“John complains”:https://www.crookedtimber.org/archives/001872.html that the version of the two-envelope paradox I give is not theologically accurate. I was trying to come up with a more theologically accurate one, but I couldn’t really. Still, the following is intended to be a little closer to theological reality.

BRIAN: Where am I?

ANGEL: Purgatory.

BRIAN: Ah, that makes sense. Hang on, does that mean I get to go to heaven one day?

ANGEL: Yep, eventually.

BRIAN: WooHoo! So how long’s the wait then? And can you turn up the heat a little, it’s kinda chilly here?

ANGEL: Well, that’s a hard question. No.

BRIAN: Why is it hard? Hasn’t The Big Guy worked out how long my penance will be?

ANGEL: Well, it turns out decisiveness isn’t one of the divine attributes. So He hasn’t yet decided.

BRIAN: That makes sense. If Bushie is our paradigm of decisiveness, it isn’t obviously a virtue.

ANGEL: Right. So he’s going to toss a coin a few times to figure out how long you stay here. If it lands heads the first time, you stay here 2 days, and we’re done. If it lands heads the second time, you stay here 4 days, and again we’re done. If it lands heads the third time, you stay here 8 days, and again we’re done. You get the picture.

BRIAN: And if it never lands heads?

ANGEL: Then you stay here a forever.

BRIAN: Hmmm, I didn’t realise I’d been that bad.

ANGEL: It was a judgment call whether you went up or down, so don’t complain too much.

BRIAN: OK then. How long will this coin tossing take?

ANGEL: It’s already done.

BRIAN: So it landed heads at least once then!

ANGEL: I wouldn’t infer that too quickly. The first toss takes a {1/2} second, the second toss a {1/4} second, the third 1/8 of a second etc, so infinitely many tosses don’t take that long.

BRIAN: I don’t like the sound of this, but what’s the verdict.

ANGEL: It’s in this envelope. Go on, open it up … WAIT! I almost forgot. Before you do that, I’m meant to offer you a deal. We can tear up that envelope, and we’ll rerun the coin flips, and this time I’ll take a day off whatever the sentence is.

BRIAN: So it would be 1 day here, or 3 days, or 7 days, or 15 days, etc.

ANGEL: You got it.

BRIAN: That sounds like a great deal. Is there a catch?

ST PETER: Funny you should ask. If you open that envelope and see it says 2 days, would you prefer to keep the envelope or take the deal?

BRIAN: Since the deal has an expected infinite sentence, I guess I’d keep the envelope.

ST PETER: And what if it says 4 days?

BRIAN: I guess, well, I guess I’d keep the envelope again.

ST PETER: And 8 days?

BRIAN: Keep the envelope.

ST PETER: See a pattern here?

BRIAN: Yeah, but the envelope was constructed by the same mechanism as the angel is using, but without the discount. How can it be better to keep the envelope?

Some quick commentary on the case.

I’ve deliberately left out of St Peter’s argument what happens if I open the first envelope and see an infinite stay in purgatory. That’s a messy case, but there’s a few things to note.

First, it isn’t obvious I should prefer the deal rather than be indifferent between the deal and keeping the envelope even in this case.

Second, it has probability zero.

Third, we can avoid even this case if we are prepared to allow something like a discontinuity. Change the case so ‘tails forever’ has the same effect as heads on the first flip, i.e. 2 days in purgatory on the first run, 1 day in purgatory on the second run. Now we have a strict conglomerability argument for keeping the original envelope. I don’t understand at all the concern about discontinuous sequences, but I didn’t include this in the original case because of those concerns.

There’s no ‘infinite swapping’ outcome here like in the original two envelope case. But I think it’s very odd that keeping the envelope, which from a neutral perspective appears to be dominated by taking the deal, can be argued against by just the same kind of reasoning in the two-envelope case.

Moreover, if we change the case so the angel doesn’t go on to do the flipping, but has a second envelope from God, we can then give a parallel argument that we really should take the deal.

Nothing in this case relies on there being an equal distribution over [10, M), which doesn’t make any sense to me. But that was never essential to the two envelope case, as John Broome pointed out in his 1995 _Analysis_ paper. Really the two envelope paradox is just a variant on the St Petersburg paradox, as illustrated here. And that paradox works with countably additive probability distributions.

