Vodoo ploppies are right, we can shove our math and sims :)

[tthree]

statistics

I think k c showed the proper statistical logic. He said open it and see what amount was inside and still had statistics that was correct. He showed the flaw in the other statistical logic. There was no flaw in his. It was simple and elegant and accurate. There is always more than one way to work a math problem. If the two ways have a different answer one or both are wrong. His statistical math came up with an answer that agrees with the other math that says the envelopes have equal likelihood of either envelope having the larger amount. The point of my extreme examples was to show the statistical logic was flawed. He got the best of me. I couldnt work out a statistical argument without the flaw. I could identify that it existed though. :grin:

 

[MangoJ]

Let me give a try with conditional probabilities, which seems appropriate in this scenario because we have unknown information.

X is the amount we see in the chosen envelope.
Y is the amount in the other envelope
A is the amount in the lower envelope value. Hence there are two envelopes A and 2A.

So we have the joint probability: P(X,Y|A) = 0.5 Delta_X,A Delta_Y,2A + 0.5 Delta_X,2A Delta_Y,A
where Delta_a,b is 1 if a=b, and 0 otherwise (Kronecker delta symbol).

We should change the envelope if the expected value of Y given X, <Y|X>, of the other envelope is larger than X (the amount we have seen).
Hence we need to calculate the probability Y given X, that is P(Y|X).

We note that P(Y|X,A) = P(X,Y|A) / P(X|A).
On the right hand side we know the marginal conditional probability P(X|A) = sum_Y P(X,Y|A) = 0.5 Delta_X,A + 0.5 Delta_X,2A.

The probability of Y given X is then, if we sum over A:
P(Y|X) = sum_A P(A) P(Y|X,A)
= sum_A P(A) [Delta_X,A Delta_Y,2A + Delta_X,2A Delta_Y,A] / [Delta_X,A + Delta_X,2A]
= [p(A=X) Delta_Y,2X + p(A=X/2) Delta_Y,X/2] / [sum_A p(A)]


Then we get for the estimated value of the other envelope - given all observations:
<Y|X> = sum_Y Y P(Y|X)
= sum_Y Y [p(A=X) Delta_Y,2X + p(A=X/2) Delta_Y,X/2] / [sum_A p(A)]
= [p(A=X) / sum_A p(A)] 2X + [p(A=X/2) / sum_A p(A)] X/2.

The dilemma people have: they somehow assume that p(A=X) = p(A=X/2), i.e. that it is equally likely that the game provider chooses $100 or $50 with same probability as the LOWER amount in the first place.
(This is different from the probability, of putting the lower amount into the chosen envelope - which is always 0.5)

As I already said in post #8 and #25, the decision depends on the probability distribution of the amount of money the game provider likes to spend.
Choosing $50 and $100 with equal probability is okay (and that indeed would justify a switch!).

But once p(A) is fixed, so is p(A=X) and likewise. The infinite switching argument fails as the following property
p(A=X) = p(A=2X) = p(A=3X) = ... = const
cannot be achieved consistently (as sum_A p(A) diverges).
So if the game provider likes spending $50 and $100 equally likely, that doesn't mean he will as likely spend $200, $400, ... to infinity. You would be EXTREMELY unlucky to catch just $100, and not billions of trillions of $.

 

[Gamblor]

I think k c showed the proper statistical logic. He said open it and see what amount was inside and still had statistics that was correct. He showed the flaw in the other statistical logic. There was no flaw in his. It was simple and elegant and accurate. There is always more than one way to work a math problem. If the two ways have a different answer one or both are wrong. His statistical math came up with an answer that agrees with the other math that says the envelopes have equal likelihood of either envelope having the larger amount. The point of my extreme examples was to show the statistical logic was flawed. He got the best of me. I couldnt work out a statistical argument without the flaw. I could identify that it existed though. :grin:

Yes the logic is right, but it seems to me its the common sense solution to the problem. What the "paradox" is getting at is that we make valid assumptions, and valid logical/mathematical inferences, that leads us to a solution that goes against common sense. Not sure if this was solved, could be missing something on my part, these kind of problems do tend to hurt my brain.

