I am of the school that "A bird in hand is worth two in the bush". As both envelopes contain some money, I am the richer no matter which choice I make and will trust my first intuition.
Universities, and elite consulting firms, are full of people made dumber by overmuch learning. The capacity to learn and the capacity to think aren’t the same thing.
I cannot be considered mathematically uneducated. PhD, EE. Back when dinosaurs roamed the earth. My dissertation was in pattern recognition, the approach I took was heavily statistical. My main interest was in signal recognition--which is statistical decision making. As I said that was long ago and I haven't done much since.
You have a 50/50 chance of the larger value envelope. Swapping envelopes is essentially repeating the experiment--your chances of getting the larger valued envelope are no different than your first choice.
Human beings don't understand conditional probability intuitively, although it isn't hard. Suppose you are flipping coins. You get a run of ten heads in a row--one chance out of 1024. The chance of 11 heads in a row--1/2048. However, that coin has no idea that it came up heads the previous ten times. Your chance that you get a head on the 11th flip is exactly (given a fair coin) 50/50.
Las Vegas has funded and adult Disneyland based almost entirely on people not understanding conditional probability.
A computer programmer would come up with an algorithm thus:
Open envelope 1.
Let A = Amount contained in envelope 1.
Open envelope 2.
Let B = Amount contained in envelope 2.
If B = 2*A, then do not open envelope 1 again and keep B.
If B = 1/2*A, then open envelope 1 again and take A.
Most of the people I know who saw through the Covid vaccine rubbish were either engineers/computer programmers or ordinary people who were suspicious of government for one reason or other.
This is interesting, and I still recall the first time I learned about the Monty Hall paradox. Mind-blowing.
However, since I rarely find myself in a situation where I'm offered my choice of 3 prize doors or two envelopes full of varying amounts of cash, can anyone tell me a practical application or two for this... principle?
You're right to suggest that the Monty Hall paradox and the envelope paradox don't really have any immediate, direct application. However, I think that the value of these examples is more conceptual.
In the case of the Monty Hall paradox, the "paradox" depends on the fact that in the experimental protocol (i.e. the rules of the game) the game show host makes his decision of which door to open based on what the contestant's initial pick was. Thus, the way that the protocol works is that the first stage influences how the second stage operates, and this introduces a subtle bias into the outcome probabilities. This sort of thing can occur in studies where you have inclusion or exclusion criteria for study participation, or when working with non-experimental data. So the Monty Hall paradox is really best thought of as a warning that subtle aspects of an experimental protocol can introduce surprising and non-intuitive biases into the data.
As for the envelope paradox, the take-away message is that probability theory is powerful, but it can be tricky to apply in practice, and you have to be very careful to get everything right. Everybody understands that there is no advantage to switching envelopes -- the only issue is explaining exactly how the argument is faulty.
anytime there is guessing and the stakes are of consequence to you and also forced to pick one... generally i could never think of the probability in number zero and one to real life situation.... there is an election coming up, and i was thinking about 3 different candidates.
i am going to pick one.
i got some new information the other day, and now it is between 2 candidates.
----------------------
for others, they would have only considered one or the other candidate.
Dr. Briggs, that was a fascinating problem, that I just spent about four hours of my life banging my head over. What I came up with was in partial agreement with your solution, but possibly a bit different from the final part of it.
What if I give you the same problem, but allow you to peek at the first envelope you pick, say envelope F? Inside, you find $8, which equals A. You know that one envelope has double the money of the other, but you do not know what the amounts are. So you now hold $8 in your hand, and you know that envelope G contains either $4 or $16.
That is, to the best of your knowledge, G can contain either 2A or A/2. The two possibilities are mutually exclusive, but both are possible within the subjective limits of your own mind. Under the logical constraints of the problem (objective), only two amounts of money are in play, but you yourself have to consider three possible amounts (subjective). From your standpoint, you can't just exclude either A/2 or 2A, even though only one of them is actually possible.
What I find sketchy in the equation given is the assumption that, when confronted with two mutually exclusive possibilities that have no other logical constraints, you can just split the difference and assign each the probability of 1/2 to get an "expected value." I don't think you can do that for this kind of problem.
