Mathematical Conundrum or Not? Number Six
Not sure I can find another conundrum that will generate as much discussion as our last one; however, it is time to move on.
So this one is called the Two Envelopes Paradox and this is how it goes:
What should you do?
So this one is called the Two Envelopes Paradox and this is how it goes:
You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other $2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?
What should you do?
Comments (1160)
That's fair enough; I wasn't actually criticising those who are participating in this thread, just expressing my surprise that the OP could be thought to be of any more interest than any kind of logical or intellectual puzzle or conundrum. I acknowledge that I may be missing something due to ignorance, though.
Probability is a big philosophical topic. It is quite tightly enmeshed with both metaphysics and epistemology. Michael Ayers wrote a lovely book, The Refutation of Determinism, which explores some of the philosophical problems associated with the concepts of probability, necessity and possibility, and also pursues some implications for the problem of free will and determinism. There is also a close connection with epistemology, Gettier examples, and some of the most puzzling paradoxes of Barn Facade County, which arise, it seems to me, from assumptions regarding the grounding of knowledge that are closely related to some of the assumption that give rise to the two-envelope paradox. Maybe I'll create a new topic about this when time permits.
A first step, which is being pursued in this thread, is to get clear on (what should be) the uncontroversial steps in the mathematical reasoning.
I'm not sure which of @JeffJo's examples you're referring to.
As for my "tree" and what it predicts -- You face a choice at the beginning between two values, and the same choice at the end between those same two values. If you flip a coin each time, then your expectation is the average of those two values both times and it is unchanged.
Opening an envelope changes things somewhat, but only somewhat. It gives more substance to the word "switch", because having opened one envelope you will never be allowed to open another. You are now choosing not between two envelopes that can be treated as having equal values inside, although they do not, but between one that has a known value and another that cannot be treated as equal in value.
But it is, for all that, exactly the same choice over again, and however many steps there are between beginning and end, your expected take is the average of the values of the two envelopes. If there's an example in which that is not true, I would be surprised.
I have come, in broad terms, to see probability as a generalization of logic, or logic as a special case of probability, take your pick. I would credit Frank Ramsey for convincing me to begin thinking this way. As my principle interest is the nature of rationality, what's of interest here -- decision theory, broadly -- is still the nature of inference.
What we have been arguing about for fifty pages is what inferences are justified and what aren't.
The OP asks, "What should you do?"
Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology.
Indeed. And just as is the case with Newcomb's problem, with the two-envelope paradox also, the dominance principle and the (maximum) expected utility principle appear to recommend inconsistent strategies when carelessly applied. Newcomb's problem is more controversial, even, than the two-envelope paradox. It is also quite rich in philosophical implications.
Definitely a good candidate for a discussion sometime!
I was not referring to a specific example but rather to his general resolution of the apparent problem. It both justifies the zero expected gain for the unconditional switching strategy and explains why the indifference principle can't be applied for inferring that the expected utility of switching is 1.25v, in the case where your envelope contains v.
I am not sure why you are saying that I am facing a choice rather than saying that I simply don't know whether my envelope is smallest or largest (within the pair that was picked). I am not facing a choice between two values. I am holding v, and I am offered to trade v for another envelope which, for all I know, might contain v/2 or 2v. One wrong inference that one might make is that just because I don't know whether the other envelope contains v/2 or 2v, and don't have any determinate means to estimate it, therefore the indifference principle applies and I can assign 1/2 to the probability of either cases. That would indeed yield an expected value of 1.25v for the second envelope. But that is a wrong inference from the fact that I don't know either P(v, v/2) or P(v, 2v). I only know that those two probabilities add up to one and their exact values are dependent on the prior distribution of possible envelope pairs. The only way for such an initial distribution to be such as to guarantee that P(v, v/2) = P(v, 2v) = 1/2 for any v would be for the initial distribution to be uniform and unbounded. For any other feasible way to randomly determine amounts of value pairs in accordance with some well behaved probability distribution, what the player can infer only is that, on average, after repeated trials, the expected value for switching tends towards zero.
I disagree. Suppose the initial distribution is, unbeknownst to you, ((5,10), (10,20), (20,40)). In that case, if you are being dealt 5, the expected value of sticking is 5. You don't know what the expected gain of switching is. But it's not the case that it is therefore zero. That would only be zero if you knew for a fact that (5, 10) is half as likely as (5, 2.5) in the prior distribution.
The way I look at the problem is that the expected gain calculation depends on an agent's knowledge. Some agents may have more information than others which they should factor into their calculation.
For example, the host may know what envelope pair has been selected (rendering the initial distribution irrelevant) but not know the amount in the player's chosen but unopened envelope. For the host (as for the player), the expected gain from switching is zero. However the host calculates this by averaging the two known amounts instead of averaging the unknown X and 2X.
When the player opens the envelope revealing the amount, the host will now know the actual gain (or loss) were the player to switch. So, from the host's perspective, the expected gain just equals the actual gain. The host now has all the relevant information.
Conversely, the expected gain that the player calculates will still be the unconditional gain of zero since she doesn't know the initial distribution or both amounts in the selected envelope pair.
A further perspective is held by those who know the chosen amount and also know the initial distribution but not which envelope pair was initially selected. So for the equiprobable ({$5,$10},{$10,$20}} distribution, the expected gains from switching for $5, $10 and $20 respectively are $5, $2.50 and -$10. The $10 amount is the only amount where not all the relevant information is known (i.e., which envelope pair it belongs to). But there is enough information to recommend switching should the player see the $10 amount.
Each expected gain calculation is justified given the information the agent has. But since that information can be different, each agent can (justifiably) reach different conclusions regarding whether the player should stick, switch or be indifferent.
This makes no sense to me. Initial distribution of what? If these are pairs of envelopes from which will be chosen the pair that the player confronts, then not only is this sample space unknown to the player, she never interacts with it. She will face the pair chosen and no other.
If that pair is {5, 10}, and she draws the 5, then she has the low value and she can only gain by switching with no risk of loss. The only case in which switching does not produce an actual gain or loss is when, contrary to the rules, the envelopes have the same value. By switching she gets the 10; but the 10 was there all along and she might have chosen it at the start. What doesn't change is your overall expectation from playing. Always Stick and Always Switch are not strategies that increase or decrease your expected take from playing.
Quoting Pierre-Normand
Why would your ignorance preclude you from facing a choice and making a decision? In the OP, you make at least two choices: which envelope to claim, and whether to keep it or trade it for the other. Whether you end up with the most or the least you could get depends on those two decisions and nothing else. What the amounts are depends on someone else.
Yes, it is tempting to say that if the game is only being played once then the shape of the initial distribution(*) isn't relevant to the definition of the player's known problem space. That's a fair point, and I may not have been sufficiently sensitive to it. This is broadly the argument @Michael and @Jeremiah are making, although they are reaching opposite conclusions regarding the expected gain from switching. (And they're, in two different ways, both right; hence the paradox). The player doesn't know what this distribution is, and, since she only is playing once, and hence only is being dealt one single envelope pair, how is it relevant what the other unrealized envelope pairs (and their probabilistic frequencies) were within this specific distribution? But it is actually relevant what the distribution might look like since there are logical constraints on the shape of that distribution that can be inferred from the assumptions that ground either strategies (i.e. switching or being indifferent to switching).
The advocate of the switching strategy is actually right in saying that if, even after she has seen that her envelope contains v, she still has no reason to assign an at least twice higher probability to the other envelope being v/2 rather than 2v, then she is justified in switching. The switching strategy can only be justified since, precisely, the prior probability distribution of this game, which is only being played once, doesn't interact with this player's problem space. When she gets to play such a game again, if ever, the prior distribution might be different.
What rather defines the problem space of this player, in accordance with the general specification of the problem that is provided to her (as described in the OP), is the range of possible envelope pairs (and their probabilistic weighs) that is being merely consistent with the general (and abstract) specification of the problem. There are however (roughly) two ways to generate such a range: bounded or unbounded. If it's unbounded, then the switching strategy is justified since the expected gain from switching is 1,25v conditionally on any v being found in the first envelope. If the range is bounded, because the player assumes that there is some finite amount of money M in the universe, however big M might be, then her problem space is such that the average expected gain from switching is zero. This is because her problem space is built up from some arbitrary weighing of all the possible bounded prior distributions, and all of those individually yield an average expected gain from switching that is zero, and so is any weighted sum of them.
(*) I am defining this initial distribution with reference to the method, unknown to the player, by means of which the initial contents of the two envelopes is effectively being determined, by means of a pseudo-random number generator, quantum device, or whatever.
Yes, absolutely, and this is specifically beyond the OP. The distributions we've been talking about have almost always been (or should have been) unknown to the player. The player doesn't even know that there is some selection process. There are values in envelopes. How they got there can be discussed, and that can be interesting when the player has, say, partial knowledge of that process, but it is not the source of the paradox, in my opinion.