 

[Gamblor]

Let me give a try with conditional probabilities, which seems appropriate in this scenario because we have unknown information.

X is the amount we see in the chosen envelope.
Y is the amount in the other envelope
A is the amount in the lower envelope value. Hence there are two envelopes A and 2A.

So we have the joint probability: P(X,Y|A) = 0.5 Delta_X,A Delta_Y,2A + 0.5 Delta_X,2A Delta_Y,A
where Delta_a,b is 1 if a=b, and 0 otherwise (Kronecker delta symbol).

We should change the envelope if the expected value of Y given X, <Y|X>, of the other envelope is larger than X (the amount we have seen).
Hence we need to calculate the probability Y given X, that is P(Y|X).

We note that P(Y|X,A) = P(X,Y|A) / P(X|A).
On the right hand side we know the marginal conditional probability P(X|A) = sum_Y P(X,Y|A) = 0.5 Delta_X,A + 0.5 Delta_X,2A.

The probability of Y given X is then, if we sum over A:
P(Y|X) = sum_A P(A) P(Y|X,A)
= sum_A P(A) [Delta_X,A Delta_Y,2A + Delta_X,2A Delta_Y,A] / [Delta_X,A + Delta_X,2A]
= [p(A=X) Delta_Y,2X + p(A=X/2) Delta_Y,X/2] / [sum_A p(A)]


Then we get for the estimated value of the other envelope - given all observations:
<Y|X> = sum_Y Y P(Y|X)
= sum_Y Y [p(A=X) Delta_Y,2X + p(A=X/2) Delta_Y,X/2] / [sum_A p(A)]
= [p(A=X) / sum_A p(A)] 2X + [p(A=X/2) / sum_A p(A)] X/2.

The dilemma people have: they somehow assume that p(A=X) = p(A=X/2), i.e. that it is equally likely that the game provider chooses $100 or $50 with same probability as the LOWER amount in the first place.
(This is different from the probability, of putting the lower amount into the chosen envelope - which is always 0.5)

As I already said in post #8 and #25, the decision depends on the probability distribution of the amount of money the game provider likes to spend.
Choosing $50 and $100 with equal probability is okay (and that indeed would justify a switch!).

But once p(A) is fixed, so is p(A=X) and likewise. The infinite switching argument fails as the following property
p(A=X) = p(A=2X) = p(A=3X) = ... = const
cannot be achieved consistently (as sum_A p(A) diverges).
So if the game provider likes spending $50 and $100 equally likely, that doesn't mean he will as likely spend $200, $400, ... to infinity. You would be EXTREMELY unlucky to catch just $100, and not billions of trillions of $.

Yes, others (mathematicians) have argued along similar lines (mentioned in the wiki article): "Chalmers suggests that decision theory generally breaks down when confronted with games having a diverging expectation", which is probably true, but there is still controversy over this "However, Clark and Shackel argue that this blaming it all on 'the strange behaviour of infinity"'doesn't resolve the paradox at all".

 

[Gamblor]

Here's my proposed solution, Mango's earlier mention of negative probability got me thinking. I will just arbitrarily add a negative probability, just to make the equation of picking the next envelope work.

1/2 * A + (-1/2) * A + 1/2 * 2A = 1

So the other envelope now has the same EV as the current one, just as we expected :laugh:

Hey before you complain I can't just arbitrarily do this, physicists and mathematicians do it all time! (that's why negative probabilities were thought up in the first place).

 

[SleightOfHand]

Yes the logic is right, but it seems to me its the common sense solution to the problem. What the "paradox" is getting at is that we make valid assumptions, and valid logical/mathematical inferences, that leads us to a solution that goes against common sense. Not sure if this was solved, could be missing something on my part, these kind of problems do tend to hurt my brain.

