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

Gamblor

Well-Known Member
#21
Its certainly not rational from a common sense perspective.

But its rational from a mathematical/logical perspective. If its not rational to do it indefinitely, it shouldn't be rational to do it once. But for some reason, statistically speaking, you have a higher EV simply by choosing to switch to the other envelope, i.e.,;

1/2 * 1/2 * X + 1/2 * 2 * X = 1.25 X

This is not rational common sense wise, but rational mathematically.
 
#22
this sounds a lot like a gamblers version of shrodingers cat. Until you open the envelope you're holding both 2A and A/2, so the best option is to never open then envelope.
 

Gamblor

Well-Known Member
#23
scooter185 said:
this sounds a lot like a gamblers version of shrodingers cat. Until you open the envelope you're holding both 2A and A/2, so the best option is to never open then envelope.
Yes very reminiscent. Also, keep in mind that QM is largely statistically based :eek:

Also, I had another thread about knowing the possible dealers next card, affected my current decision to surrender or stand. Kind of reminded me of those QM experiments where supposedly future events effect past events :p
 

MangoJ

Well-Known Member
#25
Gamblor said:
But for some reason, statistically speaking, you have a higher EV simply by choosing to switch to the other envelope, i.e.,;

1/2 * 1/2 * X + 1/2 * 2 * X = 1.25 X

This is not rational common sense wise, but rational mathematically.
Why do you think the X/2 envelope and the 2X have the same probability ? Because you have two envelopes which are symmetrical. Thats perfectly right.

Now you peek into the first, and find X=$100. Why do you think you still have 50% probability of holding the X/2, and 50% probability of holding the X2 envelope ?

Again, the key to that puzzle is the game provider. He will not put ANY amount into the envelope (from zero to infinity). So he will have some probability distribution p(Y) of chosing his "unit" which he is comfortable spending on you (for whatever reason), and prepares two envelopes with Y and 2Y.

Now with a FIXED the probability distribution p(Y), the infinite switching strategy argument is broken. After finding X=$100, one can find probabiliy p'
that the original unit Y=X and Y=X/2:

p'(X) = p(X) / (p(X) + p(X/2))
p'(X/2) = p(X/2) / (p(X) + p(X/2))


The correct strategy for the player is: Estimate p(Y), open an envelope and find X. If p'(X/2) X/2 + p'(X) 2*X > X switch envelope.
The performance of that strategy depends on the estimate, that is judging the dealer.
 

MangoJ

Well-Known Member
#26
Gamblor said:
Yes very reminiscent. Also, keep in mind that QM is largely statistically based
This is so wrong. First of all there is quantum mechanics (QM), and ordinary statistics.
Then there are people (scientists) who understand a lot of ordinary statistics, but don't understand QM (yet).

One interpretation scientists use for QM is ordinary statistics. The stupider people think that QM is just a consequence of ordinary statistics, i.e. not knowing something about a physical system and just handling with probabilities in a fancy way.

The smarter people find out that ordinary statistics is NOT sufficient to describe QM in every case. On some bizarre occasions one would need negative probabilities or probabilities > 1, which is inconsistent with ordinary statistics.

The conclusion smarter people have: QM is NOT based on ordinary statistics.
Tthere is quantum statistics (described by the principles of QM), which then yields the same ordinary statistic in special cases (which are found most often, but can be avoided).
 

Gamblor

Well-Known Member
#27
MangoJ said:
Again, the key to that puzzle is the game provider. He will not put ANY amount into the envelope (from zero to infinity). So he will have some probability distribution p(Y) of chosing his "unit" which he is comfortable spending on you (for whatever reason), and prepares two envelopes with Y and 2Y.
Lets imagine the game provider is the US Treasury and they have a license to print an indefinite amount of money.
 

Gamblor

Well-Known Member
#28
MangoJ said:
This is so wrong. First of all there is quantum mechanics (QM), and ordinary statistics.
Then there are people (scientists) who understand a lot of ordinary statistics, but don't understand QM (yet).

