you are betting only \$5 of your original \$10 on the chance of winning \$10 more.

It looks like it doesn’t it Writer4U but it can’t be really.

I’m still not clear about what has gone wrong, I really don’t know if others are or not.

Stephen

 Profile

 27 of 132 « First Prev 25 26 27 28 29 Next Last »