{ 31 comments }

dsquared 05.19.04 at 4:15 pm

Hmmmm … my intuition is that the engine of this one is a bit of fast and loose work in the joke about how long each coin flip takes. Where N = the number in the second envelope, it is not obvious to me that N actually succeeds in referring to a number, even though it looks like it does. For example, if the angel was offering the same deal to an infinite number of sinners, then we know that I were to say “let us refer to the sinner who stays the longest in purgatory as ‘Brian'”, then “Brian” does not actually name an individual.

So, in other words, it is misleading to think about the “expectation of N”; this exists if it is taken as meaning the mathematical expectation of the set of integers but I don’t think it makes sense to say that this is equal to “the expected value of N”, because I don’t think the phrase “value of N” has a referent.

q 05.19.04 at 4:17 pm

Getting into Heaven is nearly as difficult as getting tenure these days. I would say to the Angel (aka St.Peter): So God does “throw dice” after all?

The average length of purgatory in option 1 is 4 days.

The average length of purgatory in option 2 is 3 days.

The two calculations/events are unconnected, unlike the two-envelopes example.

A _non-long-stay-averse_ purgatory minimiser, who knows the value in the envelope to be 2, is therefore better off keeping the value in the envelope (2) as it is better than the average of 3 for the second option. A value greater than 3 (of 4 or more) and they will choose option 2.

Long-stay-averse purgatory minimisers will adjust the equilibrium point outwards accordingly, so that it would be possible for someone to choose a known 16 day stay in preference to chancing it with option 2. BRIAN in your example is long-stay-averse.

Brian Weatherson 05.19.04 at 4:46 pm

I don’t think we need to say that “the expected value of N” refers in order to run the argument. (Though it does make it easier to *state* the argument.) Here’s why. (To address q’s point by the way I assume the utility of each day in purgatory is constant. I should have said that somewhere.)

I’d prefer 4 days in purgatory to the following bet.

A 1/2 chance of 1 day in purgatory plus

A 1/4 chance of 3 days in purgatory plus

A 1/8 chance of 7 days in purgatory plus

A 1/16 chance of 15 days in purgatory plus

A 1/16 chance of 31 days in purgatory

There’s no infinite calculuations here by the way, just regular old expected utility calculuations. And I’d strictly prefer that bet to the deal the angel offers me. So by transitivity I’d prefer 4 days in purgatory to the deal the angel offers me. And we can give a similar argument for any value I find in the first envelope I think.

dsquared 05.19.04 at 5:13 pm

Hmmmm … I think that the reference problem here has been shifted into “for any value I find in the first envelope”. Another way of putting this is that there’s a lot of ugly innards to your remark that there is zero probability of the case where your stay in purgatory in envelope one is infinity. If that were uncomplicatedly the case, then there would be no problem in saying that the number in the first envelope is certainly finite, so the paradox disappears. But there obviously is a problem in this.

I think here that the problem is that the limit of the expectation is not the expectation of the limit. In fact, I’m sure that wih a bit of twisting, I could work this to be analogous to a couple of martingale problems in finance theory.

Bill Carone 05.19.04 at 5:19 pm

“ANGEL: I wouldn’t infer that too quickly. The first toss takes a Â½ second, the second toss a Â¼ second, the third 1/8 of a second etc, so infinitely many tosses don’t take that long.”

Again, here is an invalid “And then an infinity happens” argument. I claim this is simply an error in your use of mathematics, and therefore you shouldn’t be surprised that you get strange results; strange results happen when you add 2 and 2 and get 5 :-)

Start with the Angel saying “I flip the coin a maximum of N times.” You will see that for any N, the deal is preferred to keeping the envelope. Therefore, taking the limit as N increases indefinitely, the deal is preferred to the envelope.

Again, no paradox, just a simple intuitive answer.

You might reply that the certain equivalent of the envelope and the deal both diverge, so you can’t say which you prefer. This is also wrong; (x) and (x-1) both diverge as x increases indefinitely, but the difference between them doesn’t diverge, it converges to 1.

“ST PETER: And 8 days?

“BRIAN: Keep the envelope.

“ST PETER: See a pattern here?”

The pattern only exists when you take the limit too early; you need to take the limit as the very last step, after you have worked all the way to the conclusion.

So, for any N, the pattern is not “Keep the envelope all the time,” but “Keep it when the revealed number is less than N-1, otherwise take the deal.” As N increases, you will see that for most cases you will prefer the deal to the envelope, so your dominance “pattern” argument fails.

dsquared 05.19.04 at 5:26 pm

yeh, I think that if you take that approach to it, it sorts itself out.