One interpretation scientists use for QM is ordinary statistics. The stupider people think that QM is just a consequence of ordinary statistics, i.e. not knowing something about a physical system and just handling with probabilities in a fancy way.

The smarter people find out that ordinary statistics is NOT sufficient to describe QM in every case. On some bizarre occasions one would need negative probabilities or probabilities > 1, which is inconsistent with ordinary statistics.

The conclusion smarter people have: QM is NOT based on ordinary statistics.
Tthere is quantum statistics (described by the principles of QM), which then yields the same ordinary statistic in special cases (which are found most often, but can be avoided).
Did I ever say "QM is completely and totally based only on 'ordinary statistics'"?. I carefully kept my wording general just to avoid these hair splitting arguments. As I said QM is based on statistics, and that is blatantly true. Furthermore negative probability itself is still based on statistics, even "ordinary statistics".

My salient point being to confirm scooters post that this is "reminiscent" of QM. In interesting ways I might add. And who said you can't apply negative probabilities to this problem, its not just used in QM ;)
 

MangoJ

Well-Known Member
#29
Gamblor said:
Lets imagine the game provider is the US Treasury and they have a license to print an indefinite amount of money.
Then I have another question: How much would you spent for an "ante" to play this game ? Would you sell your house ? EV would be infinity even if you had no strategy at all (other than making sure you are taking an envelope).

In such a setting, I see no paradox situation when the infinite switching strategy proclaims an infinite EV. Any strategy would give an infinite EV.
 

Midwestern

Well-Known Member
#30
here's how i look at this problem.
you make one choice, and then never switch envelopes. here's why: i base my theory off of the solution for the monty hall problem: YOU CAN ONLY IMPROVE IF YOU HAVE NEW INFORMATION.

you're walking along the street, oblivious to the fact that soon you will be given this opportunity to choose between two envelopes. EV of this is zero.

its your lucky day and a man approaches you with the envelope game. this is essentially a gift of positive EV.
Now that you have this new information that you didnt have before you essentially can choose to PLAY the game or NOT play (i.e. choose an envelope or walk away from this offer)

if we DONT play, we walk away with Zero with 100% certainty. EV = (.5x0)
if we PLAY the game, we are gifted some arbitrary amount of either A or 2A with equal probability. EV=(.25xA and .25x2A)

the EV of our situation now is positive at .75A because of new information. as we stand, we can either make 0 with 100% certainty or 1.5 A with 100% certainty. it is obvious which choice we make: we play the envelope game for value 1.5A. Please note that regardless of the envelope you choose, each one has a "value" of 1.5A.

since A IS UNKNOWN, we replace value 1.5 A with random A again for simplicity.

Now we are offered to switch envelopes. however, the math that figures that switching yields 1.25A is a moot point because there is no more information released to us after our initial envelope selection. a rational thinker would realize that switching envelopes was -EV because of wasted time. we already had our chance to choose between 1.5 and zero in round 1, and any further switching beyond the 1st initial selection would be exchanging equal values of 1.5A (based on the math discussed in previous paragraph).

/thread :grin:
 
#31
QM and the universe

Mango I have a question for you. When I hit QM in college I performed well on the tests but didnt really think I understood the concepts like I usually do. I had to resort to memorization a robot like formula use. I feel this is not learning but rather performing. I switched to a mathematical based major after that.

You seem to project a higher understanding in this field. Now I cant remember the name of the particles involved but this is what my fuzzy memory recalls. The particle goes through the intense electromagnetic field near the nucleus of an atom. It splits into two particles with opposite spins. This is the interesting and confusing part. If a force acts on the spin of the one particle an equal but opposite effect is observed instantly in its partner particle even if the particle has traveled light years away and can't feel the force at all. It seemed to suggest a connection of all things in the universe independent of time and space. Are you familiar with this? If you are could you say whether this is speculation or fact. A little explanation might be nice.
 

MangoJ

Well-Known Member
#32
tthree said:
Mango I have a question for you. When I hit QM in college I performed well on the tests but didnt really think I understood the concepts like I usually do. I had to resort to memorization a robot like formula use. I feel this is not learning but rather performing. I switched to a mathematical based major after that.