A frequentist measure of probability lets you do that, but not probability defined by pure ignorance among equally possible options. The equation offered actually defines a different problem, which is frequentist.
The problem the equation actually describes is this. I give you envelope F, which has A amount of money in it, say $8. A machine dispenses money-filled envelopes G, which contain 2A half the time, and A/2 the other half of the time, quasi-randomly, like coin flips or dice. If we run the machine a large number of times, half the G envelopes will contain $16 and the other half will contain $4, for an average of $10. You can trade envelope F ($8) for the average of all possible G envelopes ($10), and certainly gain. Or we can offer to swap you a single unknown G envelope which will contain either $4 or $16, with an "expected value" of $10. That last option is what the equation is getting at, and it would indeed be the better choice, but it violates the original problem by allowing both 2A and A/2 objectively at the same time.
I've used the terms "objective" and "subjective" a few times here, because I think "probability" and its mathematics is an attempt to bridge that divide. Reality and determinacy are objective, and for logical problems premise may be considered so as well. Knowledge, ignorance, and prediction are of the mind, and are subjective. A frequentist measure of probability allows an objective constraint on our ignorance, so that mathematics can be applied, and we can say that a die roll has an "expected value" of 3.5 by multiplying each face of the die by 1/6 and adding them up. But in the face of alternative options with no frequency or certainties assigned, we have only our subjective ignorance, and no possible measure of probability. We just can't get an "expected value" for the two envelopes problem because that factor of 1/2 in the equation is objectively groundless and can't be used to multiply against a measured value. Using 1/2 as a pretended measure of probability when only 1 and 0 are possible and we have no way of deciding between them is perhaps the central error in this "paradox."
Your modified problem, which because of changes premises changes the probabilities (the one central lesson in the Class), can be simplied.
Here is $8. In this envelope is either $4 or $16. Would you like to swap the $8 for the envelope?
The EV of the envelope is 4*(1/2) + 16*(1/2) = 10. So if you follow EV, you switch. Should you switch back (without peeking in the envelope)? The EV of the $8 is $8. Which is still less than $10. So you don't switch back.
I've gambled for a living at times. Sometimes high on the hog, sometimes down in the dirt. $16 and $4 average out to $10, given even money odds. Why keep $8 when you can get $10? You switch every single time. Let me play this game a million times a day and I'm rich in a few days. Of course, if you only get to play once, it depends on your finances. If I'm broke, I would buy myself a beer, take the remainder, and find a game I thought had better odds.
And that's where I think I'm disagreeing. If knowing the first amount (say, $8) gets us an EV of $10 for envelope G by the logic of that equation, then we are back to the paradox. Here is an unknown, but static amount A. In this other envelope G is either A/2 or 2A. Would you like to swap A for envelope G? The EV of envelope G is A/2 * (1/2) + 2A * (1/2) = 5/4 * A. So if you follow EV, you switch. And then switch back again and again in an infinite loop because the logic is symmetrical.
The problem is not with knowing the first amount. The problem is that the equation is falsely constructed for this question. First, there is no basis for that (1/2) factor. Simply not knowing which of two possibilities is correct does not produce an objective numerical value that can be multiplied against something real. That's for known frequencies under a chaotic filter, not for pure ignorance.
Second, I think it may be wrong to try to add up multiples of the two possibilities. Subjectively, both $4 and $16 are possible, but objectively (as usable in an equation) only one of them is. We might want to tease on that a little bit, but I think there is a divide here.
If this sort of equation is possible at all, then the EV of the unknown envelope has to be the same as the value in the known envelope. If envelope F has $8, then the EV of envelope G also has to be $8, since the two envelopes are obviously symmetrical to our knowledge. That could be done by changing the two "probability" factors from (1/2) to (2/3) and (1/3) respectively. Thus, 4*(2/3) + 16*(1/3) = 8, and A/2 * (2/3) + 2A * (1/3) = A. I don't know what the logical bases for those "thirds" factors would be though.
Fascinating problem that I still don't fully understand. I'd enjoy hearing your thoughts.
There's still no paradox, because it's a completely different problem, with a different proposition and evidence.
