Talk:Two envelopes problem/Arguments

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Comment This page is devoted to discussions and arguments concerning the two envelopes problem itself. Previous discussions of this kind have been moved from the main talk page, which is now reserved for editorial discussions only.


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[edit] Simple Solution to Hardest Problem

"Let the amount in the envelope you chose be A. Then by swapping, if you gain you gain A but if you lose you lose A/2. So the amount you might gain is strictly greater than the amount you might lose."

Yes, but if you swap again, as it's the FIRST envelope that has A dollars, then by switching again, you either gain A/2 or lose A. Therefore, you can't switch forever and gain forever. It operates under the false assumpation that whatever envelope you have has A, but that A is also a constant. You can't have both.

                 -t3h 1337 r0XX0r

Oh, just noticed it's irrelevance. Sorry, won't do this again... 1337 r0XX0r 15:39, 19 January 2006 (UTC)


Its irrelevance? I'm confused. The solution to the hardest problem seems pretty simple... you are switching the amount in the envelope depending on if you look at it like it has more, or less, from Y to 2Y. You could do the same thing with Y (you gain a Y if the other envelope has more) and 100Y (you lose 50 Y if your envelope had the most). ~Tever

  • "To be rational I will thus end up swapping envelopes indefinitely" - This sentence is clearly false. After the first swap, I already know the contents of both the envelopes, and were I allowed a second swap, the decision whether on not to do it is simple. - Mike Rosoft 23:42, 24 March 2006 (UTC)

No, you never look into any envelopes in the original statement of the problem. INic 21:14, 27 March 2006 (UTC)

[[ The solution is simple. Everyone is trying to discern the probability of getting the greater amount, thus gaining the most out of the problem. However, if you were to not open either one, and continiously follow the pattern of swapping between envelopes, you would thus gain nothing. So, in order to gain the most, one would have to be content upon gaining at all, thus gaining something, and thus gaining the most out of the probablity, for without taking one, you gain nothing. ]] - Apocalyptic_Kisses, April 6th, 2006

You are right. But what rule of decision theory do you suggest should be added to allow for this behaviour? It seems to me to be a meta-rule rather than an ordinary rule. Something like "When decision theory leads to absurd results please abandon decision theory!" INic 17:43, 18 April 2006 (UTC)

[edit] What's the problem?

Can someone explain why this is such a "paradox"? It seems to me to be so much mathematical sleight of hand. Representing the payoffs as Y and 2Y or 2A and A/2 are all just misdirection. Just being allowed to play the game means you're getting paid Y. The decision involved is all about the other chunk. So you've got 1/2 chance of getting it right the first time and 1/2 of getting it wrong. Switching envelopes doesn't change that... 1/2*0 + 1/2*Y = Y/2, which, added to the guaranteed payoff of Y gives 3Y/2, which is the expectation that we had before we started playing. 68.83.216.237 03:59, 30 May 2006 (UTC)

The problem is not to find another way to calculate that doesn't lead to contradictions (that is easy), but to pinpoint the erroneous step in the presented reasoning leading to the contradiction. That includes to be able to say exactly why that step is not correct, and under what conditions it's not correct, so we can be absolutely sure we don't make this mistake in a more complicated situation where the fact that it's wrong isn't this obvious. So far none have managed to give an explanation that others haven't objected to. That some of the explanations are very mathematical in nature might indicate that at least some think that this is a subtle problem in need of a lot of mathematics to be fully understood. You are, of course, free to disagree! INic 21:17, 30 July 2006 (UTC)

[edit] The problem is logically flawed

The solution of the paradox is that it is incorrect to assume that, with just two amounts of money available, you have a fixed amount of money in your hand and still the possibility of getting a larger and smaller amount by swapping (this isn't changed at all by opening the first envelope).

As an example, assume that there are two envelopes on the table, one with $50 and one with $100. If you choose the $50 envelope first, you gain $50 by swapping, and if you choose the $100 envelope first, you lose $50 by swapping, so on average you neither gain or lose anything.