Whereas, on my view, it is the source of the paradox ;-)
Yes! This is exactly what we disagree on.
-- For a change, I'm going to take a little time and think through my response. --
On the assumption, of course, that the player takes this initial distribution to be bounded above by M for some (possibly uncknown) M; or that, if unbounded, its sum or integral converges.
Your ignorance doesn't preclude you from facing a choice and making a decision. What it precludes you from doing is basing your decision to switch (if you decide to) on a determinate expected gain, since the value of such an expected gain (conditionally on your having seen v in the first envelope) is unknown (and, in particular, is not known to be zero).
Only in the case where the distribution is uniform and unbounded can you know what the expected gain from switching is, and know it to be precisely 1.25v. (And that's because, in the unbounded case, it is precluded that you might ever hit the top of the distribution(*) and hence lose in one fell swoop all the expected gains potentially accrued from the other cases in the average time that it takes for this large loss to randomly occur). In that unbounded case, you would know that switching will earn you either v/2 or 2*v with equal probability. It merely appears paradoxical, in that unbounded case, that you are always entitled to switch but that, nevertheless, if you were to switch blindly the first time, and only open the second envelope to find some value w, then, you would still expect 1.25w from switching back. It is this apparent paradox, occurring with infinite and uniform distributions, that my Hilbert Grand Hotel analogy was meant to illustrate.
(*) I am only considering uniform probability distributions over elements of a single discrete doubling sequence, for simplicity. We can also assume it to be truncated below, at $1, say.
Here is an improvement on my earlier Hilbert Grand Hotel analogy to the two-envelope paradox. The present modification makes it into a much closer analogy.
Rather than considering an Infinite Hotel where the countably infinitely many rooms are numbered with the natural numbers, this time, they will be numbered with the (also countably infinitely many) strictly positive rational numbers. Hence, for instance, room 3/2 will be located midway between room 1 and room 2, and will be small enough to fit between them. Guests of the Hilbert Rational Hotel who want to go into rooms corresponding to rational numbers that have very large denominators will have to swallow very many magic shrinking pills, let us assume (and may have to leave their luggage behind). Initially, infinitely many guests are being randomly distributed such that there are, on average, about one hundred guests in each room. Every morning, each guest is awarded $Q, where Q is their rational room number. Every night, each guest is being given the opportunity to flip a coin to determine if she will move from room Q to either room Q/2 (if she lands tails up) or to room 2*Q (if she lands heads up). Let us suppose that all the guests are greedy and rational and, hence, all choose to flip the coin (rather than stay) in order to increase their expected earnings to 1.25*Q on the next morning. (If we take a random sample of guests within any given bounded segment of Hilbert's Rational Hotel, it is clear that they fare better, on average, than they would if they had all chosen to stay. On average, they fare 1.25 times better.)
The first thing to notice is that, after the daily reshuffling of guests, the average population of the rooms stays the same since each room is accessible from exactly two other rooms and as many guests, on average, move from room Q to room 2Q, for any Q, than the reverse. So, although any finite random sample of guests improves its fare 1.25 times, on average, the average room occupancy doesn't change and the overall population density (measured as guests per room) on any segment of the hotel doesn't vary at all.
Now, suppose that on some particular nights -- on Sunday nights, say -- all the guests swallow a pill that makes them forget what room it is that they had moved from on the previous night. They are then offered the opportunity to blindly move back to whatever room it is that they previously occupied. If they are rational and greedy, they ought to choose to move back since, if they now are in room Q, it is equally likely that they came from room Q/2 than it is that they came from room 2*Q. Hence, on average, the guests who move back to the room where they came from fare 1.25 times better than they would if they stayed.
This explains why, likewise, in the two-envelope game, with an unbounded and uniform distribution, (if such a thing is intelligible at all), it appears rational to switch envelopes in order to increase one's expected value by 1.25 and, nevertheless, after one has switched blindly, it would appear to be equally rational to switch back in order to increase one's expected value by 1.25. Such distributions, though, can no more be instantiated in real games than Hilbert's Grand Hotel can be built.
Right, the player doesn't have that perspective but we do when discussing hypotheticals with specific distributions.
Quoting Srap Tasmaner
What is the source of the paradox in your view?
As I see it, the source of the paradox is the idea that the player can calculate the expected gain conditional on the observed amount. That would require knowing the specific probabilities for {v/2,v} and {v,2v} which is just what she doesn't know.
Quoting Pierre-Normand
Yes, which I take to be a reasonable (physical) assumption.
I like this explanation. And I thought of a possibly better way explain how "unknown" is used in the TEP, by analogy:
Even if you treat the independent random variable V, for the value v in your envelope, (or the smaller value x of the pair) "as an unknown," the value y in the other envelope is still represented by the dependent random variable Y that has a very specific relationship to X, as determined by the OP and the laws of probability.
The point is that "as an unknown" is a description of your knowledge, not the role played by a value in the problem.
What you "cannot" do is assign probabilities to the cases. You can still - and my point is "must" - treat them as random variables, which have unknown probability distributions.
Quoting Srap Tasmaner
Define "interact."
Say I reach into both of my pockets, pull out a coin in each hand, but show you just one. It's a quarter, worth $0.25. (According to the US mint, their current mint set includes coins worth $0.01, $0.05, $0.10, $0.25, $0.50, and $1.00.) I'll give you either the coin you see, or the one you don't. What do you choose?
Notice I didn't say that I ever carry half-dollar or dollar coins. They are in the current mint set, but most Americans are unfamiliar with them so they won't carry them unless they were just given out as change.
In what way do you not "interact with" with the distribution of coins I carry? If I keep nothing but pennies in one pocket, does that fact not "interact with" what you will get if you take the hidden coin? Even if you do not know the distribution?
How about if I received $9.50 in dollar and half-dollar coins as change for a $10 bill from a vending machine (about the only way I get them), and put them in a different pocket so I can avoid carrying them tomorrow? Does that fact not "interact with" what you will get if you take the hidden coin? Even if you do not know the distribution?
In both the TEP, and this example, the distribution is unknown. You can't "interact" with any specific realization of a distribution, but it most definitely "interacts" with what will happen if you choose the hidden envelope or coin. That's why the answer to OP is:
No, it isn't fair to say that. No more than saying that the probability of heads is different for a single flip of a random coin, than for the flips of 100 random coins
Well, that's missing the point ;-) My purpose was to highlight a point that @Srap Tasmaner, yourself and I now seem to be agreeing on. Suppose, again, that a die throwing game will be played only once (i.e., there will be only one throw) with a die chosen at random between two oppositely biased dice as earlier described. The point was that, in the special case where the die that has been picked at random is biased towards 6, and hence the "initial distribution" is likewise biased, this fact is irrelevant to the "problem space" that the player is entitled to rely on. The initial coin picking occurred within the black box of the player's ignorance, as it were, such that the whole situation can still be treated (by her) as the single throw of a fair die.
This is what I took @Srap Tasmaner to mean when he said that "the player" doesn't "interact" with the "sample space", which is a statement that I glossed as the player's "problem space" (maybe not the best expression) not being (fully) constrained by the "initial distribution" of envelope pairs, which must be treated as being unknown but, as you've pointed out repeatedly, still belongs to a range that is subject to definite mathematical constraints that can't be ignored.
This is an important separation to keep in mind, as the conceptual difference between these approaches to probability is not the number of observations, I to fell into that misleading line of thought myself, but the real difference is in the direction in which they move. One moves from a known population to the sample, while the other moves from the sample to the unknown population. To take a "probabilist" approach to the unknown population is to go in ill equipped to do that job. I think a few of the people here have little experience in dealing with unknown populations and in that naiveness they have moved in the wrong direction.
Thanks for this explanation Pierre. It's probably a little outside my are of philosophical interest, though, which would be due to lack of time as much as anything else.
That seems to be as good a reason as any...
I do find it hard to think of logic as a form of probability. But perspectives do rely upon backgrounds, and no doubt I lack the appropriate background.
Are you saying that the problem in the Op highlights a general question as to what inferences, or rather perhaps, what kinds of inferences, we should accept?
Let [math]x_{ij}, i = 1, 2, 3,...,n, j=1,2,3 [/math] ( j is which chance mechanisms we are at) denote the observed value of the random variable [math]X[/math]. Then [math]P(X=x_{i,1})[/math], which is the sum of probabilities for the sample points that are assigned to [math]x_{i,1}[/math].
Sorry if my nested notation is a bit off, it has been awhile and I don't feel like looking it up, but basically we need another index to tell which chance mechanisms we are at, in our case I am using j.
Since [math]X[/math] is a random variable then it is a response variable of some unknown real-world function in which the unknown domain is the sample space. The domain would actually spreads between two unknowns. I know zero has often been the assumed min, but we actually have no idea what that is; as far as we know 100 bucks could be the lowest value.
Fist chance mechanism.
