That there's more to gain than there is to lose is what can be objectively measured. — Michael

I have already made enough comments on this aspect. No point going over it again.

I have no reason to believe that the former is more likely — Michael

It is precisely as likely. They are equal in probability.

Consider this,

You open envelope $A$ and you see it has 10 bucks ($x_i$).

Now you consider the other envelope, envelope $B$.

You decide $B$ could have 5 bucks ($1/2x_i$) or it could have 20 bucks ($2x_i$).

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:

$P(x_i) = P(1/2x_i)$

$P(x_i) = P(2x_i)$

Which means:

$P(2x_i) = P(x_i) = P(1/2x_i)$

$P(2x_i) = P(1/2x_i)$

They necessarily have the same chance of occurring.

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. — Pierre-Normand

But it can't be unbounded and uniform. So it is inconsistent in all possible cases.

[Exp(other) = (v/2)*Pr(picked higher) + (2v)*Pr(picked lower)] is correct in the special case where the prior distribution is uniform and unbounded, — 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.

I know it's not necessarily equally likely. But we're assuming it, hence why Srap said we should just flip a coin. — Michael

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.

It absolutely necessarily has to be equally likely as the same chance mechanism is used to decide the contents of both envelopes. The contents of both envelopes is one event from the same sample space.

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.

Let's do this, an example with limits and a defined scale

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 $X_i$ has to be the same as my cash limit. That was an assumption people made.

The problem with making assumptions about what you think is and what is not, is that you can't think of everything and there could be something you overlooked. This is why the Law Parsimony is important. Unconfirmed assumptions always carry unknown uncertainty even if they seem perfectly reasonable, that is why we should try to limit them.

Anyone want to see me make a random selection from a continuous normal distribution under the same conditions?

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.

I think people are confusing the actual cash with the selection method.

Cash limit is 15 bucks only one dollar bills.

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.

Cash limit is 15 bucks only one dollar bills.

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.

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. — JeffJo

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).

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. — JeffJo

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.

Imagine you're given £100 and are offered the choice to pay £100 to play a game with a 5/6 chance of winning (say a dice roll). If you win then you win £1,000,000 and if you lose then you lose all the money you've won up to that point.

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. — Michael

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.

The contents of A has the same exact chance of occurring as the contents of B, this relationship should hold for every single consideration of possible outcomes for $Y_i$, if it doesn't then you did something wrong.

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. — Michael

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?

This is why I think talking about average returns over repeated games is a red herring. — Michael

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.

Anyone remember the discussion of the coin flip?

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 $x_i$, and the chance of selecting that value is $P(x_i)$, 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.

Let $a$ represent the value you see when you open your envelope and let $b$ 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,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

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, — Pierre-Normand

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<x2<x4<...<xn. There is a probability function Pr(*), where Pr(xi)>=0 for all 1<=i<=n, and sum(Pr(xi), i=1 to n)=1.

From this we can prove:

• When you don't know your value v, or consider it to be fixed unknown, the expected gain from switching is 0.
• When you do know your value v, or consider it to be fixed unknown, the expected gain from switching is v*(2-3Q/2), where Q is a function of Pr(v/2) and Pr(v).
  
  • We may not know what the Pr's are, and so we also do not know Q, but we can say that 0<=Q<=1. So the expected gain, given v, is between -v/2 and +v.
  • If v=t1, then Q=0. We may not know what t1 is, but we know there is one.
  • If v=tn, then Q=1. We may not know what tn is, but we know there is one.
  • Exp(v*(2-3Q/2))=0

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.

There can be a theoretical statistical assessment even without data. There can also be a well defined Bayesian prior. In fact I could exactly define a sigma, mu and domain for $Y$ that could be used as a well justified prior. The truth is when people are considering all the possible values of $Y$, they are very much setting up a Bayesian prior.

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. — Michael

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."

There can be a statistical analysis even without data. There can also be a well defined Bayesian prior. In fact I could exactly define a sigma, mu and range for Y that could be used as a well justified prior. — Jeremiah

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 $a$ represent the value you see when you open your envelope and let $b$ 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,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

Also just a note but probability theory IS a stats course. If you are making probabilistic claims about an unknown population from a sample you are engaging in stats. $Y$, is our sample.

Trying to completely stats-block the conversation is just petty. The truth is that you can't follow me, and instead of trying you want to wave it away as "statistics".

Also just a note but probability theory IS a stats course. — Jeremiah

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 $a$ represent the value you see when you open your envelope and let $b$ 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,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

So no one can meet my challenge?