How about this extension:

BRIAN: Since we’re not exactly short of time and I’m a betting man, why don’t we gamble a bit while I’m waiting to make up my mind. You tell me, one by one, what the outcomes of God’s coin flips were and then when you say “heads”, then I will start playing you a game of “double or quits” with one of the pennies they put on my eyes. I’ll keep on playing it until the coin comes up “quits” and then I’ll be off. You’ll have to teach me that speeding-up trick obviously, but I’m willing to learn.

ANGEL: Sounds like a fair game. Wouldn’t you prefer to take me up on my kind offer of a better payoff schedule first though?

BRIAN: What do I care? In the limit, I’m outta here with probability 1 anyway.

ANGEL: You cared a heck of a lot about the limit of the expectation a couple of minutes ago …

Bill Carone 05.19.04 at 5:30 pm

“And I’d strictly prefer that bet to the deal the angel offers me.”

Watch out; the “deal the angel offers you” has a certain equivalent that diverges. You can’t make conclusions when limits diverge.

You could say that you prefer that bet to a deal where N is the maximum number of flips when N is very large; however, then you can’t make your next move: “And we can give a similar argument for any value I find in the first envelope I think.” Once you fix N, this is false. You again are taking limits too soon; they should always be the last step in the process.

Again, for any finite number of possible flips, there is no paradox, no matter how large it gets. The St. Petersburg paradox sets up a situation where limits diverge, so you can’t just “plug in infinity” but must specify your model more accurately.

q 05.19.04 at 5:41 pm

The averages I calculated in the second post were too low(!) and wrong so my conclusion does not hold, so please ignore.

Brian Weatherson 05.19.04 at 6:10 pm

I think it’s possible to both avoid the problems about the limit and any worry about there being infinitely many decisions involved if we construct certain kinds of (unnatural) radioactive elements and use their decay patterns. But the calculus involved got a little hairy and I have a million things I have to do today. So this is a promissory note for a version of the puzzle that’s meant to answer these criticisms, rather than the actual answer I hoped to give. Hopefully soon though!

Russell Arben Fox 05.19.04 at 6:14 pm

Why did St. Peter interrupt and/or take the place of the nameless angel once you asked about the catch, Brian? Was he a celestial flunky that needed to go check with the supervisor?

Keith Gaughan 05.19.04 at 7:02 pm

Heck, just open the envelope! Purgatory doesn’t suck all that much after all.

Bill Carone 05.19.04 at 7:03 pm

“I think it’s possible to both avoid the problems about the limit and any worry about there being infinitely many decisions involved if we construct certain kinds of (unnatural) radioactive elements and use their decay patterns.”

Perhaps I can save you the work; I suspect you are trying to use the “quantum uncertainty” kind of argument I’ve seen on your blog at some point. In other words, the idea that the time until the next nucleus decays has a “real” physical probability.

So, instead of flipping coins and comparing two deals, we offer two deals based on when particular nuclei decay. Then, the probability distribution on the infinite set is “real.”

This doesn’t fly; probability is just about our information, not the real world. There are no “real” probabilities, even in the quantum world. So those arguments won’t affect any conclusions above.

You aren’t a quantum physicist, and neither am I. However, Edwin Jaynes was, and he is the source of most of my comments above about probability and finite sets.

My understanding is that we should use a metaphor involving dynamics rather than uncertainty to describe e.g. the “location” of an electron. From Jaynes’s “Probability in Quantum Theory”:

“But notice that there is nothing conceptually disturbing in the statement that a vibrating bell is in a linear combination of two vibration modes with a definite relative phase; we just interpret the mode (amplitudes)-squared as energies, not probabilities. So it is the way we look at quantum theory, trying to interpret psi-squared directly as a probability density, that is causing the difficulty.”

So introducing QM doesn’t save your argument, if this is what you are trying to do.

rea 05.19.04 at 7:28 pm

The truth of the matter is that your are in Hell, condemned to an eternity of working out silly mathematical paradoxes that don’t quite make sense . . .

Matt 05.19.04 at 8:23 pm

Why stop at infinity? Infinity + 1, Infinity + 2, and so on, (and on, and on…) are perfectly good ordinal numbers.

Brian Weatherson 05.19.04 at 8:33 pm

Yep, it’s QM to the rescue. Or at least radioactivity. I don’t know anything about the physics, so I’ll defer to (what seems to be) the vast majority of experts who say there is such a thing as quantum chance. And use it as follows.