By whatever function of [math]X[/math] an [math]x_{ij}[/math] will be selected, we'll call this value [math]x_{1,1}[/math].
Second chance mechanism
We have two envelopes, lets call one [math]A_i[/math] and the second [math]B_i[/math]. We need to define [math]A_i[/math] and [math]B_i[/math] at this point. So using a simple if then statement let's define them as: If [math]A_i = \{x_{i,2}\}[/math] then [math]B_i = \{2x_{i,2}\}[/math] or if [math]A_i = \{2x_{i,2}\}[/math] then [math]B_i = \{x_{i,2}\}[/math].
At this level of the nested structure [math]x_{1,2}[/math] is put into one envelope while [math]2x_{1,2}[/math] is put into the other envelope.
The possible assignments are:
[math]A_1 = \{x_{1,2}\}[/math]
[math]B_1 = \{2x_{1,2}\}[/math]
[math]A_2 = \{2x_{1,2}\}[/math]
[math]B_2 = \{x_{1,2}\}[/math]
So the two possible cases are:
[math]E_1 = \{A_1, B_1\}[/math]
[math]E_2 = \{A_2, B_2\}[/math]
Then [math]P(E_i) = 1/2[/math], with [math]P(A_1) = 1/4 = P(x_{1,2}) = P(B_2)[/math] and [math]P(A_2) = 1/4 = P(2x_{1,2}) = P(B_1)[/math], meaning [math]P(x_{1,2}) = 1/2[/math] and [math] P(2x_{1,2}) =1/2[/math].
Third chance mechanism
The third chance mechanism is the subjective one, as it up to us to decide by what means what we want to do and I think this is the real trick to this so-called "paradox". It says, "What should you do?", and that point is really where the conflicts arises.
I still maintain that our gain/loss is: [math] P(-x_{1,3}) = 1/2, P(x_{1,3}) = 1/2 [/math]. However, I set this up, because I think it is important people understand where and how probabilities are being applied.
Quoting Andrew M
I believe there is not a paradox here but a fallacy.
Outside of being told by reliable authority "You were successful!" you need to know two things to know whether you have been successful:
That these are different, and thus that your uncertainty about one is not the same as your uncertainty about the other -- although both contribute to your overall uncertainty that you were successful -- can be readily seen in cases where probabilities attach to each and they differ.
Here are two versions of a game with cards. In both, I will have in my hand a Jack and a Queen, and there will be two Jacks and a Queen on the table. You win if you pick the same card as I do.
Game 1: the cards on the table are face down. I select a card from my hand and show it to you. If I show you the Jack, then your chances of winning are 2 in 3.
Game 2: the cards on the table are face up. I select a card from my hand but don't show it to you. You select a card. If you select a Jack, your chances of winning are 1 in 2.
In both cases, winning would be both of us selecting a Jack, but the odds of my choosing a Jack and your choosing are different. In game 1, you know the criterion of success, but until you know what you picked, you don't know whether you met it; in game 2, you know what you picked meets the criterion "Jack", but you don't know whether the winning criterion is "Jack" or "Queen".
(If you buy your kid a pack of Pokemon cards, before he rips the pack open neither of you know whether he got anything "good". If he opens it and shows you what he got, he'll know whether he got anything good but you still won't until he explains it to you at length.)
Let's define success as picking the larger-valued envelope. There is a fixed amount of money distributed between the two envelopes, so half that amount is the cutoff. Greater than that is success. One envelope has less than half and one envelope has more, so your chances of meeting that criterion, though it's value is unknown to you, are 1 in 2. After you've chosen but before opening the envelope, you could reasonably judge your chances to be 1 in 2.
You open your envelope to discover 10. Were you successful? Observing 10 is consistent with two possibilities: an average value of 7.5 and an average value of 15. 10 meets the criterion "> 7.5", but you don't know whether that's the criterion.
What are the chances that "> 7.5" is the criterion for success? Here is one answer:
We know that our chance of success was 1 in 2. Since we still don't know whether we were successful, our chance of success must still be 1 in 2. Therefore the chance that "> 7.5" is the criterion of success must be 1 in 2.
This is the fallacy. You reason from the fact that, given the criterion of success, you would have a 1 in 2 chance of picking the envelope that meets that criterion, to a 1 to 2 chance that the unknown criterion of success is the one your chosen envelope meets.
(No doubt the temptation arises because any given value is consistent with exactly two possible situations, and you are given a choice between two envelopes.)
You can criticize the conclusion that, for any value you find in your envelope, the two situations consistent with that value must be equally likely, but my criticism is of the inference.
Now since we do not know anything at all about how the amounts in the envelopes were determined, we're not in a position to say something like "Oh, no, the odds are actually 2 in 7 that '> 7.5' is the criterion." So I contend the right thing to say now is "I don't know whether I was successful" and not attach a probability to your answer at all. "I don't know" is not the same as "There is a 50% chance of each."
You can reason further that one of the two possible criteria, "> 7.5" and "> 15", must be the situation you are in, and the other the situation you are not in. Then you can look at each case separately and conclude that since the value in the unopened envelope is the same as it's always been, your choice to stick or switch is the same choice you faced at the beginning of the game.
If you switch, you will turn out to be in the lucky-unlucky or the unlucky-lucky track. If you don't, you will turn out to be in the lucky-lucky or the unlucky-unlucky track.
This is a very neat analysis and I quite agree. One way to solve a paradox, of course, is to disclose the hidden fallacy in the argument that yields one of the two inconsistent conclusions. Your explanation indeed shows that, in the case where the criterion of success is unknown, and we don't have any reason to judge that the two possible success criteria are equiprobable, then it is fallacious to infer that the expected value of switching is 1.25v (where v is the value observed in the first envelope).
This is consistent with what I (and some others) have been arguing although I have also highlighted another source of the paradox whereby the equiprobability assumption (regarding the two possible albeit unknown success criteria) would appear to be warranted, without any reliance on the fallacy that you have diagnosed, and that is what occurs in the case where the prior distribution of possible envelope pairs which the player is entitled to judge to be possible is assumed to be uniform and unbounded. This is the ideal case, physically impossible to realize in the real world, that I have illustrated by means of my Hilbert Rational Hotel analogy.
Yes, I believe it is entirely consistent with criticizing the conclusion of the faulty inference. I think we would like to believe that invalid inferences can always be shown to have undesired consequences, but that also requires agreement on (a) the further inference, and (b) the undesirability of the consequence. I suggested at one point in this thread that if told the value of the other envelope instead of your own, then you would want not to switch; I found this conclusion absurd but my interlocutor did not. Go figure.
That was a very pointed challenge, and I think it has force in the case where the analysis is assumed to apply to a game that can be instantiated in the real physical world and that can be played by finite creatures such as us. But in the ideal case where the prior distribution is assumed to be uniform and unbounded, then, the conclusion, although seemingly absurd, would be warranted.
Claiming this case is "ideal" is an entirely subjective standard pumped full of observational bias.
Now not only do we have an unknown population but we have an unknown function and that means unknown exploratory variables. Maybe Farmer Bob counted how many eggs his chickens laid the night before and that is how x was decided.
I know we are reaching an equivalent conclusion. My point is that the framework that it fits into may be different. These concepts can seem ambiguous to many, which is the fuel Bayesians, Frequentists, Subjectivists, Objectivists, Statisticians, and Probablists use to denigrate each other through misrepresentation.
My point was that the difference between "only once" and "many times" has no significance in this discussion. It can only have meaning to a statistician who is (correctly, don't think I am putting them down) trying to create a population through repetition of the game, from which he can use inference to refine his estimates of the properties of your dice.
Probability models what is unknown about a system. At the start of your dice game, we don't know which die will be rolled, or how it will land. After step 1, the gamemaster knows which, but the player does not. Since their knowledge of the system is different, they have different probability spaces. The gamemaster says, for example, that 1 is more likely than 6. The game player says they are equally likely. Both are right, within their knowledge.
Now, what if the 1-biased die is red, and the 6-biased one is green. If the player doesn't know this, only that the two biases exist, his knowledge that he has the red die does not put him in the gamemaster's position.
In the OP, and before we look in the envelope, we are in the role of the game player. The probability that the low value of the envelopes is x is the determined by the distribution of the random variable we have called X. (I'm trying to get away from calling this "initial," since that has other connotations in Probability's derivative fields.)
That distribution is an unknown function F1(x). After picking high/low with 50:50 probability, the value in our envelope is a new random variable V. Its distribution is another unknown function F2(v), but we do know something about it. Probability theory tells us that F2(v) = [F1(v)+F1(2v)]/2. But it also tells us that the distribution of the "other" envelope, random variable Y, is F3(y) = [F1(y)+F1(2y)]/2. Y is, of course, not independent of V. The point is that it isn't F3(v/2)=F3(v)=1/2, either.
