Two envelopes

by John Quiggin on May 18, 2004

Via Juan at Philosophy617 (who doesn’t think much of the proffered solutions, and probably won’t like this one) I came back to this version of the two-envelope problem put forward by Brian, a bit before I joined CT.

In this case, once you observe that Brian’s angel is giving you faulty theology, it’s easy to show that you should reject his[1] mathematics, and his offer. At the end of the problem, the angel says “It’s purgatory,” says the angel, “take all the time you want.” But the whole point of Purgatory is that it’s finite – you purge off your sins one at a time until they’re all paid off. Since we now have a finite problem, the solution is straightforward.

Recall that there are two envelopes, with numbers x and 2x representing remission of time in Purgatory, and that x is greater than 10. If your total time in Purgatory is M, we can assume that a just God is not going to give you more remission than that, so 2x is less than M, and x is less than M/2.

The trick in the problem is the apparent symmetry between the envelopes. If you pick one envelope, getting y, switching envelopes gives you y/2 or 2y with equal probability, which seems like a good bet. So it looks as though the angel can apply a Hell pump to you, with repeated offers to switch, paying a day in Hell each time.

The trick in the angel’s offer in the is that it’s not true, for any given y, that switching gives you even chances of y/2 or 2y. Suppose for example, you draw y greater than M/2. Then it’s certain that you’ve got the 2x envelope and that switching would be bad. Conversely, if you draw, say, 15 days, it’s obvious that you’ve got the x envelope and that switching would be good. Unfortunately, you can’t peek then decide whether to switch. If you could, the angel’s offer would probably be a good one. Since you can’t, and given any fixed distribution for x over the range [10,M/2], it’s easy to check that the expected gain from switching is zero.

It’s easy to extend the argument to allow for the case of a Bayesian soul, with a prior distribution that will be updated once the envelope is opened (of course, it’s too late to anything by then). You can also allow for some kinds of non-Bayesians but not too many, since the angel’s argument implicitly relies on the sure thing principle.

It’s also possible, in at least some cases, to refute the angel’s argument even when time in Purgatory may be infinite. All that’s really needed is a given probability distribution for remission time x with a finite mean.

fn1. I didn’t think angels were gendered, but the example uses male pronouns, and I’ll follow suit.

{ 25 comments }

1

Brian Weatherson 05.19.04 at 2:56 am

Yep if the theology requires that time in purgatory is finite (or at least has a finite expected mean) the argument doesn’t go through.

But why think that? Or, more the point, why let theological orthodoxy get in the way of a good puzzle?!

If you’re prepared to allow infinite expectations all sorts of puzzles arise, many of which are nicely catalogued “here”:http://philsci-archive.pitt.edu/archive/00001595/.

2

globecanvas 05.19.04 at 3:13 am

Envelopes shmenvelopes.

It’s a tenet of financial option pricing that stock prices are logarithmically distributed. Meaning, the price of a stock is as likely to double as it is to halve.

If there is equal likelihood of doubling and halving, and the current price is P, then the expected price in the future is the mean of 2P and P/2 … why, it’s 1.25*P!

Therefore all stocks are expected to rise in value. QED, the 90’s explained.

3

John Quiggin 05.19.04 at 3:28 am

Thanks for the link, Brian. It was very interesting. The countable agglomerability principle referred to there is, I think, a special case of the sure thing principle I mentioned.

I should be more on top of countable/finite additivity issues than I am, but I think Richard Chappell 05.19.04 at 9:36 am

I basically agree with Trickster. A full explanation of the paradox (and why we fall for it) is on my blog.

Do let me know if you see any flaws in my argument there!

12

q 05.19.04 at 10:29 am

This angel that is doing the “con”: how do we know he was not a fallen Angel sent from Hell (Lucifer?), as he is unreliable? It could be a trick to get you to go to Hell.

It reminds me of the old problem of the guilty prisoner told by a judge:
(1) We will execute you in the next 7 days and
(2) you won’t know the night before that you will be executed the next day.
The answer is _of course_ that the judge is possibly a sadist and cannot be not relied upon, but I leave others to work out _why_ we know this!

13

dsquared 05.19.04 at 1:28 pm

I would come at this one from measure theory and point out that the move from “the envelope is either y or 2y” to “the envelope is equally likely to be y or 2y” is only valid if y was a random draw from a well-behaved Borel set, which a fixed distribution over [10,M/2] is but the set [10,) isn’t. Same point as Bill and John really.

You can cut up an orange with a finite number of cuts and reassemble it into a sphere the size of Jupiter if you’re prepared to assume that you can define arbitrary sets on a continuous field but you can’t.

