I tried very hard to avoid replying to that behavior, but you wouldn't let me. — JeffJo

The only control I have over you is what you allow me to have.

The objective Bayesian will say that an already-flipped coin has a 50% probability of being heads, even if it's actually tails, and that my £10 envelope has a 50% probability of being the smaller amount, even if it's actually the larger amount, whereas the frequentist would deny both of these (as far as I'm aware). — Michael

Why would you think that?

In my opinion, the debate between Bayesian and Frequentist, or "objective" and "subjective," has nothing whatsoever to to with the definition of "probability." Because there isn't one. The following may be oversimplified, but I believe it addresses your concerns:

Nobody really defines probability as the ratio of successes to trials, for a simple reason. To use such a definition, one would need to perform an infinite number of trials. Since that is impossible, that definition is useless. What all of them do, however, is assume that the frequency of X should approach Pr(X) as the number of trials increases. They just disagree on what X is.

A Frequentist will assume that X is fixed property of his system. So the probability distribution is a fixed property, whether he determines it objectively or subjectively. His goal may be to use experimental data to approximate what it is, by comparing experimental results to what he expects from that fixed system.

A Bayesian's X is an idealized mental system, with a knowable probability distribution. She uses the same experimental data to change her mental system to a less idealized one. So it becomes more objective than before.

The point is that none of these distinction have anything to do with the experiments "I will now flip a coin," "I just flipped a coin," or the Two Envelope Problem. You may believe that your past experience with coin flips (objectively) confirms your (both Frequentist and Bayesian) impression that unflipped coin will land Heads with a 50% probability. But you are really just using the subjective application of the Principle of Indifference, and then accepting that it coens't contradict your experience. There are two cases, neither is preferred over the other either subjectively or objectively (the weak PoI), and there is a subjective similarity between the two (the strong PoI).

When the coin has been flipped, but is not yet revealed, all of these terms apply the same way. It is still a potential result and the same probability factors - whatever you want to claim they are - apply. The coin is not in a state of being 50% heads and 50% tails. It is in an unknown state that has a 50% chance to be either, just like before the flip it had a 50% chance to end up in either. (Many forget that the point of Schroedinger's Cat was that this mixed state is paradoxical by a classical definition of "existence").

The same applies to the Two Envelope Problem. Regardless of which adjectives you want to apply to the definition of probability, the chance that the other one has v/2, given that yours has v, is conditional. It depends on how the (insert the same adjectives here) probability that the pair contained (v/2,v) AND you picked high compared to the (insert the same adjectives here) probability that the pair contained (v,2v) AND you picked low.

You can't assume that both are 50%, because you'd need to assume that it applies to all possible values of v. A true Bayesian would see that such a distribution is impossible.

This is not well-defined. It needs re-stating to make it unambiguous. — andrewk

The last thing that really needs to be addressed is this. @andrewk was correct that X needs to be defined, he was just wrong in how that needs to be done. It needs to be defined with notation with an index to represent the unknown limit and unknown distribution.

If you want your equations to be complete @Srap Tasmaner you need to address Andrewk's standards here. I don't want to sort the notation myself, but you seem to enjoy that aspect.

This may help:

https://www.encyclopediaofmath.org/index.php/Random_variable

I'm not really sure that this addresses my main question. You say "the coin is not in a state of being 50% heads and 50% tails. It is in an unknown state that has a 50% chance to be either, just like before the flip it had a 50% chance to end up in either." Would you say the same about the envelope? There's a 50% chance of picking the lower-value envelope, and so after having picked an envelope it's in an "unknown state" that has a 50% chance of being either the lower- or the higher-value envelope?

There's a 50% chance of picking the lower-value envelope, and so after having picked an envelope it's in an "unknown state" that has a 50% chance of being either the lower- or the higher-value envelope? — Michael

Let's leave the envelopes aside for a moment.