Looking in the envelope [correction:] not does change our role from that of the game player, to the gamemaster. Just like seeing the color of your die does not. Simply "knowing" v (and I use quotes because "treat it as an unknown" really means "treat it as if you know the value is v, where v can be any *single* value in the range of V") does not change increase our knowledge in any way.
Exactly.
Maybe I need to explain Simpson's "Paradox." It is a very similar, not-paradoxical fallacy. It just seems to be a paradox if you use probability naively.
Say there is a fatal disease that 1 in 1,000 have, without showing symptoms. But there is a test for it that is 99.9% accurate. That means it will give a false positive with probability 0.1%, and a false negative with probability 0.1%. Sounds very accurate, right?
Say you are one of 1,000,000 people who take the test. You test positive, and think that there is a 99.9% chance that you have the disease. But...
[*]999,000 of these people do not have the disease.
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So fully half of the people who test positive - 999 out of 1,998 - do not have the disease. Yes, you should be worried, but not to the degree suggested by the 99.9% accuracy. The good news is that if you test negative, there is only a 0.0001% chance you have it.
The fallacy is confusing the probability of a correct result in a specific circumstance, with the probability that a specific result is correct. It sounds like the two should be the same thing, but they are not. The latter is influenced by the population who take the test. The 99.9% false negative figure applies only to the population that has the disease, not to the population who test positive. The 99.9% false positive figure applies to the population that does not have the disease, not to the population who test negative.
The exact same thing happens in the TEP. The fallacy is confusing the 50% chance to pick the lower value, with a 50% chance that a specific value is the lower.
The OP doesn't specify that the thought experiment must have physically possible instantiations. The only ideal, here, consists in pursuing strictly the mathematical/logical consequences of the principle of indifference even when it is interpreted such as to entails that the relevant distribution of possible pairs of envelope contents must be assumed by the player to be uniform and unbounded.
I agree with this, and with much of what you wrote before. I command you for the clarity and rigor of your explanation.
I am not so sure about that. The game master does have prior exact knowledge of the function F1(x), which I possibly misleadingly earlier called the "initial distribution". According to the OP specification, the player doesn't necessarily know what this function is (although, under one interpretation of the problem, it must be assumed to be uniform and unbounded). When the player opens her envelope her epistemic position remains distinct from that of the gamemaster since she still is ignorant of F1(x).
My earlier point, which is a concession to some of the things @Jeremiah and @Michael have said, is that the nature of the "system" whereby some probability distribution is being generated, and which F1(x) represents, only is relevant to a player who plays this specific game, either once or several times. But what it is that I may not have conveyed sufficiently clearly, is that when the player not only plays this game only once, but also, isn't seeking to maximize her expected value with respect to the specific probability space defined by this "game" of "system", but rather with respect to the more general probability space being generated by the distribution of all the possible games that satisfy the OP general specification, the specific F1(x) that is being known to the game master is irrelevant to the player at any step. The mutual dependences of F1, F2 and F3, though, as you correctly point out, are indeed relevant to the player's decision. They are relevant to constraining the player's relevant probability space. This is the point that @Michael may have missed.
That's a bit like saying that a geometrical circle is defined by a real world cheese wheel.
This is the part I'm still struggling with a bit.
Even if I were to convince Michael that he had mistakenly assumed the chances for each criterion of success were equal, his response will still be something like this:
There are still exactly two possibilities, and by stipulation I don't know anything about the odds for each, so by the principle of indifference I will still treat them as having even odds.
So there's this backdoor 50% tries to sneak back in through.
I know we have the argument about the sample space for X being uniform and unbounded. I just think there must be something else.
Without a sample distribution, then your assumptions of the unknown distribution carry additional uncertainty that cannot be quantified. These additional assumptions come at a cost.
I am not sure how you would define a normal prior for this problem since it is being assumed, is it not, that the amount of money that can be found in the envelopes is positive? If negative values are allowed, and the player can incur a debt, then, of course, a normal prior centered on zero yields no expected gain from switching. Maybe a Poisson distribution would be a better fit for the player's prior credence in a real world situation where negative amounts are precluded but no more information is explicitly given. But such a prior credence would also fail to satisfy the principle of indifference as applied to the 'v is lowest' and 'v is highest' possibilities, conditionally on most values v being observed in the first envelope.
A normal distribution does not have to have a mid of 0, nor do they need negative values.
I did not say that it has to be centered on zero. Normal distributions are unbounded on both sides, however. They assign positive probability densities to all real values.
It'll be fine once we use a MCMC and get the HDI. You forgot the part of my argument where we need a sample distribution.
Yes, everything will be fine and dandy. How did I not think of that...
The HDI is 95% of the area under the curve. It is a cut off range.
Highest Posterior Density (HPD) Interval
It sounds to me like you're trying to figure out what would be a good prior for what amounts to "Pick a number." I mean, you could do research, see what people tend to pick and in what range, but what can you do with just math here?
People have been assuming a prior that fits their expectations the best; however, if you are gonna model off such assumptions you need to at least try to mitigate observational bias, which is why I think you should use a normal distribution. The point is to lessen the impact of false assumptions, and due to robustness of the normal distribution, I feel this makes it a better choice. Also because of the empirical rule one can better estimate the relative position of [math]Y[/math] on the distribution and the normal distribution has a common occurrence in nature. For all these reasons I feel it is a good choice for a prior.
If you assume a normal distribution (already transformed by chance mechanism number 2) as the prior then when you see Y you could say by the empirical rule:
[math]P(\sigma+\mu \geq Y \geq \mu) = 34\%[/math]
[math]P(\sigma-\mu \leq Y \leq \mu) = 34\%[/math]
[math]P(2\sigma+\mu \geq Y > \sigma+\mu) = 13.5\%[/math]
[math]P(2\sigma-\mu \leq Y < \sigma-\mu) = 13.5\%[/math]
[math]P(3\sigma+\mu \geq Y > 2\sigma+\mu) = 2.4\%[/math]
[math]P(3\sigma-\mu \leq Y < 2\sigma-\mu) = 2.4\%[/math]
Here is a picture.
Now you just need an estimator for [math]\sigma[/math].
On the one hand, my description was too general, a description of the case where the higher and lower values are separated by an arbitrary interval rather than a prescribed one. Since that interval is determined, there is this "bucket" effect, where every amount above k is normalized as, or equivalent to, 2k. And it works in the other direction too. Anyway, there is this pressure here to empty the space of all values except k and 2k: it's impossible to be less than k or more than 2k. But in between there's conflict: On the one hand k wants to treat every value greater than itself as 2k; 2k wants to treat every value less than itself as k, which means they cannot agree on how to treat the interval between them. The only way to maintain consistency is to empty it completely, so the space becomes as discrete as it possibly can while presenting two choices: it narrows to a set of 2.
It occurred to me when trying to pick a criterion for success, that the other envelope is it -- and that creates some weirdness about what picking is -- but I shrugged that off on the grounds that knowing its value would merely allow you to deduce the average value. But there's an element of truth to the idea. And I want to say there is something strangely not number-like, or very minimally number-like, going on. In effect, there are only the two elements, and an order defined on the set that says one is greater than the other. But that turns out to be arbitrary. Could be "red" and "blue", or "heads" and "tails", or "0" and "1", so long as one of them designated as better (and one worse).
When I was trying to think of analogies for the "criterion of success" argument I thought of things like picking red and blue marbles, but labeling them in a language you don't know. With my playing cards, there is a standard order which I just tossed out: instead of saying you need to get the high card, you're supposed to get whatever I say. And I want now to say there is something in the problem that encourages this, because when you see the value of your choice you have to make lots of assumptions even to guess whether that's a good value or a bad, much less know it's good or bad. Numbers don't usually act like that. We usually have some context for saying whether a number is big or small. (Why I had to toss the standard order for playing cards.)
All of which makes doing calculations of any kind quite strange, because there is so much about the problem that makes it impossible to treat what are manifestly numbers as numbers. Once you have complete knowledge, everything becomes normal again and you can say "10 < 20" or "10 > 5". But with anything less than complete knowledge, the game acts like something else entirely.
Just some musing on waking up in the middle of the night. Does this make any sense to anyone else?
The problem here is that at the first chance mechanism [math]x_{i,1}[/math] is an independent observation. However, when you get to the second chance mechanism, when the envelopes are filled, this could be thought of as another function which transforms the distribution into one where the values have a relationship. It is not as simple as trying to determine where on the distribution I am, as I also have to account for that relationship.
Using my simulation from before for an example,
Here is a plot of the values after they have been put into envelopes :
https://ibb.co/nGsEOz
Now for comparison:
Here is the selection before the envelopes are filled.
https://ibb.co/m0uuOz
The second chance mechanism transforms the distribution into one where now we have a relationship to account for.
There are actually two random variables in the OP, which may need two models.
And I'm sure I intended to have a "not" in there somewhere. I'll fix it.
I'm not sure what "real world" has to do with anything. But...