I am not qualified to solve this problem, but it seems to me that the flaw in kc's solution is that he is not using all the information. He is saying the ev before looking at the envelope and before switching is the same. But he didn't mention that we now know that one envelope contains $100. This gives us new information that the other envelope contains $50 or $200, which still brings us to the paradox

 

[tthree]

Back to the original "paradox"

Maybe I didn't explain as clearly, to avoid a long'ish post, but here is how it is formulated. With apparently valid premises, and apparently valid logical and mathemical deductions, it leads to the absurd conclusion that simply by switching envelopes, you increase EV. This is a paradox, and the question is, is there a flaw with this argument?


The switching argument: Now suppose you reason as follows:

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.
7. So the expected value of the money in the other envelope is

1/2 * 2A + 1/2 * A/2 = 5/4 * A

8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

At step 9:
Envelop 1 contains A
envelop 2 contains on the average 1.25*A = B
envelop 1 contains on the average A = 1.25*B

The average sum of both envelopes expressed in terms of A is 2.25*A
The average sum of both envelopes expressed in terms of B is 2.25*B
since they must be equal 2.25*A = 2.25*B
therefore (A = B AND EITHER (A = 2*B 0R B = 2*A))

Therefore: EITHER both envelopes contain $0
OR the switching logic has been proven false

EUREKA we have a winner paradox solved by mathematical proof
YOU ARE NO BETTER OFF SWITCHING AT STEP 8 IN EITHER CASE

*If you argue for the first switch being mathematically sound then the second switch must be false.

 

[tthree]

There is a flaw

I didnt really solve the paradox. Who will spot the flaw in the proof first.

 

[k_c]

I am not qualified to solve this problem, but it seems to me that the flaw in kc's solution is that he is not using all the information. He is saying the ev before looking at the envelope and before switching is the same. But he didn't mention that we now know that one envelope contains $100. This gives us new information that the other envelope contains $50 or $200, which still brings us to the paradox

To me it seems simple to prove there is no advantage in switching envelopes after one of them is chosen and opened and the amount in it revealed.

If the envelope containing A dollars was chosen then switching results in a profit of A dollars.
If the envelope containing 2*A dollars was chosen then switching results in a loss of A dollars.

EV(switching) = .5*A + .5*(-A) = 0

btw any argument that results in winding up with A/2 dollars is flawed since there are only 2 possible amounts to end up with, A or 2*A.

Also the EV of the game is 1.5*A for any choices made before the final outcome is revealed.

 

[tthree]

Here is the corrected proof and what it proves

At step 9:
Envelope 1 contains A
envelope 2 contains B
envelope 1 contains on the average 1.25*B
envelope 2 contains on the average 1.25*A
The average sum of both envelopes expressed in terms of A is 2.25*A
The average sum of both envelopes expressed in terms of B is 2.25*B
since they must be equal 2.25*A = 2.25*B
therefore (A = B AND EITHER (A = 2*B 0R B = 2*A))

Therefore: EITHER both envelopes contain $0
OR the switching logic has been proven false

EITHER YOU ARE NO BETTER OFF SWITCHING AT STEP 8 IN EITHER CASE ABOVE, OR NO BETTER OFF SWITCHING IN STEP 9, OR BOTH(when they both have $0).

*If you argue for the first switch being mathematically sound then the second switch must be false. So in this example you either dont switch or switch exactly once as proven here. The paradox is actually solved. It is common sense but here is the proof.

This is exactly like shrodingers cat. Opening an envelope collapses the equations, whether it is your envelope or the other, at which point you choose the unopened envelope.

 

[tthree]

To me it seems simple to prove there is no advantage in switching envelopes after one of them is chosen and opened and the amount in it revealed.

If the envelope containing A dollars was chosen then switching results in a profit of A dollars.
If the envelope containing 2*A dollars was chosen then switching results in a loss of A dollars.

EV(switching) = .5*A + .5*(-A) = 0

btw any argument that results in winding up with A/2 dollars is flawed since there are only 2 possible amounts to end up with, A or 2*A.

Also the EV of the game is 1.5*A for any choices made before the final outcome is revealed.