Imagine an interval [0, L] for some positive real number L. Now let u and v be unequal real numbers in that interval. What is the chance that u < v? Intuitively, that's just v/L. Given a choice between u and v, what is the chance of picking u or v? 1/2. Given that you picked one of u and v, what is the chance that the one you picked is less than the other? We'll call your choice S and the other T (and abuse that notation):

P(S < T | S = u) = v/L, P(S < T | S = v) = u/L

Not only do we not know that those are equal, we know that they aren't, because u and v are unequal. But we can say

P(S < T | S ∊ {u, v}) = (u + v)/2L

because the chances of picking u and picking v are equal. Clearly,

* P(S < T | S ∊ {u, v}) = 1/2

if only if

* (u + v)/2L = 1/2
* u + v = L

But there's no reason at all to think that u + v = L. All we know is that u + v ≤ 2L. (We could go on to ask what P(u + v = L), but I'm not sure what the point of that would be.)

So the answer to this question, "What is the chance that the one you picked is smaller?" may or may not be 1/2, even though your chance of picking the smaller is 1/2. (And if it turns out u + v = L, that's sheerest coincidence and has nothing to do with your choice.)

"My chance of picking the smaller" is just not the same as "the chance of what I picked being smaller", as I've been saying ineffectually for like 3 weeks.

"My chance of picking the smaller" is just not the same as "the chance of what I picked being smaller", as I've been saying ineffectually for like 3 weeks. — Srap Tasmaner

That's because I disagree with your interpretation of probability. Your reasoning would seem to suggest that there's a 50% chance of a coin flip landing heads, but that after a flip, but before looking, we can't say that there's a 50% chance that it is heads. I think that we can say that.

Before picking an envelope I will say that there's a 50% chance that I will pick the smaller envelope. After picking an envelope but before looking I will say that there's a 50% chance that I have picked the smaller envelope. After picking an envelope and after looking I will say that there's a 50% chance that I have picked the smaller envelope. Picking an envelope and looking inside doesn't provide me with information that allows me to reassess the initial probability that there's a 50% chance of picking the smaller envelope.

As an example with the coin, say you flip it and if it's heads put it in a red envelope and if it's tails put it in a blue envelope. I don't know that the colour of the envelope you give me is determined by the result; I only know that you will give me an envelope containing the result of a coin toss. If you give me a red envelope then I will assess the probability that it's heads as 0.5, and I am right to do so, even though a red envelope always means a toss of heads.

That's because I disagree with your interpretation of probability. Your reasoning would seem to suggest that there's a 50% chance of a coin flip landing heads, but that after a flip, but before looking, we can't say that there's a 50% chance that it is heads. I think that we can say that. — Michael

Admittedly, in strident moments I have said things like this.

But look at my last post. It's not about interpretations of probability. It's about how conditional probability works, and it can be a little counter-intuitive.

Admittedly, in strident moments I have said things like this.

But look at my last post. It's not about interpretations of probability. It's about how conditional probability works, and it can be a little counter-intuitive. — Srap Tasmaner

Take my example of the coin toss and the red or blue envelope. The person with full knowledge of the rules will say that P(H | red envelope) = 1, but the person who doesn't know how the colour of the envelope relates to the result of the coin toss will say that P(H | red envelope) = 0.5. The condition of it being a red envelope isn't one that he can use to reassess the initial probability that P(H) = 0.5. It's right for him to say after receiving a red envelope that the probability that the coin landed heads is 0.5.

And the same with the case of the money. The person with full knowledge will say that P(B = X | A = 20) = 1 if he knows that £20 is never selected as the lower value, but the person who doesn't know of the distribution will say that P(B = X | A = 20) = 0.5 because the condition of his envelope containing £20 isn't one that he can use to reassess the initial probability that P(B = X) = 0.5. It's right for him to say after opening his envelope that the probability that the other envelope contains twice as much is 0.5.

Your reasoning would seem to suggest that there's a 50% chance of a coin flip landing heads, but that after a flip, but before looking, we can't say that there's a 50% chance that it is heads. I think that we can say that. — Michael

Your 50% applies to your uncertainty about the state of the coin. You flip a coin it has a 50% chance of T or H, after that lands that part is done. Without seeing the coin you guess there is 50% chance it is heads. That 50% is about your guess. One is subjective the other is objective, but I agree with your general direction.