The formula resulting in a gain of 5/4 of the original amount applies only if there are three amounts of money available (with the ratios 4/2/1) and you have the middle amount in the first place. So if you have $50 and there are $100 and $25 on the table you gain (0.5*$100 +0.5*$25 -$50) =$12.50.
If on the other hand you have initially $25, you gain (0.5*$100 +0.5*$50 -$25) =$50 ,
whereas if you have the $100 initially, you lose (0.5*$50 +0.5*$25 -$100) = -$62.50 ,
so on average you wouldn't gain or lose anything either with this 'three envelope' situation.

Thomas

[edit] The problem of measuring gain/loss

I suggest that the problem exists with the fact that the "gain" or "loss" of the envelope holder is being considered. Assuming A is the smaller amount in one of the envelopes, then the average amount of money to be holding after picking the first envelope at random is:

{1 \over 2} A + {1 \over 2} 2A = {3 \over 2}A

Contrarily, assuming A is the larger amount in one of the envelopes, the average amount of money to be holding after picking the first envelope at random is:

{1 \over 2} A + {1 \over 2}{A \over 2} = {3 \over 4}A

In both cases, after having picked the first envelope, the envelope holder is unable to determine whether the amount in his/her envelope is the greater or lesser amount. If he/she were to assume that he/she had the average amount, in both cases, he/she would realise that he/she had the same amount to gain/lose in switching envelopes ({A \over 2} and {A \over 4} respectively).
--TechnocratiK 16:56, 22 September 2006 (UTC)

[edit] Another proposed solution

Motion to include this in the main article... its proposed by me. All in favour say "I"... seriously though, please give me feedback (rmessenger@gmail.com if you like), and please read the whole thing:

First, take A to be the quantity of money in the envelope chosen first, and B to be that of the other envelope. Then take A and B to be multiples of a quantity Q. Even if the quantity in A is known, Q is not. In this situation, there are two possible outcomes:

(I) A = Q, B = 2Q
(II) A = 2Q, B = Q

The question is whether we would stand to benefit, on average, by taking the money in B instead of our random first choice A. Both situations (I) and (II) are equally likely. In situation (I), we would gain Q by choosing B. In situation (II), we would lose Q by choosing B. Thus, the average gain would be zero:

{1 \over 2} Q + {1 \over 2} (-Q) = 0

The average proportion of B to A is irrelevant in the determination of whether to choose A or B. The entire line of reasoning in the problem is a red herring. It is true that the average proportion of B to A is greater than one, but it does not follow from that determination that the better option is to choose B.

For Example: Assume one of the envelopes contains $100 and the other $50. The two possibilities are:

(I) A = $50, B = $100
(II) A = $100, B = $50

If you repeat the above event many, many times, each time recording the following:

{R_i}={{B_i} \over {A_i}}, {A_i}, {B_i}

then \bar{R} will approach {5 \over 4}, and both \bar{A} and \bar{B} will approach $75. The value of \bar{R} is totally irrelevant. What is relevant is that \bar{A} = \bar{B}. This means that it makes no different which envelope you choose. On average, you will end up with the same amount of money, which is the expected result.

Step 8 assumes that because \bar{R} \neq 1, one will gain on average by choosing B. This is simply and plainly false, and represents the source of the "paradox."