The way the punishment is determined is as follows. We create an atom with a half-life of a second, and wait for a second. Let

x = The time from the atom’s creation until its decay, if it does decay, or 1 second otherwise.

y = The probability that the atom would decay in time less than or equal to x.

The sentence is then determined as follows.

If y > 0.5 or y = 0 then 2 days in purgatory

If 0.25 < y <= 0.5 then 4 days in purgatory If 0.125 < y <= 0.25 then 8 days in purgatory If 0.0625 < y <= 0.125 then 16 days in purgatory etc The sentence is then written out, placed in the envelope and given to you. The angel also doesn't flip coins. She just creates another such atom and measures its decay time (or non-decay), except the sentences are always 1 day less, i.e. 1 day or 3 days or 7 days etc. We still get the very odd result that if you open the envelope, you should prefer sticking with what you've got to taking the angel's deal, no matter what you see. But it looks like you should be indifferent to, or perhaps even prefer, the deal before you look in the envelope. And that's what's at the core of the two-envelope paradox.

Brian Weatherson 05.19.04 at 8:35 pm

PS: I do assume here that God, and the angel, have infinitely fine powers of observation. In God’s case that follows from omniscience, and probably from omnipotence. In the angel’s case I assume she is given those powers on loan while my sentence is being worked out.

And I assume, contra perhaps material world physics, that there is a fact of the matter about precisely the atom decays. That’s a bit of a stretch, but we can’t really expect the afterlife to have just the same physics as this world.

Dave 05.19.04 at 9:31 pm

Let’s try to do a little mathematical analysis of the issue.

First, the angel’s statement about the time required to do the flips is correct in the sense that arbitrarily many throws can be made in one second (Sum n from 1 to infinity of (1/2)^n = 1). At least in colloquial speech, that’s what the angel is saying, so I have to problem with the statement.

Now it’s not so much an issue of expected time in purgatory, but rather the expected utility of that stay. We will say we have some function F(t) where F(t) is the utility of staying t days in purgatory before going to heaven. If F(t) = -t, then = Sum n from 1 to infinity of -((1/2)*2)^n = -infinity. So clearly we would pick any result from the first envelope, because odds are the second envelope is very, very bad (note that the second envelope is never infinite, but can be arbitrarily large).

Now, any function F that decreases linearly or superlinearly (in other words, each day in purgatory is at least as bad as the day before it) is going to make very bad. But if F(t) decreases at a less than linear rate, the expectation might be finite, and then we might pick the second envelope. For example, if F(t) = -Sqrt(t), then = Sum n from 1 to infinity of 2^-(n/2) ~= -3.41. Note that in this case, if you got anything more than 16 days on the first try, you might consider switching. If F(t) = -ln(t), then is somewhere between -1.3 and -1.4. In this case, you probably want to switch if you got 4 or more days.

In general, I don’t think I’d gamble on the second envelope. My F(t) probably decreases superlinearly (as I would guess does most people’s).

Dave 05.19.04 at 9:34 pm

And because HTML is allowed in comments, my expected utility value, <F(t)>, didn’t show up. But you know what I mean.

Bill Carone 05.19.04 at 10:48 pm

I must point this out: to produce a decision theory paradox, you need direct computation on infinite sets, the Copenhagen interpretation of QM, angels, and God.

I do like your style :-).

However, since all of these things are rife with paradoxes of their own, it is hard to “blame” decision theory. Furthermore, all paradoxes disappear with two simple ideas:

1) infinite sets are limits, and 2) probability represents information (or even beliefs).

“I’ll defer to (what seems to be) the vast majority of experts who say there is such a thing as quantum chance.”

Perhaps some current sources? My understanding is that the Copenhagen interpretation has been on the way out for a while now, but I’m no expert (this is from my quantum physicist friend). In other words, you are assuming that the wave functions are probability functions, and I don’t think that is the majority view any more.

To repeat and summarize:

1) You use infinite sets directly and get paradoxes.

2) I model the exact same situation using limits, and don’t get paradoxes. In fact, I get intuitively correct answers.

3)I conclude that the paradoxes are simply artifacts of your infinite sets. Remove them, and you remove the paradoxes.

Nicholas Weininger 05.20.04 at 1:46 am

To continue matt’s theme of set-theoretic analysis: what happens if the angel is really an unreachable cardinal?