14

des 05.19.04 at 2:08 pm

Daniel “Plus-sign” Davies alleges:

You can cut up an orange with a finite number of cuts and reassemble it into a sphere the size of Jupiter if you’re prepared to assume that you can define arbitrary sets on a continuous field but you can’t.

Oh, yes I can. But it doesn’t work with an orange because I can’t cut it fine enough with the knife that it is that I have. Now, Zorn’s lemon, on the other hand…

(runs away and hides)

15

Richard Bellamy 05.19.04 at 3:35 pm

The best way to visualize it intuitively is to consider an alternate hypothetical where first the Angel gives you an envelope with “X” written on it, and then gives you the opportunity to choose between two SEPARATE envelopes, on of which is guaranteed to be 2X, and the other of which is guaranteed to be X/2.

In this case, it is mathematically worthwhile to make the switch.

The faulty math in the actual problem is in trying to convince you that you are really in this 3 envelope world, when you are really in the 2 envelope world.

16

Bill Carone 05.19.04 at 3:41 pm

“You can cut up an orange with a finite number of cuts and reassemble it into a sphere the size of Jupiter if you’re prepared to assume that you can define arbitrary sets on a continuous field”

Notice that you can’t do the above if you treat your continuous field as a limit of a finite set.

Measure theory is a great tool for dealing with complicated probability calculations, but it suffers from the same nonsense that happens when you start with infinite sets without specifying the limiting procedure used to create them.

17

dsquared 05.19.04 at 3:45 pm

“Zorn’s lemon”, jesus. I begin to see how it is that people get hunted down and killed for message board posts :-)

I think that at its deepest level, though, this paradox has a lot in common with the Banach-Tarski one. You’re being encouraged to cut up the real number line in a way which gives you a measure that looks like it ought to work as a probability measure but doesn’t, in the same way in which B-T invites you to divide up a continous 3-space with a measure that looks like it ought to have a volume but doesn’t.

18

Scott Martens 05.19.04 at 3:57 pm

This is a sucker fallacy, but it’s a math geek’s fallacy. You’d never catch the check-out girl at your local grocery store in this one.

Consider the money in the envelope version of the problem, since it eliminates infinities: Either you have x in your hand or you have 2x in your hand. Average return from not switching: 1.5x. If you switch envelopes, you have either x or 2x. Average return from switching: 1.5x. I imagine some stereotyped Ms Trailer Park would come to that conclusion quickly enough: “Now, lemme get this straight. If I open this here envelope, either I get $100 or I get $200. If I open that there envelope, I get either $100 or $200. An’ you wanna know if I wanna change envelopes? You tryin’ t’test me for ESP er sumpthin’?”

19

Bill Carone 05.19.04 at 5:04 pm

““Now, lemme get this straight. If I open this here envelope, either I get $100 or I get $200. If I open that there envelope, I get either $100 or $200.”

That isn’t what we know in the problem, though. If we did, then you would be right.

What we know is that one envelope has twice the amount as the other, and we’ve opened one and seen, say $100. This isn’t the same situation as you describe.

Both the answers “Always switch” and “It doesn’t matter” are wrong. It is more complicated than that.

Define x as the “small” amount in the envelope (so the two envelopes have x and 2x in them). You don’t know what x is, so you assign a probability distribution to describe your information about x.

For example, my distribution for x would be different if I were playing with Bill Gates than if I were playing with my professor.

You now take one envelope and look inside. Define y as the amount you see.

Now, either y=x or y=2x. The probability isn’t necessarily 50%; you need to calculate it using your initial distribution for x.

For example, if you know that I have decided to limit my losses to $100, your probability for x will be zero for any x>$50. If you observe y=$75, then you won’t switch, since seeing the $75 has told you that you have the higher envelope for sure (since I wouldn’t risk $150, the other must have $37.50).

After you see y, you can use standard probability calculations to find the probability that the other envelope has y/2 (call it p1). You should switch only when p1 is less than (2/3).

In practice, here is how it works: if I open the envelope and see $100, I think “Before I saw this, what were my probabilities for x=$50 and for x=$100? If the former was less than twice the latter, I should switch.”

So, depending on your initial distribution for x, you might want to switch or you might not, depending on what you see in your envelope.

20

Anarch 05.21.04 at 8:51 am

Some will say that, in the limit, there are zero balls in the urn, since each ball is taken out. This is incorrect.

I was with you on everything else but you have to be more careful here: you’re using an English formulation which implicitly implies commutativity of cardinality and limits. Or, to look at it another way, you’re talking about a limit without specifiying what your limiting process really is.