I like your reasoning! The only reason I deleted your addition of it was that it was original research, and that is strictly forbidden. However, it's not forbidden here at the talk page (but not encouraged either). I think you should elaborate on your idea some and try to get it published somewhere. What I lack in your explanation right now is a clear statement of the scope of your reasoning; how can I know in some other context that the expected value is of no use? I don't think you say it's always of no use, right? And another thing that is interesting with your solution is that you say it's correct to calculate the expected value as in step 7, only that it's of no use. But if it's of no use, in what sense is it correct? iNic 22:46, 13 November 2006 (UTC)
No, I was wrong about that. The expected value calculation is done wrong. This is because we are supposed to be comparing the expected values of two options. One option is to keep A, the other is to go with B. If we do two different EV calculations for both, we will see that we can expect the same value for each. Take Q to be the smallest value between the two envelopes. There are two equally likely possibilities for each option: we hold the greater amount, or we don't. If we hold the greater amount, we hold 2Q, if not, we hold Q; as such:
E_A = {1 \over 2}Q + {1 \over 2}2Q = {3 \over 2}Q
E_B = {1 \over 2}Q + {1 \over 2}2Q = {3 \over 2}Q
E_A = E_B = {3 \over 2}Q
They are equal, so we can expect the same amount of money regardless of whether we keep A or go with B. Amen.
And, their equation for EB is wrong for one simple reason: the value of A is different in both of the two possibilities. As soon as the two envelopes are on the table, the value of Q never changes. Since we don't know Q from opening one envelope, A must take on two different values in the two terms of their EV equation. A is not a constant in the equation even if we know what it is. Here's their equation:
E_B = {1 \over 2}{1 \over 2}A + {1 \over 2}2A
In the first term, A=Q, in the second, A=2Q. Since Q is constant, A must change. An EV equation must be written in terms of constants. Since theirs isn't, it's wrong.
Think of it this way: since there are two different possibilities, their are two different A's. You don't know which one you have:
Possibility 1: A1 = 2Q,B1 = Q
Possibility 2: A2 = Q,B2 = 2Q
So their EV equation would look like this:
E_B = {1 \over 2}{1 \over 2}A_1 + {1 \over 2}2A_2
OK, so this means that you now agree with all other authors that step 7 and not step 8 is the erroneous step in the first case? And you seem to claim that step 7 is the culprit even in the next case where we look in the envelope, as you say that "A is not a constant in the equation even if we know what it is." Does this mean that you disagree with the common opinion that it's step 2 that is the erroneous step there? And what about the next variant? Is the expected value calculation not valid there either? And how can i know, in some other context, that a number, say 1, is not really a constant? You still have to specify how we can avoid this paradox in general. As long as you haven't generalized your idea to a general rule you haven't really proposed any new solution, I'm afraid. iNic 02:26, 15 November 2006 (UTC)
Your right. But its simple: if you forget probability for a second, and imagine that as soon as the envelopes are on the table, you repeat the possible outcomes of the event millions of times. Each time the event takes place, there is some Ai and Bi which are the A and B values you got that time. These are different each time you do it. So as a simple rule, they can play no part in an expected value calculation; unless A actually was a constant. Here's where it gets interesting: It seems like the same situation, but its not. If we imagine A is a constant, and if the same event were repeated, A would not change, as is explained in steps 1-6. If you model the situation this way (I have done computer models for both), such that the experimenter, if you will (the computer), first chooses a value for A, then randomly selects a value for B as a function of A based on steps 1-6, then the expected value of B actually does end up equaling five fourths of the EV of A!
If instead you tell the experimenter to first choose a value to be the smallest value, then choose which envelope to put it in, and put two times that in the other: The result is as you would expect for the real life scenario. So while the possibility of B>A is the same for each instance of both situations... and the two possible values of Bi with respect to Ai are the same, 'tis not the same situation.
I can make it even clearer how there are two different situations being represented by separating out the probabilistic variable:
Starting with the true event: the experimenter chooses a smallest value, call it Q, and puts it in one envelope, and puts 2Q in the other. Then, lets say you flip a coin: if it's heads, Zi = 0, if tails, Zi = 1 for instance... now observe:
Ai = (1 + Zi)Q
Bi = (2 − Zi)Q
B_i = {{2-Z_i} \over {1+Z_i}}A_i
As you can see, now we have an unchanging algebraic expression for all the variables. Its easy to see the results of the coin toss simply dictates which envelope you pick up. Please note that (referring to the last expression) Bi is either (1/2)A or 2A with equal probability. Its also easy to write our expected value expression! There is only one (well understood) probabilistic event: a coin toss. We know the expected value of Z is exactly 1/2. Plug it in and we get what we would expect: EB = EA = (3 / 2)Q, which holds up regardless of whether Q is the average of a constantly changing value, or simply a constant through all instances. NOW, the "OTHER EVENT":