This 2nd problem:
I give you an amount of money, A. I put a 2nd amount of money in an envelope, and it will either be 2 A or 1/2 A.
If you follow EV logic here, then EV of switching = A + 1/4A = 5/4 A. EV of staying with A is already known. The logic is NOT symmetrical. You don't 'switch back' if you are following EV decision, since the EV of switching back is the guaranteed amount, A.
This second problem is completely different from the original problem. In the original, you're given two envelopes and told that one envelope has 'A' money in it, the other has '2A' money in it, and you select one. You do not know anything about your selection except that you either have A or 2A money. You can also say that you know that you did not choose an envelope with "0.5A" in it. Remember, A is an arbitrary number that was already assigned at the start, and it can't be reassigned after switching.
As for the idea that you cannot assign numerical probabilities to a set of known, limited outcomes, see Briggs' book Uncertainty, chapter 4.3 and 4.4, on the symmetry of logical constants (he cites Stove). Probability is not a "real thing," it's a measure of our knowledge, and if all we know is there are 2 states, then each state's probability is 1/2.
We're partly on the same page. The second problem is completely different from the original problem, but the equation offered as the paradoxical solution to the original envelope problem actually describes the second problem, not the original.
By the original problem, the logic is symmetrical. By the equation, the logic is not symmetrical, and favors switching. That's where the paradox seems to be coming in. For the original problem, the equation is simply wrong.
You're using the term "A" differently than Briggs did in the article. There, it was defined as the amount in the envelope you first chose, say, F. You are using it as the base value of the two possibilities, which I'll designate as "x". So there are two objective possibilities for the value of A: x and 2x.
Subjectively, there are two possibilities for the value of the money in alternative envelope G: A/2 and 2A. These differ by a factor of 4, while the objective possibilities differ by only a factor of 2.
There is actually nothing wrong with using EV here; the problem is that the equation calculates it incorrectly for the original puzzle. We can't really get EV directly from A, but in terms of x, EV is simply the average of x and 2x, or (3/2)x. We can also say that x = (2/3) * EV. So if the two wads of money in the two envelopes are $4 and $8, then the expected value of the money in either F or G is $6.
A has a 1/2 chance of being x, and a 1/2 chance of being 2x. If we now want to calculate EV from A, we have: (1/2)*x + (1/2)*2x = EV. Substituting (2/3) * EV for x, we get (1/2)*(2/3) * EV + (1/2)*2*(2/3) * EV = EV, or (1/3) * EV + (2/3) * EV = EV. That may be tautological, but the equation works, and it gives us the right factors, (1/3) and (2/3), for keeping expected value the same for both envelopes.
I agree with you and Briggs that probability is not a "real thing," and that it is a measure of our knowledge. But I stand by my claim there is a critical distinction between frequentist probability that can be run through a chaotic filter numerous times to converge on a definite number, and "probability" drawn from pure ignorance between limited alternatives.
Suppose you have $8, and I offer to swap you for a randomly chosen envelope of money picked out from a hundred on my desk. All envelopes on my desk contain either $4 or $16. If you assume that each state's probability is 1/2, then it makes sense to make the trade according to the equation given in the article. What I haven't told you, though, is that only two of the hundred envelopes contain $16, and the rest contain only $4. If you mistake the simple fact of 2 states and nothing else known for a 1/2 probability for each state, you will most likely lose your money. You will calculate a $10 EV for the swap, when your actual EV is more like $4.24.
The last problem is not probability, its *decision*. p(pick a 4| E1) where E1 is "all envelopes have either a 4 or a 16 in them" is different from p(pick a 4| E2) where E2 is "you tell me there are 99 4s and 1 16s." Change the evidence, change the probability.
You can enumerate the possible outcomes of the 100 envelopes given your evidence (which you always assume is true), and literally just count. The probability ends up as 1/2 with the first evidence. It is 99/100 with the second evidence, since that additional evidence eliminated all the other possible outcomes.