You seem to project a higher understanding in this field. Now I cant remember the name of the particles involved but this is what my fuzzy memory recalls. The particle goes through the intense electromagnetic field near the nucleus of an atom. It splits into two particles with opposite spins. This is the interesting and confusing part. If a force acts on the spin of the one particle an equal but opposite effect is observed instantly in its partner particle even if the particle has traveled light years away and can't feel the force at all. It seemed to suggest a connection of all things in the universe independent of time and space. Are you familiar with this? If you are could you say whether this is speculation or fact. A little explanation might be nice.
I cannot guess from your description which particle or experiment you mean in detail. However from your described effect of two particles with opposite spin, which turns out to be opposite even after manipulation of only one of those particles - I guess you mean experiments on entanglement (i.e. a pair of photons).

Let me clarify the scenario. You have an experiment where the output are two photons. Both photons have opposite spin if you measure them. You can choose one of the photon, and make all kinds of experiments with it, it behaves like a normal photon. Now you make two experiments:
#1: You measure the spin of both photons. They are always opposite.
#2: You choose one photon, and manipulate it's spin reversibly (i.e. by rotating it). You then measure the spin of both photons. They are still opposite.

Now we come to the interpretation: The naive observer would think that the spin is opposite in the beginning (this is the outcome of #1), and since we rotate the spin of a single photon, and after rotation we compare it with the spin of the other photon (still being opposite as #2), the naive observer concludes that the rotation of the single photon must have instantly rotated the other photon as well.
Now the naive observer tries to prove his theory by direct experiment #3: He measures the spin of both photons (they are opposite as in #1), then he rotates the spin of one photon alone. In the final step he compares the spin of both photons again: They are now equal (not opposite)!

Obviously the naive theory of having two photons, which are connected by some instantaneous force, must be wrong.

But where is the flaw ? There are no two photons from the beginning. All there is is a single "Twophoton", which looks from either "side" as a normal photon, but when you rotate the Twophoton, both sides change simultaneous as it is the very same object. Once you measure the spin of this Twophoton (on either side), the Twophoton decays into two individual photons (with opposite spin).

Since this is a card forum, let me give an analogy. You have a complete single deck of cards, and you replace a single card with a joker. Then you shuffle the cards, and cut then in two piles. Obviously one of the piles will contain the joker. If you look through one pile and find the joker, you immediatly know that the other pile of cards doesn't contain the joker (as there is only 1 of them in the deck). You can even go to a remote location, look through the pile and "instantly" manipulate the other pile to contain a joker (or not).
At least for that part of the instant interaction, there clearly isn't something like a mystical force between both piles interacting over long ranges instantly. You would never say you had 2 piles of a half deck (2 photons). What you had was the half of a full deck (one side of a Twophoton).

This is as far as always measuring opposite spin. But what about manipulation of spin without measuring it ?
Say we did't replace a card with the joker before cutting the deck. We deal with an original deck, but a third party writes a hidden note that any specific card (i.e. the Seven of Spades) will be considered as the joker.
Now we can do the the experiment again, of course always find that only one pile will contain the joker (which is written on the hidden note).
After splitting the deck, you can even now manipulate the "jokerness" of your pile, by simply declaring that "all Spades on the cards should be considered as Hearts, all Hearts as Clubs etc." This you can do in the remote room also.
After you made the declaration, you may still hold the joker in your pile - or the joker will be in the other pile.

If you then look through the deck (and make the declared adjustments for the suits), and asking for the hidden note - you then will know if you have the joker in your pile or if the joker is in the other pile. Of course the jokerness of both piles will still be opposite.
For the observer it looks like you somehow instantly moved the joker from the remote room to the other pile (or vice versa). In fact you didn't.

Quantum mechanics is similar - which the "minor" detail that there is no hidden note which card the joker will be. You have a tool for measuring the jokerness of the pile you hold in your hand, and you have the ability to rotate the jokerness of your very pile (if you would know the joker note).

The observed results are the same as the photon experiments. The only difference - and that is fundamental - there is no hidden note).
 