For definiteness’ sake let A_i be the set of balls in the urn at the i’th stage, let |X| be the cardinality function, and let A be the “limit” of the A_i, i.e. A = { x | there is an N such that x \in A_i for all i > N }, that being the most natural notion of a “limit” of a countable sequence of sets.

There are then two ways of parsing the statement “in the limit there are zero balls in the urn”:

1) lim_{i \to \infty} |A_i| = 0
2) | lim_{i \to \infty} A_i | = 0

The first translation is incorrect by exactly the argument you gave (either by the standard or non-standard approach). The problem is that the second translation is *not* incorrect, because lim_{i \to \infty} A_i = A = 0, and I’d argue that your English formulation of the problem could be appropriately translated in either way.

[In fact, my original post was going to be a screed to the effect that #2 was the only way to translate your English statement until I, uh, woke up.]

Usual disclaimers apply, of course, the most important of which is YMMV.

21

Anarch 05.21.04 at 9:13 am

You’re being encouraged to cut up the real number line in a way which gives you a measure that looks like it ought to work as a probability measure but doesn’t, in the same way in which B-T invites you to divide up a continous 3-space with a measure that looks like it ought to have a volume but doesn’t.

IIRC, the whole point of B-T is that it relies on cutting up R^3 into something that isn’t a measure. [My knowledge of group actions and orbifolds is somewhat, well, nonexistent, however, so take that with a grain of salt.] The two paradoxes are thus, IMO, unrelated: the two-envelope problem shows the perils of treating infinite-valued [i.e. extended] measures as if they were finite-valued, while the B-T paradox shows the wackiness that can ensue from the existence of non-measurable sets — or more specifically, from the application of choice to the continuum.

Put another way, if one works in the Solovay model of ZF + DC + “Every set of reals is nice” (IIRC, you get that every set of reals is measurable, has the perfect set property and is Baire) then the B-T paradox disappears, but the two-envelope problems retains its full force in any model of ZF, let alone ZFC.

22

Bill Carone 05.21.04 at 5:06 pm

Anarch, thanks for the response.

“There are then two ways of parsing the statement “in the limit there are zero balls in the urn”:”

I agree; the limit of the sum vs. the sum of the limits.

My understanding of Gauss’s dictum is always to take limits as late as possible. So I instinctively chose to take the limit of the sum.

Is there a reason why I wouldn’t want to follow this advice?

23

Anarch 05.21.04 at 11:14 pm

No problem, Bill.

The short answer is that you’d not want to follow Gauss’ dictum when the behavior of the
“infinitary object” isn’t well-approximated by the behavior of the “finitary objects”. This comes up all the time in logic, a lot of times in topology, infrequently in algebra and almost never (that I know of) in analysis — the latter because infinitary gadgets in analysis are almost always defined as their finitary approximations.

[Extensive examples available upon request, but I should warn you that the post I wrote to that effect is, uh, long.]

The other way to answer your question, btw, is “whichever way the problem asks you to do it.” If it’s talking about something inherently infinite (see my response to you in the next thread) then you have to deal with the full object, using finitary approximations only when explicitly licensed by theorems; if it’s talking about the finitary approximations then you have to use those, regardless of the behavior of the infinitary object.

All that should be festooned with IMO and YMMV, as appropriate.

24

Bill Carone 05.23.04 at 5:01 am

“[Extensive examples available upon request, but I should warn you that the post I wrote to that effect is, uh, long.]”

I certainly would like to hear them, if you have the time and inclination.

“The other way to answer your question, btw, is “whichever way the problem asks you to do it.””

I certainly agree with that.

However, I am finding more and more that, when I take limits as late as possible, paradoxes tend to disappear. Is this something that others have found? Or am I just lucky in my choice of problems?

25

Bill Carone 05.23.04 at 5:17 am

Anarch,

“If it’s talking about something inherently infinite (see my response to you in the next thread)”

I may be several kinds of idiot, but I can’t seem to find the response referred to here. Could you repost, please?

My current position would be that there is no such thing as something that is inherently infinite except when it can be defined as a limit of a sequence of finite things. I don’t think “infinity” should be allowed in mathematics, except when used as a figure of speech that really means a limit. Gauss agrees, Cantor doesn’t (I think).

For example, I find the infinite set of natural numbers quite useful, but I think of it as a limit of a finite set {1,..N} as N goes to infinity.

I think of real numbers as very useful, but I think of them as limits of rational numbers with finite decimal expansions.

Both of these things are “inherently infinite”, no?

So, it would be great if you could provide examples of infinite objects that cannot be represented as such limits, or that we would grossly misunderstand if we did look at them merely as limits.

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