What if the envelope containing A/2 is chosen. You are defining A as the smaller envelope not a random envelope. The EV for your random choice of A is 1.125*A. .5*A + .25*A/2 + .25*2*A

 

[iCountNTrack]

What if the envelope containing A/2 is chosen. You are defining A as the smaller envelope not a random envelope. The EV for your random choice of A is 1.125*A. .5*A + .25*A/2 + .25*2*A

Sorry your EV equation is not true,

I use X instead of A for clarity

KC did not assume anything:

I chose an amount X,
if the amount i chose was the smaller amount, and if i decide to switch, i gain +X dollars, the probability of this happening is 50%

if the amount i chose was the larger amount, and if i decide to switch, i lose -X dollars, the probability of this happening is 50%.

Therefore the net gain for switching envelopes is

0.5*(+X)+0.5(-X)=0

The error in the previous analysis is that you cant assume that the amount X represents at the same time the larger amount and the smaller amount, you have to look at each case separately .

It amazes me sometimes how people turn such a simple problem to a more philosophical problem involving quantum mechanics.

 

[tthree]

Sorry your EV equation is not true,

I use X instead of A for clarity

KC did not assume anything:

I chose an amount X,
if the amount i chose was the smaller amount, and if i decide to switch, i gain +X dollars, the probability of this happening is 50%

if the amount i chose was the larger amount, and if i decide to switch, i lose -X dollars, the probability of this happening is 50%.

Therefore the net gain for switching envelopes is

0.5*(+X)+0.5(-X)=0

The error in the previous analysis is that you cant assume that the amount X represents at the same time the larger amount and the smaller amount, you have to look at each case separately .

It amazes me sometimes how people turn such a simple problem to a more philosophical problem involving quantum mechanics.

If you choose the larger amount X and switch you lose X/2. You dont open an envelope with zero(X - X) in it but half of X in it. Solving a differeent way than in my previous post.



EV(for the game) = .5*X + .5*(.5*(X-X/2) + .5*(X+X)) = .5*X + .25*X/2 + .25*2*X
= (.5 +.125 + .5)*X = 1.125*X

EV(for switching) = .5*(X-X/2) + .5*2*X = X/4 + X = 5*X/4 = 1.25*X

The parallel with Shrodingers cat is that until you open an envelope each state of envelope 1 exists with equal likelihood, better to switch to envelope 2 or better to keep envelope 1, and could be viewed as existing simultaneously. Only when an envelope is opened can a non paradoxical statistical argument for switching be made. That is when the paradox collapses indicating which envelope has the higher EV.

 

[iCountNTrack]

If you choose the larger amount X and switch you lose X/2. You dont open an envelope with zero(X - X) in it but half of X in it. Solving a differeent way than in my previous post.



EV(for the game) = .5*X + .5*(.5*(X-X/2) + .5*(X+X)) = .5*X + .25*X/2 + .25*2*X
= (.5 +.125 + .5)*X = 1.125*X

EV(for switching) = .5*(X-X/2) + .5*2*X = X/4 + X = 5*X/4 = 1.25*X

The parallel with Shrodingers cat is that until you open an envelope each state of envelope 1 exists with equal likelihood, better to switch to envelope 2 or better to keep envelope 1, and could be viewed as existing simultaneously. Only when an envelope is opened can a non paradoxical statistical argument for switching be made. That is when the paradox collapses indicating which envelope has the higher EV.

Let me try again,
We have defined our problem and our variables as such:

Small amount in the envelope is X
Large amount in the envelope is 2X.

Let Y represent the amount found in the envelope when it is opened.
Y=X (50% frequency)
or
Y=2X (50% frequency)

if Y=X and i switch envelopes I gain +X this will happen 50% of the times
if Y=2X and i switch envelopes I gain -X this will happen 50% of the times.

net gain from switching=0.5*(+X)+0.5(-X)=0

As i mentioned before we cannot redefine our variables as we please while solving the problem, in the beginning we defined X to be the small amount and 2X to be the large amount, we have to stick with that till the end. since we defined the possible choices to be X and 2X, it is impossible to have X/2.