The experimenter first chooses a value for A, then chooses Bi based on steps 1-6. If we accommodate the coin toss, we get:
B_i = ({1 \over 2} + {3 \over 2}Z_i)A
First, notice that if Z=0, Bi = (1 / 2)A, and if Z=1, Bi = 2A, and since the probability of either is equal, we would expect that this is identical to the true event, but it's not! If we try to solve for the expected value of B, we plug in our known expected value for Z and what do we get? five fourths. What a surprise! So its easy to see that, while these too situations appear similar, the similarities break down when Z assumes value other than 0 or 1. It's like saying: I'm a function with roots at 2, 4 and 7.. what function am I? There can be many seemingly similar situations at first glance, with the primitive tools used to analyze this problem. But they behave differently.
So the moral: If it changes from instance to instance AND its expected value isn't already known, it CAN'T be used as a term in an expected value calculation! Just as you'd expect! We can't even use Bi to find the expected value of B! because if an EV equation was written in terms of a variable that changes all the time, then the expected value would always be changing!
And specifically addressing the problem: Step one fine IF we remember that they have set A to equal the money in the envelope THIS TIME. Step 2 is testably TRUE. Steps 3-5 are stating the obvious. Step 6 is the problem. It is true within the parameters of step 1: that is, AT EACH INSTANCE, half of the time, B=2A, and the other half, B=(1/2)A. But this relates specifically to values of Bi and Ai. As I stated, if we want to calculate expected value in this type of situation, It must be written in terms of ALREADY KNOWN expected values! We don't already know the expected values of either A or B, hence we can't squeeze blood from a turnip!
In the true situation, there are two truely random probabilistic variables: Q and Z. A and B are simply functions of Q and Z (Z being the binary decision, Q being the random number). In their situation, they took the random probabilistic variables to be A and Z, and let B be a function of those. This is different because it assumes that the probability distribution of A is totally random, when in fact it depends on that of the true independent variables (the ones that are actually chosen at random!!!) A was never chosen at random! only Q and Z are!
When determining the expected value of a variable: Express the variable at any instance in terms of the random decisions or numbers that compose it. This equation will always hold true when relating the expected value of the variable to the known expected values of the random decisions or numbers.
i.e. suppose I said I was giving away between 0-5 dollars to one of two people, AND between 10-15 dollars to the other. You're one of those people, and.. say, Joe is the other. so Y is the amount of money you just got, and J is the amount Joe got. Is it worth you're while to switch with Joe?
First, identify the variables: i am choosing between two people, so there is one binary decision Z (independent). There is the amount of money between 0-5 dollars, call it q (independent), and the other quantity between 0-5 dollars (to be added to 10), call it Q (independent). Y and J are dependent upon these as such:
Yi = Ziqi + (1 − Zi)(10 + Qi)
Ji = (1 − Zi)qi + Zi(10 + Qi)
And write expected value expressions as follow:
EY = EZEq + (1 − EZ)(10 + EQ)
EJ = (1 − EZ)Eq + EZ(10 + EQ)
We know EQ = Eq = 2.5;EZ = 1 / 2; As such:
EY = EJ = $7.50
The point is, had you assumed your money Yi was an independent variable and somehow tried to define poor Joe's money Ji in a given instance in terms of your own, you would have gotten it wrong! Neither J nor Y are independent variables, hence we know nothing of their expected values, SO: if we had an expression for J in terms of Y, we wouldn't be able to fill in the expected value of Y, because its dependent! We would be forced to do something tragic.. assume Y is a constant, and define Joe's money in terms of this magical made up constant! Therein lies the mistake! Note: When I give you your money, you know Yi. This doesn't mean you know anything about the infinitely many other possible values of Y!

[edit] History of the problem

Someone correct me if I'm wrong here, but in the original problem

Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "Suppose that I have the amount A in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2A. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man?

the mistake in reasoning is two-fold. The first mistake is that the probability of winning is not 0.5. If they both are equally rich, then they both have a total wealth of W. The amount in one wallet would be any amount A such that 0 <= A <= W. So the probablity that A is greater than the amount in the other wallet is A/W, which may or may not be 0.5.

Second, if A is greater than B, then A + B < 2A and the amount in his possession at the end of the game cannot be more than 2A.

Fryede 20:42, 11 December 2006 (UTC)

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