The question is not about the probabilities, but on whether you should make your decision by using this specific probability model. If i have 8 dollars, do I risk it and assume that this model truly and accurately describes this real situation? Or do I add to E and make a new model - "you're not the kind of guy to lose money" which changes the probability, but probably can't be explicitly quantified. Or perhaps you let me pick 20 and look first, then mix them up again. Now I have more information, which means I can make a new model. But my choice is not about what my model says, its about whether i want to take the risk that my model both accurately describes the world, and I can afford to possibly lose my 8 dollars.
Say I'm given the choice to play Russian roulette, one round. If i win, i get 1 billion dollars. If i don't , i keep my 0 dollars. My EV of playing is 166 million bucks and 5/6 "being alive"...
I think your EV is actually much better than that: 833 million bucks and 5/6 "being alive," assuming 6 chambers and only one bullet. I like you being 6/6 alive though, so I won't encourage you to take that offer. :)
I agree with your distinction between probability and decision, though not with where you draw the line. If you think you're not the kind of guy to lose money, then that may increase the aggressiveness of your investment decisions, but it will not affect the probability that they will work.
Decision is ultimately subjective, but probability is partly objective.
The last problem I offered illustrates this difference. The objective fact is that there are only two out of 100 envelopes that have $16, and the rest have $4. The probability that you will draw an envelope containing $16 is 1/50. All you know is that some envelopes contain $16 and the others contain $4. You may decide to make the swap based on the blind hope that there are as many with $16 as with $4, but that does not change the fact that the results of your swap will reflect a 49/50 chance of drawing a $4 envelope.
I am of the school that "A bird in hand is worth two in the bush". As both envelopes contain some money, I am the richer no matter which choice I make and will trust my first intuition.
This is Yoda wisdom
The paradoxes will continue until the grant money runs out.
But let us suppose the grant money is in one of two envelopes....
Oh, it is. No need to “suppose” that.
Universities, and elite consulting firms, are full of people made dumber by overmuch learning. The capacity to learn and the capacity to think aren’t the same thing.
There is a paradox, but the paradox is in the model not in reality. The lesson is to distinguish the model from reality. Nicely done, sir.
OK, I have to admit I don't get it. At all.
I cannot be considered mathematically uneducated. PhD, EE. Back when dinosaurs roamed the earth. My dissertation was in pattern recognition, the approach I took was heavily statistical. My main interest was in signal recognition--which is statistical decision making. As I said that was long ago and I haven't done much since.
You have a 50/50 chance of the larger value envelope. Swapping envelopes is essentially repeating the experiment--your chances of getting the larger valued envelope are no different than your first choice.
Human beings don't understand conditional probability intuitively, although it isn't hard. Suppose you are flipping coins. You get a run of ten heads in a row--one chance out of 1024. The chance of 11 heads in a row--1/2048. However, that coin has no idea that it came up heads the previous ten times. Your chance that you get a head on the 11th flip is exactly (given a fair coin) 50/50.
Las Vegas has funded and adult Disneyland based almost entirely on people not understanding conditional probability.
A computer programmer would come up with an algorithm thus:
Open envelope 1.
Let A = Amount contained in envelope 1.
Open envelope 2.
Let B = Amount contained in envelope 2.
If B = 2*A, then do not open envelope 1 again and keep B.
If B = 1/2*A, then open envelope 1 again and take A.
Most of the people I know who saw through the Covid vaccine rubbish were either engineers/computer programmers or ordinary people who were suspicious of government for one reason or other.
Dumbfounded to discover the solution to this long-running paradox is "just be more explicit in writing out the outcomes."
This is interesting, and I still recall the first time I learned about the Monty Hall paradox. Mind-blowing.
However, since I rarely find myself in a situation where I'm offered my choice of 3 prize doors or two envelopes full of varying amounts of cash, can anyone tell me a practical application or two for this... principle?
You're right to suggest that the Monty Hall paradox and the envelope paradox don't really have any immediate, direct application. However, I think that the value of these examples is more conceptual.
In the case of the Monty Hall paradox, the "paradox" depends on the fact that in the experimental protocol (i.e. the rules of the game) the game show host makes his decision of which door to open based on what the contestant's initial pick was. Thus, the way that the protocol works is that the first stage influences how the second stage operates, and this introduces a subtle bias into the outcome probabilities. This sort of thing can occur in studies where you have inclusion or exclusion criteria for study participation, or when working with non-experimental data. So the Monty Hall paradox is really best thought of as a warning that subtle aspects of an experimental protocol can introduce surprising and non-intuitive biases into the data.