 

[tthree]

Understanding the proof

I can find no fault with your logic. It is what the proof showed but if you understood what the proof actually showed you wouldnt follow that logic. The proof showed that the analysis of unknown quantities was meaningless. It wasnt until you quantified an envelope that it became meaningful. At that point an envelope is opened you find $100. Now you know that 1 envelope contains $100 and the other either contains $50 or $200. Switching now losses $50 or gains $100. X is still unknown and useless in the analysis of this positive EV switching situation. When you open the other envelope you find out if you lost X ($50) or gained X ($100). X is not a constant.

 

[tthree]

Mangoj should share the credit

I would like to give credit to Mangoj. It wasnt until I read his solution that I got the inspiration for the proof. I got 3/4 of the way through his analysis and thought, he is going to solve this, anticipating where he was heading. He went in a different direction which also showed that you cannt argue for a second switch. The proof showed the argument for the first switch was equally flawed unless an envelope was quantified.

 

[MangoJ]

I would like to give credit to Mangoj.

I must decline that. It is always difficult to disprove statements (there is EV in switching) which are wrong by symmetry (both envelopes are equal) and inconsistency (infinite switching strategy) - other than using that same symmetry or inconsistency.

The short statement of k_c was most beautiful to read. The amount you win or lose with switching is the amount chosen by the other party - not the amount (or half of that) you see in the envelope. This didn't occurred to me but is so obvious when said. Shame it didn't occurred to me in the first place - I instead ran into the direction of obtaining additional knowledge. That still might be true. In the trivial case where you have perfect knowledge of the contents of both envelopes, EV from that knowledge is maximal. Hence EV must depend on knowledge of the situation.

It didn't occurred to me that even in the absence of additional knowledge a decision whether switching or staying is favorable (of course both are equal) could be made other than by simple symmetry.

 

[sagefr0g]

...... .

It amazes me sometimes how people turn such a simple problem to a more philosophical problem involving quantum mechanics.

so, errhh well i think you are pretty much an expert in the qm area, so in your opinion is it ever worth while to ponder how qm might relate to such problems?

 

[MangoJ]

so, errhh well i think you are pretty much an expert in the qm area, so in your opinion is it ever worth while to ponder how qm might relate to such problems?

There is no relation between quantum mechanics and the envelope problem.

 

[k_c]

I can find no fault with your logic. It is what the proof showed but if you understood what the proof actually showed you wouldnt follow that logic. The proof showed that the analysis of unknown quantities was meaningless. It wasnt until you quantified an envelope that it became meaningful. At that point an envelope is opened you find $100. Now you know that 1 envelope contains $100 and the other either contains $50 or $200. Switching now losses $50 or gains $100. X is still unknown and useless in the analysis of this positive EV switching situation. When you open the other envelope you find out if you lost X ($50) or gained X ($100). X is not a constant.

You are descibing a different problem, which is fine if that's what you want to solve.

A=amount in 1st envelope selected
amount in 2nd envelope is either A/2 or 2*A, each with 50% probability
if amount in 2nd envelope = A/2 then loss = A/2 for switching envelopes
if amount in 2nd envelope = 2*A then gain = A for switching envelopes
EV(switching) = .5*(-A/2) + .5*A = .25*A

Basically this is just similar to saying that if you flip a coin betting 1 unit that heads occurs and are paid even money if you win but only lose half your bet if you lose then this is positive EV = +.25.

However, there's also the possibility of selecting the envelope that does not contain A first.

The complete set of possiblities

Select A first
[u]envelope 1[/u]	[u]envelope 2[/u]
A		A/2
A		2*A
EV(switching) = .5*(-A/2) + .5*A = .25*A

Select not A first
[u]envelope 1[/u]	[u]envelope 2[/u]
2*A		A
A/2		A
EV(switching) = .5*(-A) + .5*A/2 = -.25*A

Overall EV(switching) = .5*(.25) + .5*(-.25) = 0