“Well, I

triedto call him up and tell him I’d decided to keep the envelope, but the secretary says he not only couldn’t be found, he couldn’t even be constructed– guess I’ll have to assume his existence axiomatically…”Jason 05.20.04 at 1:55 am

Ummm, I am getting confused by all the comments, but I always thought this paradox was a Bayesian’s poke at Classical Statistics.

It really begs the question – what do you mean by “random” in this context? You can’t have a uniform distribution over the reals (this is not just because it is “infinite” as so is the interval [0,1], but because its measure is infinite).

With any actual distribution and a risk averse decision maker, you get that they shouldn’t switch. Your decay example is one such distribution, but it doesn’t matter which one you used (your one does have the additional nice property you described of never switching even after opening the envelope).

Jason 05.20.04 at 2:01 am

Forget my previous comment, I didn’t read the follow on post clearly enough.

dsquared 05.20.04 at 7:09 am

I still think you’re pushing around a fundamental referential failure into ever cleverer hiding places. I’d now say that y doesn’t name a number, because as x tends to 1 second, the limit of y looks like it can be cut up to give the appropriate measure but can’t. I still think that this version of the paradox relies on cutting up time in the same way the Banach-Tarski paradox cuts up space; abusing a continuity assumption to make us think it has a straightforward physical interpretation.

Bill Carone 05.20.04 at 3:28 pm

Brian,

We seem to be talking past each other; your new example has exactly the same problem as your others. Let me see if I can spell out the problem more clearly.

“If y 0.5 or y = 0 then 2 days in purgatory

If 0.25 y 0.5 then 4 days in purgatory

If 0.125 y 0.25 then 8 days in purgatory

If 0.0625 y 0.125 then 16 days in purgatory

etc.”

Note the “etc.” This is a typical signal of this type of problem (along with the dreaded “…”). You have created an infinite set in the middle of your calculations; the infinite set here is the set of all possible sentences. Your conclusion is based directly on this infinite set. Any paradox must be checked to see if it disappears when you model this correctly, as a limit.

The correct way to handle this is to start with a finite set of N possible sentences (2 days, 4 days, 8 days, … 2-to-the-N days), calculate through all the way to your conclusion, then see what happens as N increases indefinitely (at the end of the problem, not the middle).

If you had done so, you would see that your conclusion is false: “We still get the very odd result that if you open the envelope, you should prefer sticking with what you’ve got to taking the angel’s deal, no matter what you see.”

Note that for any N, no matter how indefinitely large it gets, this isn’t true. So, there is no “odd result”; it is simply an artifact of your error of directly calculating on an infinite set.

Matt Weiner 05.20.04 at 9:14 pm

Bill, do you think that it is impossible to define a mathematical function on an infinite set? One of my favorite functions is the one that, given x, returns:

2 if x = 1

4 if x = 2

6 if x = 3

etc.

If I couldn’t calculate using that, I would be very much distressed.

Also, on your account of what we should do, why should we stick with one value of N for both distributions? If we cap the original verdict at 1 flip, and then take the limit as the cap on the second verdict goes to infinity, then we get the result that you shouldn’t take the deal. If we cap the original verdict at 2 flips, and then take the limit as the cap on the second goes to infinity, we get the result that you shouldn’t take the deal. In fact, if we repeat this procedure as the cap on the original verdict goes to an infinite number of flips, we always get the result that you shouldn’t take the deal. Of course, we can create the opposite result by exchanging the order of the limits.

So I don’t see how your approach guarantees the intuitive result, or any result whatsoever.

Bill Carone 05.21.04 at 2:22 am

Matt, thanks for the response.

“Also, on your account of what we should do, why should we stick with one value of N for both distributions?”

I actually was going to harp on that originally: the problem is ill-posed if different limits give different answers.

However, I decided that, to keep to the spirit of Brian’s problem, I should keep the same N for both. Here’s why.

In Brian’s problem, the two deals are exactly the same except one is one day less than the other. The source of the paradox is preferring the former to the latter. If we start messing around with different Ns and Ms for each deal, that part of the problem is destroyed. So I decided, as a modeling decision, to use one N for both.

(Aside: If we were using hyperreals, we would use the same infinite N for both as well, or the problem wouldn’t look paradoxical at all; the two deals would be completely different, not just off by a day).

So I think Brian would agree that, if we are to use limits, we should use the same Ns for both.

More later, about your favorite function. Don’t distress! :-)

Tricky infinites 05.21.04 at 3:24 am

It is a problem with the infinites.