#33
Thanx Mango

I guess I was being taught by a "naive" observer. Unfortunately that is all to often the case. The ones that really understand the concepts tend to work in the field. The ones that merely performed on the tests tend to teach.
 

MangoJ

Well-Known Member
#34
tthree said:
I guess I was being taught by a "naive" observer. Unfortunately that is all to often the case. The ones that really understand the concepts tend to work in the field. The ones that merely performed on the tests tend to teach.
Naive is not a judgement. I guess we all are naive with respect to quantum mechanics.
QM tells us the rules of the game. It doesn't mean we know how to play it.
 
#35
MangoJ said:
Naive is not a judgement. I guess we all are naive with respect to quantum mechanics.
QM tells us the rules of the game. It doesn't mean we know how to play it.
I didnt mean to make it sound like you were being judgemental. Only that that was what my professor taught us and he was clearly didnt understand. I had one professor whose lectures I always chose to miss after sitting through a dozen or so. He either didnt know anything about what he was teaching or couldnt communicate at all. I felt so sorry for the other students that were hoping the professor would spoon feed them the subject. If I didnt know he was wrong by already having read the material he was covering I would have been lost. I ran a Q&A session for any students that wanted to be helped the evening of his lectures.
 

k_c

Well-Known Member
#38
There is no paradox.

Gamblor said:
Say somebody offers you a game. There are 2 envelopes, one has X money, the other has 2X the amount.

You randomly choose an envelope, open it up, and it contains $100. Your given the choice to take whats in the other envelope instead. Probability theory tells us that we should switch to other envelope because the expected EV is:

0.5 x $50 + 0.5 x $200 = $125

WTF? This completely defies common sense, you would think it should not matter. Taken to extreme, lets say the envelope has X or 2X money again. You randomly choose envelope A (and not peak this time). The EV of envelope is B is 1.25 * X. So you take envelope B. But then, given the choice to take envelope A, its EV is 1.25 * X * 1.25, ad infinitum.

This is the "two envelop paradox" http://en.wikipedia.org/wiki/Two_envelopes_problem. There is no widely agreed upon consensus on how to solve this problem. At the very least there is no simple explanation, which you would think there would be.
1 envelope contains A dollars, the other contains 2*A dollars
Choose 1 envelope: EV before checking amount = .5*A + .5*(2*A) = 1.5*A

Open an envelope and see with 100% certainty the amount it contains. 100% of the time you know that 50% of the time you have chosen the envelope containing A dollars and 50% of the time you have chosen the envelope with 2*A dollars.
EV = 1*(.5*A + .5*(2*A)) = 1.5*A for either switching or not switching.
 
#39
statistics can say anything

k_c said:
1 envelope contains A dollars, the other contains 2*A dollars
Choose 1 envelope: EV before checking amount = .5*A + .5*(2*A) = 1.5*A

Open an envelope and see with 100% certainty the amount it contains. 100% of the time you know that 50% of the time you have chosen the envelope containing A dollars and 50% of the time you have chosen the envelope with 2*A dollars.
EV = 1*(.5*A + .5*(2*A)) = 1.5*A for either switching or not switching.
Here is someone who can make statistics say the truth. Good analysis. Im still not sure if you were trying to be funny but who can find fault with that analysis.
 

Gamblor

Well-Known Member
#40
k_c said:
1 envelope contains A dollars, the other contains 2*A dollars
Choose 1 envelope: EV before checking amount = .5*A + .5*(2*A) = 1.5*A

Open an envelope and see with 100% certainty the amount it contains. 100% of the time you know that 50% of the time you have chosen the envelope containing A dollars and 50% of the time you have chosen the envelope with 2*A dollars.
EV = 1*(.5*A + .5*(2*A)) = 1.5*A for either switching or not switching.
Well yes, we very well know before we choose either envelope, if we define one envelope as having A and the other 2A, either envelopes have an EV of 1.5 * A, common sense tells us so. The point being, lets say you choose an envelope, and find $100 dollars in it. The other contains either $50 or $200. Thus, the other envelope "magically" has a higher EV ($125) which would seemingly compel us to pointlessly choose the other envelope.
 
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