## Two envelopes

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 over at Crooked Timber last year.

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.

I really shouldn’t have wasted any time on this, but I can’t resist a riddle, damn you, and so had to trawl through the Crooked Timber website until I found the most logical response:

“This seems pretty simple to me. You’ve got a choice of two envelopes. Once you’ve picked one, the angel is essentially allowing you to make the identical choice again. In no way does making the choice twice change the odds, since you don’t learn anything about the second choice from having made the first choice. Therefore, the choice the angel is giving you is this:

1) Take a chance between x and 2x in heaven or

2) Take a chance between x and 2x in heaven and also spend a day in hell.

Unless you are curious about hell, there’s no reason to take the angel’s deal—it gets you exactly the same thing as the original deal, except you also have to spend time in hell, maybe as punishment for 2nd guessing yourself.

Posted by Jonathan · November 6, 2003 09:35 PM”

There’s no real puzzle, here, just the angel basing his argument on a false assumption: i.e. that one envelope has y and the other 2y or y/2, when we know from the first premiss, that it’s a choice of x or 2x. One option is twice the other, so it’s not possible to face a choice between 2y or y/2.

On the angelic gender issue, I’ve always assumed angels were all male because of their names (can you name an angel with a female name?).

Or you could put it like this:

y=x or y=2x with equal probability. So, to switch is to gain 2x-x or x-2x, i.e. gain x or lose x with equal probability. Therfore the expected gain from a switch is 0.5(x) + 0.5(-x).

I can’t for the life of me see why this generated such debate on Crooked Timber, but then again it is a bit of a windbag’s forum (you’ve mentioned the low signal-to-noise ratio yourself, John).

The whole deal is no good. Whilst most angels in general seem to be truthful it is possible that it is one of the fallen angels (Lucifer and his bros / homies) and not to be trusted.

The real problem is that,as far as I can remember from the nuns, that once in hell there is NO getting out. So a day in hell would end up being for infinity (a bit like waiting on the Telstra help line).

As well I think that those few days in heaven would in fact delay the ultimate entry into heaven. Purgatory is finite and a bit like prison where you pay off your sins day by day. Having a “day release” with 10 days in heaven would only add to the time it took to get out of P to H. Seeing that P is compulsary you wouldn’t need the 10 days as a break to build up strength to complete P. You must complete P anyway – there is no choice.

Best tell the angel – “thanks but no thanks” and get back to doing the time. (Me I’d try to bargain for Library and net access in P)

Here is a reference to (almost) all you need to know about angels. I can’t see any mention of gender.

http://www.newadvent.org/cathen/01476d.htm

farrel is right.

the angel’s faulty logic has nothing to do with infinity. it has nothing to do with the fact that x could be any number greater than 10.

the riddle works if x is always 10, its just that its harder to spot the fault when you use x instead of a fixed number.

there are two envelopes, 10 and 20 days in heaven.

the angel says you have y so changing either gives you y/2 or 2y.

thus 0.5*2y + 0.5*y/2 = 1.25y

so you should change.

but as turnbull (at crooked timber) and farrel here point out, y is not the same, its either x or 2x.

all the other comments about infinities (including john’s post) is just fluff. (its talking about a potential different problem, but one that doesnt need to be present for the argument)

Contrary to the commentators, I think the paradox works if you suppose any integer > 10 is equally likely to be chosen for x.

The lesson I draw is that problems requiring an equal probability for each element of a countably infinite set are not properly posed.

yes that is my “potential different problem” but its not why this riddle exists.

the original riddle obscures the x and 2x with y, and thus you are misdirected.

you (and others at crooked timber) being infinitely refined mathematicians have spotted another problem…but thats not where the extra .25 comes from that tempts you to swap envelopes. (as i show conclusively above)