Probability theory does not tell us how to define outcomes. The outcomes of a coin toss could be called {"Heads", "Tails", "Edge"} or {"Win", "Lose", "Draw"}. But you can't use those in an expectation calculation, can you?
So measure-theoretic Probability Theory requires a way to express outcomes with numbers. So its strict definition of a "random variable" is a function whose argument is an outcome, in whatever form you choose to use, but whose result is a number. So G() might be the function for your gain for a bet on Heads, so G("Heads")=1, G("Tails")=-1, and G("Edge")=0.
Personally, I prefer to call the qualities I use to describe outcomes "random variables." You can still think of them as functions whose arguments are abstract outcomes,whether or not the results are numbers. Especially in problems like the OP, where there is no need to be so formal, and no "real world" significance whatsoever.
I think this illustrates the issue you are struggling with:
In the second case, if you had to bet on "odd" or "even," you might flip a coin and so have a 50:50 chance of either. That's as good an option as it gets. But that doesn't mean you expect the wager be fair.
There are still exactly two possibilities, but that doesn't mean the chances are the same for each. The Principle of Indifference requires that you make some assessment about the equivalence of the outcomes.
If you look in your envelope and see $10, the question in the OP requires you to make an assessment about expectation. You don't have any information that allows you to do so. And it makes no sense to talk about a "prior" - which actually refers to something else - in this case. The only point in doing so, is if you have the means to update it.
A normal distribution refers to a random variable whose range is (-inf,inf), and is continuous. The first cannot apply to the TEP, and the second is impractical.
*Edit, sometimes I use my phone, and it just turns out a mess.
It has to do with the sorts of inferences that are warranted on the ground of the assumption that the player still "doesn't know" whether her envelope is or isn't the largest one whatever value she finds in it. And this, in turn, depends on how "doesn't know" is meant to be interpreted. Is that meant to imply that the the player is entitled to apply the principle of indifference and therefore assign a 50% probability (exactly) to each one of the two possibilities irrespective of the value that she finds in her envelope? This is what I would take to entail that the prior distribution is uniform, and not to be consistent with a prior belief that the amount of money in the universe is finite.
When I say that the prior distribution is uniform, I mean this to represent the player's prior expectation (or credence) that whatever real positive value v it is that she will find in her envelope, it remains equally likely, from her point of view, that the other envelope might contain v/2 or 2*v. This would also entail that the prior expectation that the player would find a value v in her first envelope that is lower than some upper bound M, however large M might be, is infinitesimal. That such uniform and unbounded prior expectations don't apply to rational agents being faced with "real world" problems is what I meant.
The sample distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. Since we have no such sampling, let alone a statistic, there is no sample distribution, or use for one. Period. This just isn't a statistics problem.
A theoretical model of the probability problem can be set up, just not as a statistics problem. But doing so does not address the OP, where we have no information that would allow us to set one up.
The only things we can say about the OP are:
[*]If you look and see value v, you need two probabilities from that distribution: Pr(X=x/2) and Pr(X=x).
[*]The only point in mentioning them, is that the expectation calculation requires both.
[*]The expectation for the other envelope is v*[Pr(X=x/2)/2 + 2*Pr(X=2x)]/[Pr(X=x/2) + Pr(X=2x)].
[*]This is 5v/4 if, and only if, we know that Pr(X=x/2) = Pr(X=2x).
[*]This must be greater than v for at least one value of x.
[*]This must be less than v for at least one value of x.
[*]The expectation of this formula, over the range of V, is the same as the expectation of v over that range. See "If you don't look ...".
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And since the OP does not include information relating to this, it does not reside in this "real world."
That's fine with me. In that case, one must be open to embracing both horns of the dilemma, and realize that there being an unconditional expectation 1.25v for switching, whatever value v one might find in the first envelope, isn't logically inconsistent with there being an unconditional expectation 1.25w for sticking, whatever value w one might be shown to be in the other envelope (as @Srap Tasmaner had suggested, by way of reductio of the unconditional switching strategy). The situation would therefore be analogous to the one that I illustrated with my Hilbert's Infinite Hotel thought experiment.
Edited: I had posted an objection that doesn't apply to what you said since I overlooked that you were only here considering the case where both envelopes remain sealed. I agree with your post.
Andrewk had a completely valid point about things being ill-defined and one that has never been fully addressed. I don't agree with the methods so far that have been used to try and map this out.
Quoting andrewk
Let me give you an example of what I mean.
The OP has two random variables that have yet to be defined, [math]X[/math] and [math]Y[/math].
Since [math]X[/math] is undefined, for the moment I will stick with my previous definition. Although I am not fully satisfied it.
For [math]Y[/math] maybe something like this:
[s][math] y_i = \beta_0 +I_1\beta_1A_i+I_2\beta_2B_i+\varepsilon_{ij}[/math][/s]
This would just be an example of what I mean, I am still hammering it out. Something to think about during the dull points in my day.
I will work at defining everything as long as my interest hold. Mapping everything out is good practice and it helps one to understand everything much better.
How does this relate to what I am doing? As far as I can tell it doesn't.
I will not consider "this is not a statistics problem" as valid or thoughtful criticism and I will continue to evaluate this by the methods I consider most appropriate.
But it isn't logically consistent. With anything. That's what I keep trying to say over and over.
1.25v is based on the demonstrably-false assumption that Pr(X=v/2)=Pr(X=v) regardless of what v is. It's like saying that the hypotenuse of every right triangle is 5 because, if the legs were 3 and 4, the hypotenuse would be 5.
The only thing we need to understand to "embrace the dilemma," is why this is so. It isn't simply that one is conditional and one is unconditional, it is that the events used in the first do not represent a value.
[math]\cfrac12(a-b)+\cfrac12(b-a)=0[/math]
Since a and b are both greater than 0, a/b and b/a are both defined. We can, if we like, also say
[math]\cfrac12(a-a\cfrac{b}{a})+\cfrac12(a\cfrac{b}{a}-a)=0[/math]
or
[math]\cfrac12(b\cfrac{a}{b}-b)+\cfrac12(b-b\cfrac{a}{b})=0[/math]
for whatever reason.
For instance, if a/b = 1/2, then
Then
[math]\cfrac12(a-2a)+\cfrac12(2a-a)=0[/math]
or
[math]\cfrac12(\cfrac{b}{2}-b)+\cfrac12(b-\cfrac{b}{2})=0[/math]
Isn't all of this true whichever of a and b is larger, and whatever their ratio?
The issue is that we’re approaching the problem in two different ways. See here. Both statements are true. Our concern isn’t with defending one or the other statement but in showing that one or the other is appropriate given the context. Which is the rational position for a single play of the game after opening the envelope?
If you know only one value, you don't have enough information to prefer either sticking or switching. Flip a coin. That's what you did the first time you chose an envelope, and that's what you're doing now.
Certainly. It is a complicated way of saying that, before you choose, the expected values of the two envelopes are the same. There is even a simpler way to get there: the total amount of money is (a+b), so the expected value of either envelope is (a+b)/2.
But this applies only if we don't look inside one.
If we do look, and see v, then we are considering two possible pairs of envelopes, not the one pair you described. The pair is either (a,v) or (v,b), where a
Let Q be the probability that the pair was (a,v), so the probability that it was (v,b) is (1-Q). Before we look:
[*]There is a (1-Q)/2 probability that our envelope contains value b.
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If our envelope has a, we are in case 1. If our envelope has b, we are in case 3. But if it has v, we know that we are in case 2 BUT IT COULD BE case 2.1. or case 2.2. This still works out as a 50:50 chance that we picked high or low before we look; that is, cases 1 and 2.2 add up to 1/2, as do cases 2.1and 3.
But what if we look?
The expectation for the other envelope, given that one contains v, is a*Q + b*(1-Q). If a=v/2 and b=2v, this reduces to 2v-3vQ/2. And finally, it is only if Q=1/2 that it reduced further to 5v/4.
What you are saying is correct in any case (most cases?) where the prior probability distribution of the envelope values isn't unbounded and uniform. In the case where it is, then there is no inconsistency between the expected value of the unseen envelope being 1.25v conditionally on the value of the seen envelope being v, and this being the case regardless of which one of the two envelopes has been seen.
Actually, the formula is correct in the special case where the prior distribution is uniform and unbounded, since, in that special case, the conditional probabilities Pr(picked higher|V=v) and Pr(picked lower|V=v) remain exactly 1/2 whatever v might be. In the more general case, the formula should rather be written:
Exp(other) = (v/2)*Pr(picked higher|V=v) + (2v)*Pr(picked lower|V=v)
Are you sure you didn't rather mean to write "Exp(other) = (v/2)*Pr(picked higher|V=v) + (2v)*Pr(picked lower|V=v)"?