As for the envelope paradox, the take-away message is that probability theory is powerful, but it can be tricky to apply in practice, and you have to be very careful to get everything right. Everybody understands that there is no advantage to switching envelopes -- the only issue is explaining exactly how the argument is faulty.
anytime there is guessing and the stakes are of consequence to you and also forced to pick one... generally i could never think of the probability in number zero and one to real life situation.... there is an election coming up, and i was thinking about 3 different candidates.
i am going to pick one.
i got some new information the other day, and now it is between 2 candidates.
----------------------
for others, they would have only considered one or the other candidate.
If I solve this will I be able to levitate or have mind-blowing kung fu powers?
Of course.
🤣
This reminds me of the Fermi "Paradox." It's not a paradox if the assumptions are wrong.
The real paradox is how arrogant midwits manage to walk and chew gum at the same time
If there’s a jackass behind your door (and there is)…switch! Oops wrong problem. Never mind.
Should I mention https://en.wikipedia.org/wiki/Monty_Hall_problem
Wouldn't the best answer to be to swap once and assess whether the new envelope is heavier than the first one? Zero mathematics layman's answer 😁
Sure, but that of course changes the premises. Suppose there are only checks inside.
Or 1 envelope has a 5 dollar bill, and the other has a 10 dollar bill.
Or if one wanted to be really cheeky - one envelope has fifty, $1 while the other has a single $100.
Dr. Briggs, that was a fascinating problem, that I just spent about four hours of my life banging my head over. What I came up with was in partial agreement with your solution, but possibly a bit different from the final part of it.
What if I give you the same problem, but allow you to peek at the first envelope you pick, say envelope F? Inside, you find $8, which equals A. You know that one envelope has double the money of the other, but you do not know what the amounts are. So you now hold $8 in your hand, and you know that envelope G contains either $4 or $16.
That is, to the best of your knowledge, G can contain either 2A or A/2. The two possibilities are mutually exclusive, but both are possible within the subjective limits of your own mind. Under the logical constraints of the problem (objective), only two amounts of money are in play, but you yourself have to consider three possible amounts (subjective). From your standpoint, you can't just exclude either A/2 or 2A, even though only one of them is actually possible.
What I find sketchy in the equation given is the assumption that, when confronted with two mutually exclusive possibilities that have no other logical constraints, you can just split the difference and assign each the probability of 1/2 to get an "expected value." I don't think you can do that for this kind of problem.
A frequentist measure of probability lets you do that, but not probability defined by pure ignorance among equally possible options. The equation offered actually defines a different problem, which is frequentist.
The problem the equation actually describes is this. I give you envelope F, which has A amount of money in it, say $8. A machine dispenses money-filled envelopes G, which contain 2A half the time, and A/2 the other half of the time, quasi-randomly, like coin flips or dice. If we run the machine a large number of times, half the G envelopes will contain $16 and the other half will contain $4, for an average of $10. You can trade envelope F ($8) for the average of all possible G envelopes ($10), and certainly gain. Or we can offer to swap you a single unknown G envelope which will contain either $4 or $16, with an "expected value" of $10. That last option is what the equation is getting at, and it would indeed be the better choice, but it violates the original problem by allowing both 2A and A/2 objectively at the same time.
I've used the terms "objective" and "subjective" a few times here, because I think "probability" and its mathematics is an attempt to bridge that divide. Reality and determinacy are objective, and for logical problems premise may be considered so as well. Knowledge, ignorance, and prediction are of the mind, and are subjective. A frequentist measure of probability allows an objective constraint on our ignorance, so that mathematics can be applied, and we can say that a die roll has an "expected value" of 3.5 by multiplying each face of the die by 1/6 and adding them up. But in the face of alternative options with no frequency or certainties assigned, we have only our subjective ignorance, and no possible measure of probability. We just can't get an "expected value" for the two envelopes problem because that factor of 1/2 in the equation is objectively groundless and can't be used to multiply against a measured value. Using 1/2 as a pretended measure of probability when only 1 and 0 are possible and we have no way of deciding between them is perhaps the central error in this "paradox."