Coin flips and coin flips with colored envelopes are just the wrong kind of example to look at, because (a) you have categorical instead of numeric data, which means you're going to be tempted to substitute the probability of an event for the event, and (b) coin flips have magic numbers built in, magic numbers that happen to match up to the magic numbers you're trying to distinguish (the chances associated with choosing). This is just bad methodology. When you're trying to figure out some bit of math, you should go out of your way to avoid these magic numbers, and only bring them back in as easy-to-solve special cases of the general problem.

I gave you an example about as general as I could think of. Look at how that example works.

There is no reason you can't look at the OP and consider it in a categorical case.

True.

It's also slightly more complicated than I wanted because of the "reference" problem. If you don't designate either u or v as the reference variable, it all goes to hell.

The amount in the envelopes is already set, just like the coin was already flipped, the uncertainty we have been considering is subjective in nature.

Consider this: Say I have an unfair coin, on average it flips H 9 out of 10 times. You don't know this; however, all you see is a coin and without knowing it is unfair you give H a 50% chance. That is the difference between subjective and objective probability.

The only way for you to know that the coin flips H 9 out of 10 times is to flip the coin several times. Maybe I flip it for you a few times, say I get four heads in a row, and you are starting to doubt your 50/50 assumption. Then I flip it more and get two more heads, now you no longer believe it is 50/50.

That's Bayesian inference in a nutshell.

Imagine u around .75L and v around .9L. They're just randomly selected values in [0, L]. We can't say at the same time that P(u < v) = .9 and P(v < u) = .75. Oops.

Instead you have to say something like P(u < V | V = v, u in [0, L]) = v/L. And then P(u > V | V = v, u in [0, L]) = 1 - v/L. Anyway that's closer.

(The other way, you might get, as with the values above, P(u > v) + P(v > u) > 1 or you demand that u + v = L, which is silly.)

I feel bad about how messy I'm still being with variables.

Reinventing math step-by-step is interesting, and I'm gaining insight by making every possible mistake, and doing so in public, but it would be far more efficient just to study more.

If I'm told that one envelope contains twice as much as the other, and if I pick one at random, am I right in saying before I open it that there's a 50% chance that my envelope contains the smaller amount?1 If so, I must also be right in saying after I open it and see the amount that there's a 50% chance that my envelope contains the smaller amount (assuming I don't know how the values are selected). — Michael

You are referring to two different probabilities. The first is P(lower) which is 50%, the second is P(lower|amount) which depends on the specifics of the distribution.

To see this, suppose the distribution is {{5,10},{5,10},{5,10},{5,10},{10,20}} with an equal probability of the host selecting any one of those envelope pairs.

There is a 50% chance of the player choosing the lower envelope from the envelope pair that the host selected. When the amount is unknown, there is also a 50% chance that the selected envelope is the lower envelope.

However if the player observes the amount to be 10, P(higher|10) = 4/5 and P(lower|10) = 1/5. The expected value from switching is 10/2 * P({5,10}) + 10*2 + P({10,20}) = 5 * 4/5 + 20 * 1/5 = 8. So there is a negative expected gain from switching.

If the player does not know the initial distribution and simply assumes P(lower|10) = P(higher|10) = 1/2 then the calculated expected value from switching will be 12.50 which is wrong. However it will still be the case that P(lower) = 1/2 since that does not depend on their mistaken assumption.

So the difference depends on whether one is conditioning on the specific amount or not. Without knowing the distribution, P(lower|amount) cannot be calculated, only P(lower) can. Opening the envelope and learning the amount constitutes a kind of context switch where the player switches from considering P(lower) to considering P(lower|amount).

Edit: fixed math

P(lower) = P(lower|5) + P(lower|10) + P(lower|20) = 4/10 + 1/10 + 0/10 = 1/2. — Andrew M

This isn't what you mean, is it?

P(lower | 5) = 4/4 = 1, P(lower | 10) = 1/5, P(lower | 20) = 0/1.

You just telescoped the step of multiplying by the chance of picking that number.

Could put & where you have |.

You just telescoped the step of multiplying by the chance of picking that number.

Could put & where you have |. — Srap Tasmaner

Yes, I messed up the math. Have edited...