Yes, this is true of the unconditional expected values of sticking or switching. But those are not the same as the conditional values of sticking or switching (conditional, that is, on the specific value of one of the two envelopes). In the case where the prior distribution isn't uniform and unbounded, then, the unconditional values of sticking and switching are equal to each other, and they are finite. In the case where the prior distribution is uniform and unbounded, the unconditional values still are equal to each other since they are both infinite. And also, the conditional values of switching or sticking both are equal to 1.25v, conditional on v being the value of any one of the two envelopes. (The trick, of course, is to see why this doesn't lead to a contradiction in the case where the prior distribution is uniform and unbounded. It's the same reason why the guests of the Hilbert Rational Hotel are rationally justified to blindly move to new rooms, and, if they have already moved, are rationally justified to blindly move back to their previous rooms).
It's not necessarily equally likely. It may be equally likely, conditionally on £10 being the value of the first envelope, but it may also be morel likely that the other envelope contains £20, or more likely that it contains £5. In the first two cases, you may fare better if you switch. In the third case, you fare worse. (It ought to be twice as likely that the other envelope contains £5 rather than £20 for the switching strategy to break even). On (weighted) average, over all the possible values that you can be dealt initially, you don't fare any better by switching. Only in the case where the prior distribution of possible envelope values is uniform and unbounded do you have a positive expected gain from switching conditionally on any value v being seen in your envelope.
We have [math]X=f(x_i)[/math] for some unknown population. Where [math]x_i[/math] is some positive real number on a countable domain.
Then you have two more variables, the envelopes [math]A[/math] and [math]B[/math], which the contents of have a relationship.
Which can be displayed with this makeshift table.
1 and 2 represent which case you are in.
It is important to understand the significance of this relationship,
as [math] P(A_i=x_i) = P(B_i=2x_i)[/math ]
and [math]P(A_i=2x_i) = P(B_i=x_i)[/math]
which means [math]P(Y_a)=P(Y_b)[/math]
Knowledge of the distribution is not needed, nor is a prior, as the value in envelope [math]B[/math], has the same exact chance of occurring as the value in envelope [math]A[/math].
Consider an example were we know the contents of both envlopes
[math] A=10 =x_i[/math] and [math]B=20 = 2x_i[/math]
You get [math]A[/math] and you see the 10, what you didn't know, however, that there was only a 2% of getting this 10 bucks, which means, due to the relationship there was only a 2% chance that [math]B[/math] ended up with the 20 it did. This equality will be true in every single case for every possible distribution.
Now you might be tempted to think well what about the chance of 5 bucks? I have ten the other one may have 5. It won't matter, the equality would hold, it would just be 2% chance for 10 and likewise 2% chance for 5. Whatever the case may be and whatever you think might be in the other envelope that equality will always be true: That is one thing you know and can count on.
This means [math]P(a_i) = P(b_i)[/math] and since probability must sum to one and you only have two choices then it is a 50/50 split. Some people have been ignoring this relationship and treating the envelopes as merely two different amounts, but that is not congruent with the OP and it is an error when considering the probability.
Whatever the chance of the value you see when opening your envelope it has exactly the same chance as the contents of the other envelope.
People come up with an idea of how they think it works, then they model their belief and when their model matches their beliefs they decide that confirms their beliefs, it turns into a type of confirmation bias. Obviously if you model your subjective beliefs then your model will confirm your subjective beliefs, that is no paradox. You have to separate the objective process from the subjective process.
The secret is to realize that [math]Y[/math] is a trap.
If there's no reason to believe that we're more likely to lose than win on switching, i.e. if there's no reason to prefer sticking, and if we can afford to lose, then switching is a good gamble for a single game, even if not a winning strategy over many games. I either lose £5 or I gain £10. That's a bet worth making for me, and so if I were to play this game and find £10 in my envelope then I would switch.
I would say that, if it's not a winning strategy over many games of the same kind, then it's not a rational strategy over just one game of that kind; unless you aren't averse to playing money games with negative or null expected value. Playing the game only once merely increases the variance. It changes nothing to the expected value. (Although, as andrewk earlier noted, what choice you ought to make also depends on what your personal utility valuations are; here I am simply assuming that the player's only goal is to act such as to maximize expected value).
What would make the switching choice worth making would be if the chance of your losing £5 isn't at least twice as large as your chance of winning £10 is. But you don't know that to be the case either. If you naively rely on the principle of indifference, this will lead you to make a mistake in every case where you are playing this game, are being dealt an envelope with value v, and, conditional on the amount in the envelope dealt to you being v, the chance of your losing £5 is more than twice as large as your chance of winning £10. In the other cases, your choice to switch yields a null or positive expected value. The only thing that you know for sure is that, over the long run, such mistakes would exactly cancel out your gains. So, the expected gain from switching, when you have no clue at all where the value v that you are holding is located within the probability distribution of the possible envelope values, is exactly zero. It is not 1.25v.
Lastly, if you have no clue at all what the distribution is, and you expect any possible distribution to be equally likely with no constraint at all on the maximum or minimum amounts possibly (and plausibly) being contained in the envelopes, then, yes, the expected value of switching, conditionally on v being the value of your envelope, is 1.25v. But that can only happen in a world where the game master has an infinite fund of money.
Quoting Pierre-Normand
Right, and if I stick then I'm guaranteed to walk away with £10, whereas if I switch then I either walk away with £5 or £20. £20 is greater than £10, and a greater difference than that between £5 and £10, and so unless I have a reason to believe that £20 is less likely than £5, why wouldn't I switch? Because if I play the game repeatedly (with different values in my chosen envelope) then I don't gain over the person who never switches? That seems like a non sequitur.
I don't see how the expected value of either always switching or always sticking over repeated games has any bearing on this.
That only makes sense if you favor taking the chance of gaining a larger amount A than the amount B that you can possibly lose irrespective of their relative probabilities and, hence, irrespective of the expected value of the outcome.
Suppose I offer you to play a game with two dice. You throw them once and sum them up. If you roll any value from 1 to 11, you must give me £5. If you roll 12 then I must give you £10. Let us assume that we only are going to play this game once. Would you also say, in this case, that you are willing to risk losing £5 for the chance to win £10?
No, because I know the probabilities aren't in my favour. If I know that they're not in my favour then I won't play. If I know that they're in my favour then I will play. If I don't know the odds then I will play.
In the two envelope case, you don't know the odds of winning. But you do know (or ought to be able to deduce) that the odds aren't either in your favor, neither in your disfavor. The expected gains from either switching or sticking both are zero. That is the case, anyway, on the assumption that the game master doesn't have access to infinite funds.
So to make this a better analogy, let's say that some third party asks us both to play the game. He will roll two dice, and if I win then you give me £10 and if I lose then I give you £5. He doesn't tell us what result counts as a win for me and what counts as a win for you. It could be that 1-11 is a win for you, or it could be that 1-11 is a win for me, or it could be that 1-6 is a win for me.
I would be willing to play, as I have more to gain than I have to lose. You, presumably, wouldn't be willing to play, as you have more to lose than you have to gain.
How is this any different to saying that I'm equally likely to win as lose?
It is obviously different since on the assumption that you are equally likely to win as lose it follows that the expected value of switching is 0.25*v whereas saying that the odds neither are nor aren't in your favor is equivalent to saying that the average expected value of switching is v.
(I'll come back to this conversation tomorrow)
I don't disagree that the average expected value (from repeated games) is v. But that's not what I'm talking about. What I'm talking about is what I said here:
The average return for repeated games is 0, as you're almost certain to lose at some point. But playing it just once? That's worth it.
This is why I think talking about average returns over repeated games is a red herring.
What would you do if you opened envelope A and saw 1000 bucks?
Quoting Michael
Actually it is. Whatever is in the other envelope had the same chance to occur as what is in your envelope. If you have 10 and that had 75% chance to occur then any possible value in the other envelope must also have a 75% chance to occur. If the other envelop has 5 bucks then it has 75% chance to occur. If it is 20 then it has a 75% chance to occur. The envelopes will always share this relationship, always. So whatever you consider as the possible outcome, all considerations are necessarily equally likely
I already said that you have to be willing to lose. If I knew for a fact that I had a 10% chance of losing £1,000,000 and a 90% chance of winning an extra £1,000,000 then I wouldn't bet, despite the greater probability and greater expected value, because I need that first million far more than the second million.
That there's more to gain than there is to lose is what can be objectively measured. Either the envelopes contain £5 and £10 and I lose £5 by switching or the envelopes contain £10 and £20 and I gain £10 by switching.
That I'm willing to bet that it's the latter is, of course, a subjective matter. I have no reason to believe that the former is more likely, and as a risk of £5 is acceptable to me, switching is a no-brainer.
I have already made enough comments on this aspect. No point going over it again.
Quoting Michael
It is precisely as likely. They are equal in probability.
You open envelope [math]A[/math] and you see it has 10 bucks ([math]x_i[/math]).
Now you consider the other envelope, envelope [math]B[/math].