Your modified problem, which because of changes premises changes the probabilities (the one central lesson in the Class), can be simplied.
Here is $8. In this envelope is either $4 or $16. Would you like to swap the $8 for the envelope?
The EV of the envelope is 4*(1/2) + 16*(1/2) = 10. So if you follow EV, you switch. Should you switch back (without peeking in the envelope)? The EV of the $8 is $8. Which is still less than $10. So you don't switch back.
Knowing the first amount changes everything!
I've gambled for a living at times. Sometimes high on the hog, sometimes down in the dirt. $16 and $4 average out to $10, given even money odds. Why keep $8 when you can get $10? You switch every single time. Let me play this game a million times a day and I'm rich in a few days. Of course, if you only get to play once, it depends on your finances. If I'm broke, I would buy myself a beer, take the remainder, and find a game I thought had better odds.
And that's where I think I'm disagreeing. If knowing the first amount (say, $8) gets us an EV of $10 for envelope G by the logic of that equation, then we are back to the paradox. Here is an unknown, but static amount A. In this other envelope G is either A/2 or 2A. Would you like to swap A for envelope G? The EV of envelope G is A/2 * (1/2) + 2A * (1/2) = 5/4 * A. So if you follow EV, you switch. And then switch back again and again in an infinite loop because the logic is symmetrical.
The problem is not with knowing the first amount. The problem is that the equation is falsely constructed for this question. First, there is no basis for that (1/2) factor. Simply not knowing which of two possibilities is correct does not produce an objective numerical value that can be multiplied against something real. That's for known frequencies under a chaotic filter, not for pure ignorance.
Second, I think it may be wrong to try to add up multiples of the two possibilities. Subjectively, both $4 and $16 are possible, but objectively (as usable in an equation) only one of them is. We might want to tease on that a little bit, but I think there is a divide here.
If this sort of equation is possible at all, then the EV of the unknown envelope has to be the same as the value in the known envelope. If envelope F has $8, then the EV of envelope G also has to be $8, since the two envelopes are obviously symmetrical to our knowledge. That could be done by changing the two "probability" factors from (1/2) to (2/3) and (1/3) respectively. Thus, 4*(2/3) + 16*(1/3) = 8, and A/2 * (2/3) + 2A * (1/3) = A. I don't know what the logical bases for those "thirds" factors would be though.
Fascinating problem that I still don't fully understand. I'd enjoy hearing your thoughts.
There's still no paradox, because it's a completely different problem, with a different proposition and evidence.
This 2nd problem:
I give you an amount of money, A. I put a 2nd amount of money in an envelope, and it will either be 2 A or 1/2 A.
If you follow EV logic here, then EV of switching = A + 1/4A = 5/4 A. EV of staying with A is already known. The logic is NOT symmetrical. You don't 'switch back' if you are following EV decision, since the EV of switching back is the guaranteed amount, A.
This second problem is completely different from the original problem. In the original, you're given two envelopes and told that one envelope has 'A' money in it, the other has '2A' money in it, and you select one. You do not know anything about your selection except that you either have A or 2A money. You can also say that you know that you did not choose an envelope with "0.5A" in it. Remember, A is an arbitrary number that was already assigned at the start, and it can't be reassigned after switching.
As for the idea that you cannot assign numerical probabilities to a set of known, limited outcomes, see Briggs' book Uncertainty, chapter 4.3 and 4.4, on the symmetry of logical constants (he cites Stove). Probability is not a "real thing," it's a measure of our knowledge, and if all we know is there are 2 states, then each state's probability is 1/2.
We're partly on the same page. The second problem is completely different from the original problem, but the equation offered as the paradoxical solution to the original envelope problem actually describes the second problem, not the original.
By the original problem, the logic is symmetrical. By the equation, the logic is not symmetrical, and favors switching. That's where the paradox seems to be coming in. For the original problem, the equation is simply wrong.