The probability of an infinite is not zero but infinitely small. And when Brian encounters an infinite he should take the deal because then the odds for a smaller number are infinitely close to 1.

For there to be a paradox there must be a chance of an infinite in the envelope, because if there wasn’t, there would be a maximum, which would imply that there is no paradox.

But if there is a possibility for an infinite there is no paradox either, because in that case Brian would take the deal.

So unless the paradox is hidden somewhere between finite and infinite there is none.

Matt Weiner 05.21.04 at 6:32 pm

Bill, thanks for the response back.

I don’t think the problem is ill-posed as such. The mathematical calculations are ill-formulated, since they lead to paradoxical conclusions, but the situation Brian proposes is conceivable (if rather absurd)–God really could create such an atom, and I suppose He could put you in Purgatory for arbitrarily long stays if He wanted. (Note that in Brian’s latest variant, there’s never a question of an infinite stay, only of arbitrarily long stays.)

In Brianâ€™s problem, the two deals are exactly the same except one is one day less than the other.On the most obvious reading, that’s false; deal 2 isn’t calculated by subtracting 1 from deal 1. Rather, each of them is calculated by an independent process.

If deal 2 were calculated by subtracting 1 from deal 1, it would be appropriate to use the same N for both limits. It would also be straightforwardly false that, no matter what you see in the first envelope, you would prefer it to the second envelope; if you saw 8 in the first envelope, you would know that the second envelope contains 7, and so on for every other number. That’s more or less the case Brian discussed here; here’s my variant of it.

Sorry if this seems obvious–just wanted to make clear that the independence of the numbers in the envelopes is necessary for the appearance of paradox.

Luc 05.22.04 at 1:03 pm

We still get the very odd result that if you open the envelope, you should prefer sticking with what youâ€™ve got to taking the angelâ€™s deal, no matter what you see. But it looks like you should be indifferent to, or perhaps even prefer, the deal before you look in the envelope. And thatâ€™s whatâ€™s at the core of the two-envelope paradox.

I still don’t get it, i suppose.

If you calculate the expectation, with or without the deal, it is infinite. The envelope is supposed not to contain an infinite number, so the content of an envelope is always less than the expectation (!).

When comparing the expectation with and without the deal, there is no difference.

Comparing an envelope with an expectation, selects the envelope.

And only when comparing without using expectation, you can intuitively select the deal.

The opening of the envelope doesn’t change anything, and doesn’t change the preference for the deal or not (?).

Bill Carone 05.23.04 at 5:05 am

“In Brian’s problem, the two deals are exactly the same except one is one day less than the other.

On the most obvious reading, that’s false;”

You are clearly right; the deals have the same probabilities and payoffs, except that the payoffs of the second deal are always one less than the corresponding payoffs in the first. It isn’t that, e.g., if the first payout is 8, then the second is 7.

This doesn’t change the argument that, if we are going to use limits, we should use the same N for both, right?

Bill Carone 05.23.04 at 6:44 am

“I don’t think the problem is ill-posed as such.”

All I mean by ill-posed is that

– we can formulate different mathematical models that don’t conflict with the situation, and

– those models give different answers.

The idea is that anything left out of a problem must be irrelevant to the answer, or the problem is ill-posed.

Similar to the treatment of the “random chord” problem in most probability texts. (However, that isn’t really ill-posed; see Jaynes’s “The Well-Posed Problem”).

I don’t know what people do when ill-posed problems are tested by experiment.

I don’t think Brian’s problem is ill-posed, as I think there are strong symmetry arguments to model the two envelopes as two finite deals with N possible outcomes each and then take N to infinity. So I think this is the one way to model the problem, and it gives the intuitive answer (that you should take the second “1 day smaller” deal).

My position is that all the other ways suggested of looking at this problem use infinite sets directly without specifying limiting procedures, and I therefore reject them as models.

Brian’s formulations have an infinite set of possible outcomes, and his calculations are based on that; specifically when he says that no matter what you see in the first deal, the other envelope still has an infinite expected value, so you should keep it. You can’t get the infinite expected value without calculating directly on the infinite set.

And as you said before, you can use the same argument the other way, showing that no matter what the second one pays out, the first has an infinite expected value, so you should go with the second deal. This is often the case with infinite sets that aren’t limits; they can often prove A and not-A. This doesn’t happen with limits; as long as the limits are specified in advance, there will be no paradoxes.

Of course, I might be wrong…

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