I messed up the math — Andrew M

You're not in my league at messing up the math!

It is a nice clear argument, using @JeffJo's multiple sets of envelopes, and makes the point I keep failing to make. P(lower)=1/2, but P(lower | 10) = 1/5.

I have a feeling though that @Michael will still think that absent knowledge of the distribution, he can turn back to 50% as an assumption.

I have a feeling though that Michael will still think that absent knowledge of the distribution, he can turn back to 50% as an assumption. — Srap Tasmaner

The best way to counter that assumption, it seems to me, just is to remind oneself that even though one may not know what the distribution is, assuming only that it is not uniform and unbounded, then, whatever this unknown distribution might be, the raised expectation from switching always is zero. So, it doesn't matter that the actual distribution is unknown. It's still known that it must be such as to make the raised expectation from switching zero, for the reason illustrated by @Andrew M in one specific case.

On edit: "the raised expectation from switching" is meant to refer to the average expectation of the always-switching strategy. The specific expectation of switching, conditionally on having found some determinate amount X in the first envelope, can only be guessed and can't be assumed to be zero.

I have a feeling though that Michael will still think that absent knowledge of the distribution, he can turn back to 50% as an assumption. — Srap Tasmaner

You have not really proven he can't. You yourself are making your own assumptions when considering expected gain over the two possible cases.

If a loaded coin flips H 9 out 10 times, without that knowledge, an uninformative 50/50 prior is completely justified.

You have not really proven he can't. — Jeremiah

I know! It's why I'm still here.

Yet you assume he is wrong?

In this case, it's the fact that the Hotel has countably infinitely many rooms that enables the assumption of equiprobability to hold. — Pierre-Normand

I was rather careless in this post where I had made use of Hilbert's Grand Hotel for illustrative purpose. In my thought experiment the equiprobability assumption didn't actually hold because, after the guests have moved to their new rooms, they can often deduce from the room number they end up in where it is that they came from. Initially, the guests are equally distributed in the rooms numbered 2,4,6, ... Then, after flipping a coin and moving, accordingly, from room X to either room Y = X/2 or 2*X, there are three cases to consider. They may end up (case A) in a room within the range (1,3,5,...); (case B) within the range (2,6,10,...); or, (case C) within the range (4,8,12,...).

In case A, the players can deduce from the number Y of the room where they moved being odd that they moved there from room 2*Y and hence will want to move back if given the opportunity.

In case B also, they can deduce the same thing since, although Y/2 is an integer, it is an odd number and hence not in the initial range. They will also want to move back if given the opportunity.

Only in case C will they not be able to deduce where it is that they came from. About half of them, on average, will have moved to room Y from room Y/2 and the other half from room 2*Y. Their expectation for moving back to the room where they came from, conditional on them being in room Y, is 1.25Y (whereas, in cases A and B, the guaranteed outcome of moving back was 2Y)

So, even though the expectation that any guest who starts up in room X, from the initial range (2,4,8,...), is 1.25X after 'switching', the conditional expectations for moving back from Y to the initial room is either 2Y or 1.25Y. Rational (and greedy) guests should always accept the option to move, and then, wherever they end up, provided only that they don't know where they came from, they also should always accept the option to move back. This only makes sense in Hilbert's Grand Hotel, though, and not in any hotel that has a merely finite number of rooms.

It has always been the same error, making assumptions based on Y. Seeing Y does not actually change the contents of the envelopes.

Yes.

I accept that the expectation of gain would apply whether you looked in the envelope or not, and thus there are symmetrical expectations that each envelope is worth more than the other. I also believe that always switching is equivalent to always sticking in multiple trials. From both of these reasons, I conclude either:

1. You cannot talk about expectations here at all (which I find troubling); or
2. The argument is fallacious.

If a loaded coin flips H 9 out 10 times, without that knowledge, an uninformative of 50/50 prior is completely justified. — Jeremiah

Yes, for sure, but if someone hands me, as a gift, an envelope containing some unknown amount of dollars for me to keep, an 'uninformative' prior that is a uniform distribution (or continuous probability density function) from zero to infinity isn't reasonable for me to use for almost any purpose.