You decide [math]B[/math] could have 5 bucks ([math]1/2x_i[/math]) or it could have 20 bucks ([math]2x_i[/math]).
You don't know what the chance of either 5 or 20 bucks is but you do that the probability of 20 is equal to the probability of 10 and the probability of 5 is equal to the probability of 10, which means the probability of 5 is equal to the probability of 20.
In other terms:
[math]P(x_i) = P(1/2x_i)[/math]
[math]P(x_i) = P(2x_i)[/math]
Which means:
[math]P(2x_i) = P(x_i) = P(1/2x_i)[/math]
[math]P(2x_i) = P(1/2x_i)[/math]
They necessarily have the same chance of occurring.
But it can't be unbounded and uniform. So it is inconsistent in all possible cases.
Quoting Pierre-Normand
What you are saying, is that if you postulate a distribution where (yes, I did reverse it) Pr(picked higher)=Pr(picked higher|V=v) and Pr(picked lower)=Pr(picked lower|V=v), then the results of the two conceptually-different formulas are the same. What I am saying is that one is conceptually incorrect, and one is conceptually correct. And I keep repeating this, because it is the error people make in the TEP when they perceive a paradox.
It cannot be equally likely without postulating a benefactor with (A) an infinite supply of money, (B) the capability to give you an arbitrarily-small amount of money, and (C) a way to select a random number uniformly from the set of all integers from -inf to inf.
All three of which are impossible.
But the reason you should reject the solution you use is because it is not a correctly-formed expectation. You are using the probability of picking the smaller value, where you should use the probability that the pair of values is (v,2v) *AND* you picked the smaller, given that you picked v.
If a certain someone here had a greater understanding of statistics they would realize why this is must be true. You have to consider the whole process from start to finish and model the whole thing.
Say I have a 15 bucks, that is my cash limit and it is all in one dollar bills.
There now we know what the game master has.
I decide to use my computer to randomly select a value,
It selects 3.
I stuff 3 bucks in one envelop and 6 bucks in the other other. Both envelopes are filled from the same chance mechanism and sample sample space; it is one event.
I don't think I need to to follow through on the rest. No one ever said the domain of [math]X_i[/math] has to be the same as my cash limit. That was an assumption people made.
Cash limit is 15 bucks only one dollar bills.
2 bucks.
So I put 2 in A and 4 in B. Once again they are the same event pulled from the same sample space at the same time.
I don't know why people keep making all these assumptions about the unknown domain, distribution, and selection methods when there are an uncountable number of ways to accomplish this task.
Heck, I could have reached into my wallet and just pulled out 10 ones realized that I don't have enough for 20 bucks and just then arbitrarily decided to stuff 4 bucks in one and 8 in the other.
No random selection at all, but the contents of both envelopes still came from one event from the same sample space at the same time. Whatever the probability of such a method was they both share it.
I could have consulted the Ouija board, examined the start charts, went on a deep spiritual journey and only after downing an insane amount of LSD shoved some numbers into a hat and drew the number 3. Then jammed 3 bucks in A and 6 bucks in B.
Anyone want to guess at the probability of getting 3 in A in such a case ? I can tell you exactly what it is: It is the same probability of getting 6 bucks in B. As they are both selected in the same manner, at the same time from the same sample space.
I just want to note that we seem to be in agreement on everything. The only reason why we seemingly disagreed in our recent exchange is because you objected to my stated requirement that the game be conceived as a "real world" problem, and hence that the possibility of a uniform and unbounded prior distribution ought to be precluded, in order that the switching strategy could be shown to yield a zero conditional expected gain rather than a 0.25*v conditional expected gain. I was thus merely expressing the caveat that you are now making explicit in your first paragraph. There is an abundance of discussion of the ideal and impractical case where this "real world" constraint doesn't apply in the literature about the two envelope paradox, and this is the case which, unlike the "real world" case where the prior distribution is well behaved, still is controversial. (See the Chalmers' paper, mentioned earlier in the thread).
I'm not saying that for every value in the chosen envelope it is equally likely that the other envelope contains twice as much as half as much. I'm saying that, given the £10 in my envelope, I will assume that the other envelope is equally likely to contain £20 as £5. I accept that there is an upper (and lower) bound. I'm just assuming that my £10 isn't it.
This is a nice example but you seem to be offering it as a purported counterexample to the principle that what makes it rational to play a game (and apply a given strategy) only once is for that game (and specific strategy) not to yield a negative expected value and hence not to tend to average negative expected gains when played repeatedly.
But your example is defective since if you are offered the options to "play just once" or play until you lose everything, then what you are comparing are two strategies applied to a single game and it is quite clear that the first strategy has a very large expected value (namely, £5,000,500/6) while the second strategy has a null expected value. Those are the amounts that you can expect to gain, on average, while playing the game repeatedly while applying those strategies. The best strategy, in order to maximize your expected value, would be to play until (and if) your total earnings exceed £5,000,000 and then stop. Past that point, your expected gain from rolling the die once more becomes negative. What dictates your choice of strategy still is its average return, or expected value, even if the game only is being played once.
If you are mixing case one and case two you already confounded your unknown limits. Whatever you consider for the possible contents of the unopened envelope the probability of its occurrence must equal the probability of the occurrence in your envelope, which means every consideration of possible outcomes should be consider as equally likely because that relations holds in every single case. If your model cannot fit to that reality then your model is wrong, as it does not reflect the problem in the OP.
Indeed, if you assume it to be equally likely that the odds of winning (irrespective of the amount) are stacked in your favor as that they are stacked in my favor, then, with this specific and asymmetrical payoff ratio, your overall expected value is positive while mine is negative. But this problem is importantly disanalogous to the two envelope problem.
To make this example more relevantly analogous, the game master would need to hand out to each of us an envelope while only informing us that one envelope contains twice the amount of the other envelope. She would then roll two dice, both players would reveal their envelope contents, and whoever wins will be entitled to switch envelopes just in case she doesn't already have the larger amount. The odds of winning, as before, are unknown. In this version of the game, each player, who initially knows only the content of her own envelope, still stands to win twice as much as she stands to lose. So, you would still seem to be committed to conclude that it is rationally mandated that they should chose to throw the dice (after having been dealt their envelopes and seen its content). And this is true for both of them. Does that make sense?
Repetition is actually built into the game. You choose between a pair of envelopes, then you choose again between that same pair of envelopes. More repetitions could readily be added.
Your view is precisely that your chances of gain are better on the second choice than the first.
You flip a coin, before it lands it has a 50/50 chance for H or T. After it lands what is its chance for H or T?
Remember that?
You flip a coin, it has a 50/50 chance of being H or T, it lands on H that event is over and done with; physically the coin cannot be T. However, you don't look at it, so you guess that it could be H or T, and you give each a 50/50 chance. That is your guess and the probability applies to your guess. You guess can never change the actual value on the coin.
You randomly selected [math]x_i[/math], and the chance of selecting that value is [math]P(x_i)[/math], which is unknown, but after it is selected what is its chance of being selected? 0, as the event is already decided. The contents of the envelopes cannot change no matter what you guess at, that event is over and done with and both envelopes have been filled.
By the time you are looking at your 10 bucks the contents of both envelopes has already been decided and is set in stone. Since the contents of both envelopes is determined at the same time by the same method then the probability of both must always be equal. If your model does not conform to this truth then you are wrong.
This means,
[math]P(a) = P(b)[/math]
Always!
You consider the possible values for [math]b[/math],
[math]b=x_{i,1}[/math]
[math]b=2x_{i,1}[/math]
[math]b=x_{i.2}[/math]
[math]b=2x_{i,2}[/math]
Then the probability of each is,
[math]P(a)=P(b=x_{i,1})[/math]
[math]P(a)=P(b=2x_{i,1})[/math]
[math]P(a)=P(b=x_{i,2})[/math]
[math]P(a)=P(b=2x_{i,2})[/math]
Since they are all equal to [math]P(a)[/math] then this must be true:
[math]P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})[/math]
If your model does not meet this requirement then you are wrong.
No, I agree it has to be constrained to operate in the real world. That's why there has to be a real-world maximum value, you can't have an arbitrarily-small real-world value, and you can't choose a uniformly-distributed integer in the range (-inf,inf) (which is not the same thing as choosing a uniformly-distributed real number in [0,1) and back-calculating a gaussian random value from it).
What I'm saying is that there is no real-world component present in the OP. You can use real-world examples to illustrate some properties, but that is all you can do. Illustrate. The OP itself is purely theoretical.
All you can do conclusively, is determine what set of properties must apply to any possible real-world distribution that describes the problem, apply the laws of probability to it in a manner that is consistent with the OP, and then draw general conclusions.
Ignoring continuous distributions (which can be made discrete by considering the range A<=v<2A as the same outcome), any instance of the game can be described by the random variable X representing the smaller value in the envelopes (its outcome needs another, for which gets picked). The values x come from the set {x1,x2,x3,...xn}, where x1
From this we can prove:
[/list]
For Michael: It is true that in any game, the potential gain is bigger than the potential loss. But the possibility of, say, +$12 is always counterbalanced by the possibility of -$12 with the exact same possibility.