You're using the term "A" differently than Briggs did in the article. There, it was defined as the amount in the envelope you first chose, say, F. You are using it as the base value of the two possibilities, which I'll designate as "x". So there are two objective possibilities for the value of A: x and 2x.
Subjectively, there are two possibilities for the value of the money in alternative envelope G: A/2 and 2A. These differ by a factor of 4, while the objective possibilities differ by only a factor of 2.
There is actually nothing wrong with using EV here; the problem is that the equation calculates it incorrectly for the original puzzle. We can't really get EV directly from A, but in terms of x, EV is simply the average of x and 2x, or (3/2)x. We can also say that x = (2/3) * EV. So if the two wads of money in the two envelopes are $4 and $8, then the expected value of the money in either F or G is $6.
A has a 1/2 chance of being x, and a 1/2 chance of being 2x. If we now want to calculate EV from A, we have: (1/2)*x + (1/2)*2x = EV. Substituting (2/3) * EV for x, we get (1/2)*(2/3) * EV + (1/2)*2*(2/3) * EV = EV, or (1/3) * EV + (2/3) * EV = EV. That may be tautological, but the equation works, and it gives us the right factors, (1/3) and (2/3), for keeping expected value the same for both envelopes.
I agree with you and Briggs that probability is not a "real thing," and that it is a measure of our knowledge. But I stand by my claim there is a critical distinction between frequentist probability that can be run through a chaotic filter numerous times to converge on a definite number, and "probability" drawn from pure ignorance between limited alternatives.
Suppose you have $8, and I offer to swap you for a randomly chosen envelope of money picked out from a hundred on my desk. All envelopes on my desk contain either $4 or $16. If you assume that each state's probability is 1/2, then it makes sense to make the trade according to the equation given in the article. What I haven't told you, though, is that only two of the hundred envelopes contain $16, and the rest contain only $4. If you mistake the simple fact of 2 states and nothing else known for a 1/2 probability for each state, you will most likely lose your money. You will calculate a $10 EV for the swap, when your actual EV is more like $4.24.
The last problem is not probability, its *decision*. p(pick a 4| E1) where E1 is "all envelopes have either a 4 or a 16 in them" is different from p(pick a 4| E2) where E2 is "you tell me there are 99 4s and 1 16s." Change the evidence, change the probability.
You can enumerate the possible outcomes of the 100 envelopes given your evidence (which you always assume is true), and literally just count. The probability ends up as 1/2 with the first evidence. It is 99/100 with the second evidence, since that additional evidence eliminated all the other possible outcomes.
The question is not about the probabilities, but on whether you should make your decision by using this specific probability model. If i have 8 dollars, do I risk it and assume that this model truly and accurately describes this real situation? Or do I add to E and make a new model - "you're not the kind of guy to lose money" which changes the probability, but probably can't be explicitly quantified. Or perhaps you let me pick 20 and look first, then mix them up again. Now I have more information, which means I can make a new model. But my choice is not about what my model says, its about whether i want to take the risk that my model both accurately describes the world, and I can afford to possibly lose my 8 dollars.
Say I'm given the choice to play Russian roulette, one round. If i win, i get 1 billion dollars. If i don't , i keep my 0 dollars. My EV of playing is 166 million bucks and 5/6 "being alive"...
I think your EV is actually much better than that: 833 million bucks and 5/6 "being alive," assuming 6 chambers and only one bullet. I like you being 6/6 alive though, so I won't encourage you to take that offer. :)
I agree with your distinction between probability and decision, though not with where you draw the line. If you think you're not the kind of guy to lose money, then that may increase the aggressiveness of your investment decisions, but it will not affect the probability that they will work.
Decision is ultimately subjective, but probability is partly objective.
The last problem I offered illustrates this difference. The objective fact is that there are only two out of 100 envelopes that have $16, and the rest have $4. The probability that you will draw an envelope containing $16 is 1/50. All you know is that some envelopes contain $16 and the others contain $4. You may decide to make the swap based on the blind hope that there are as many with $16 as with $4, but that does not change the fact that the results of your swap will reflect a 49/50 chance of drawing a $4 envelope.
from my perspective NO new information. so probability never changed.