THIS IS ALL WE CAN CONCLUDE ABOUT THE OP. There can be no statistical analysis, which includes Bayesian Inference, because they require a population of games played in the real world.
You seem to think that it is only the highest-possible v where you have an expected loss. Maybe you are confused by the fact that it was the easiest example that shows it.
It isn't so. Each value you see in your envelope can have a different value of Q. (Recall that the expectation is v*(2-3Q/2).) So "assuming my £10 isn't [the highest] " does not justify "assum[ing] that the other envelope is equally likely to contain £20 as 5."
Then please, show us one that applies to the problem. And explain how it is a "well justified" anything, and not just a hypothetical.
Let [math]a[/math] represent the value you see when you open your envelope and let [math]b[/math] represent the value in the other envelope. They where both decided at the same time, by the same method. They are one outcome.
This means,
[math]P(a) = P(b)[/math]
Always!
You consider the possible values for [math]b[/math],
[math]b=x_{i,1}[/math]
[math]b=2x_{i,1}[/math]
[math]b=x_{i.2}[/math]
[math]b=2x_{i,2}[/math]
Then the probability of each is,
[math]P(a)=P(b=x_{i,1})[/math]
[math]P(a)=P(b=2x_{i,1})[/math]
[math]P(a)=P(b=x_{i,2})[/math]
[math]P(a)=P(b=2x_{i,2})[/math]
Since they are all equal to [math]P(a)[/math] then this must be true:
[math]P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})[/math]
If your model does not meet this requirement then you are wrong.
Statistics is a branch of applied mathematics that uses probability theory to analyze, and draw inferences from, data. That's why probability theory is a course required in a statistics curriculum.
This doesn't make it a "stats course", any more than arithmetic is a "calculus course." Even if you do add, subtract, multiply, and divide in calculus.
Nothing in this thread requires concepts that were not taught in your probability theory course, and nothing taught outside of it is appropriate here. So please, stop being petty.
You are of course entitled to your opinions.
Now,
Let [math]a[/math] represent the value you see when you open your envelope and let [math]b[/math] represent the value in the other envelope. They where both decided at the same time, by the same method. They are one outcome.
This means,
[math]P(a) = P(b)[/math]
Always!
You consider the possible values for [math]b[/math],
[math]b=x_{i,1}[/math]
[math]b=2x_{i,1}[/math]
[math]b=x_{i.2}[/math]
[math]b=2x_{i,2}[/math]
Then the probability of each is,
[math]P(a)=P(b=x_{i,1})[/math]
[math]P(a)=P(b=2x_{i,1})[/math]
[math]P(a)=P(b=x_{i,2})[/math]
[math]P(a)=P(b=2x_{i,2})[/math]
Since they are all equal to [math]P(a)[/math] then this must be true:
[math]P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})[/math]
If your model does not meet this requirement then you are wrong.
If you add a guarantee that there's at least one envelope worth more than the price of the envelope, you're on your way to reinventing either the lottery or the raffle.
Bingo.
Saw what you did there.
Of course. Someone about to buy such an envelope on the street ought to hope a friendly and helpful philosopher would be walking by to point out that
? [s]p ? q[/s] p ? q ? P(p) + P(q) = 1
but that you cannot infer further any of these:
edit: meant "exclusive or"
*On my phone sorry for any typos.
First you're presented with two sealed envelopes; you can't tell which is bigger.
You open one and observe its value; you still can't tell which is bigger.
If you had an estimate for how much money would be in play, observing a given value may or may not lead you to revise that estimate. Depends.
The conditional probabilities also change when you know the value of one of the envelopes, but you don't know what they are anyway.
The value you observe is no help making the one decision you have to make, so it's reasonable to treat the second choice you face as a repeat of the first, meaning it's just a coin flip: you're either going to gain or lose, relative to the first choice, but you don't have any idea which, not even probabilistically, and the amount you might gain is the same as the amount you might lose.
If it would be your own money at stake here, you shouldn't be playing at all.
Clarifying a little. Suppose X=5. This is the sort of thing you don't know, but choice being what it is, you can still be confident that P(Y=X | X=?) = P(Y=2X | X=?).
P(Y=X | X=5, Y=10) = 0, but for you that's P(Y=X | X=?, Y=10) = ??. 0 or 1. Changing that to a pair of terms multiplied by P(X=...) just moves your ignorance around, though it's better formally.
You can recognize that conditioning on the value of your pick changes the probabilities, even though knowing exactly how they change amounts to knowing the value of X. And you don't.
One way to adjust the game so that your own money is at stake would be to merely write down the two amounts in the envelopes. The cost for playing is the value v that's written in your envelope. If you chose to play, and switch, then you must pay this amount v upfront and the game master must give you back the amount that's written down in the second envelope. On the assumption that you gain no knowledge at all (not even probabilistic knowledge) about the the probability that your cost is smaller than the potential reward, then the paradox ensues since if we make no assumption at all regarding the prior distribution being either bounded, or unbounded and uniform, then the arguments that the expected value of switching is v or 1.25v seem to be equally valid.
That's what I'm saying too.
The general problem would be something like this: can you improve your performance even in situations where you are unable to evaluate your past performance?
I think the answer to this turns out to be [i]yes[/I], and I would consider that a result of some importance.
***
There is more information in the second round -- it's unclear.
I don't quite understand what you mean. What are you referring to as one's "past performance"? Is that the amount of money in one's envelope before one has been offered the opportunity to switch?
Right, that was the idea. Whether it might be possible to make a better second (or later) choice without knowing how good the previous choice (or choices) was.
I accidentally addressed this before, I think, when I noted that any arbitrary cutoff that helps you by falling in the [x, 2x] interval, works as a criterion of success (for having chosen 2x).
I'm going to mull it over some more. The natural answer is [I] of course not! [/I] but I'm not so sure. Any information you might use to improve future efforts is also information you could use to evaluate past ones... [That's an argument against the idea.]
I'm just wondering if there's an alternative to the usual predict-test-revise cycle in cases where you can't get confirmation or disconfirmation, or at least don't get it straight off. In our case, that might be, for the iterated case, using a function or a cutoff value you revise as you go, without ever finding out how well your earlier picks went. Within the single game, finding out the value of one envelope is not enough to be confirmation of your guess, but you might still use it to make your next guess better.
Is this any clearer? Apo and I have talked about the desirability of getting binary answers. I'm just thinking about how you might proceed without them. I expect such issues have been thoroughly chewed over already by people who actually do science!
[ small clarification ]
I suppose I'll have to have another look at the McDonnell & Abbott paper, because I think that's kind of what I'm talking about.
To put it relatively starkly: are there cases in which you can know your performance is improving not because you can test that directly but because you know the procedure you're following to improve is theoretically sound? Imagine an iterated version of our game in which your winnings each round go into an account you don't have access to. You do want to increase your winnings, but you'll only know how much they've grown when you stick. In the meantime, you gather data from each round. You could say that it's just a different perspective on what the experiment is, and maybe that's all there is to it.
I thought of one analogy: suppose you're firing artillery at an enemy position but you have no spotters. (Sometime in the past, no aerial surveillance or anything.) You could use your knowledge of how the enemy troops are usually disposed on this type of terrain, and also how they're likely to respond to bombardment. You could, for instance, attempt to fire at the near side of their line first -- using only physics and an estimate of how far away they are -- and then fire slightly farther, if you expect them to retreat a bit to get out of range. If you had some such general knowledge, you might be able to systematically increase the effectiveness of your strikes without ever seeing where they hit and how effective you were on target.
It strikes me as a deep problem because so much of what we do is guided by faith in a process. Many of those processes have been tested and honed over time, but induction is still a leap of faith. Some things we do get to test directly, but some of the very big things we don't. I'm thinking not just of science here, but also our moral, social, and political decisions. I'd like to think certain acts of kindness or courage, small though they may be, might improve a society over time, but I don't expect to know that by experiment. Anyhow that's part of what's been in the back of my mind here.
Cover's very clever (IMHO) strategy is to randomise the decision to switch, as a Bernoulli RV with probability p that is a monotonic decreasing function of the observed amount Y. Whereas the threshold strategy will deliver positive expected gains for some distributions of X and zero expected gains otherwise, the Cover strategy delivers a positive expected gain for every possible distribution of X.
I do enjoy a clever idea, and that idea of making the decision to switch itself a random variable strikes me as very clever indeed. I especially like the fact that it delivers expected gains even with no knowledge whatsoever about the possible distribution of X, and even if X is a fixed amount (a 'point distribution'.
It sounds like the initial idea was from TM Cover at Stanford, in this 1996 paper, but I've only read the later paper, by McDonnell and Abbott.