Chris’s post generated such an interesting comments thread that I feel I have to hop on the bandwagon. The following is a theologically revised version of a puzzle that’s been doing the rounds the past decade or so.

You, a rational agent, are in Purgatory, for good it seems. Things could be worse – you’ve heard horror stories about hell, but they could be better – you hear great things about heaven. These two possibilities seem to be equally valuable, in opposite directions. You’d be indifferent between your current state of affairs and a gamble with a 50% chance of a day in heaven and a 50% chance of a day in hell. (Purgatory is a lot like earth, so this kind of gambling is highly encouraged.)

One day an angel appears with a nice offer. God will give you some time in heaven for good behaviour. But He decided to play a little game to figure out how much time you’ll get. He wrote down two numbers, x and 2x, on slips of paper, and dropped them into identical envelopes. You will get one of the envelopes, it’s your choice which, and the slip will be good for the number of days in heaven that is written on it. The angel doesn’t know which envelope is which, and he doesn’t know what x is, except that it’s over 10. (They don’t send angels down for smaller missions than that.)

So you pick an envelope, and are about to see how long you’ll get in heaven when…

The angel makes you an offer. If you’ll do a day in Hell to cover for a friend of his that got caught up in a little scandal, he’ll give you the other envelope. He argues that this must be a good deal, as follows. Let y be the number written on your slip. The other envelope has either 2y or y/2 written on it. Each is equally likely, so your expected number of days in heaven if you switch envelopes is 0.5 * 2y + 0.5 * y/2 = 1.25y. So the expected gain from swapping is 1.25y – y = y/4. Since we know y > 10, y/4 > 2.5, so this is more valuable to you than a day in Hell.

The reasoning starts to sound attractive, until you worry about what other offers the angel has in mind if you accept. So you ask for some time to think about it. “It’s purgatory,” says the angel, “take all the time you want.”

{ 29 comments }

Doug Turnbull 11.03.03 at 8:46 pm

Of course, the reasoning there is bad, obviously. Switching the envelopes logically can’t increase your expected payoff, since it’s still a random choice.

The trick is the “y” is not the same in the “2y” and “y/2″ cases, so you can’t average them and get a good answer. Kind of a sneaky angel, really.

Ducki 11.03.03 at 8:59 pm

I think that the interpretation is supposed to be “he’ll give you the other envelope _also_,” giving you both envelopes, which contain 2 “coupons” for 3 times X days in heaven in return for a day in hell.

Instead of having to choose blindly between 1X and 2X, you have to choose between Arrangement A(the original quandary) and Arrangement B(3X days in Heaven in return for 1 day in Hell) where X >= 10 days.

At least that’s how I read it. The angel makes it an AND instead of an OR.

Ted Barlow 11.03.03 at 8:59 pm

If the angel was right, you could dramatically increase your time in heaven simply by taking half an hour and handing the envelopes back and forth. You can’t trust angels in logic puzzles.

sidereal 11.03.03 at 9:01 pm

Typical angel. Always spinning the numbers to trick you into doing what they want. “Oh no, the 3 are equivalent to 1 and vice versa. Alpha and Omega are concurrent. No, trust me. The math is good”

Ted K. 11.03.03 at 9:10 pm

Correct me if I am wrong, but in many of these puzzles if you want the optimum answer you are better off throwing out the math and making a little box of either or.

If I got x the first time, then the second envelope is 2x, minus the day in Hell, gives me 2×-1 in expected value. Since I would have had x if I had stood pat, changing the envelope gives me 2×-1-x, or x-1, in marginal value.

If I got 2x the first time, then the second envelope is x. Minus the day in Hell, that gives x-1 in expected value. Since I would have had 2x if I had stood pat, changing the envelope gives x-1-2x, or -1-x in marginal value.

To use averages then. If I stand pat I will get either x or 2x, for an average result of 1.5x.

If I switch I will get ((x-1)+(2×-1))/2 or 1.5×-1 in expected value.

The average result of standing pat is better than the average result of switching. I could have come to the same conclusion by looking at the marginal change in switching and noticing that it adds up to a negative number.

But, this does not tell us where the fault in the angel’s logic lies. I agree with the earlier folks that it involved playing fast and loose with the averages, but I don’t see it off the top of my head.

Ted K.

Mats 11.03.03 at 9:44 pm

This is also called the “grass is greener paradox”. It is the type of paradox that is created by confusing infinite and finite numbers. The Greeks didn’t see through them, but they were solved with the advent of calculus (differential/integral analysis).

The “grass is greener” paradox can hence be stripped down to: pick at random a whole positive (greater than 0) number x . Tell me what you picked. Now pick another random one. This has almost certainly to be greater than x, since there is infinetely many more positive numbers greater than x than there are positive numbers lower than x. How can the second of two randomly chosen numbers almost certainly be the greatest?

And the original greek paradox, think it was Xenon’s: To close a door you must firt shut it halfways. But then you have to shut it half of the rest of the way. And then you have another half. So there are infinitely many steps you have to take in order to fully shut the door. But you don’t have an infinite amount of time. So you can’t shut a door. All movements have to be imaginary.

Brian Weatherson 11.03.03 at 9:56 pm

I didn’t mean to imply 3x is an option – the choice is one envelope or the other.

And the puzzle doesn’t just rely on being sloppy with infinities, though it’s easier to state it that way. If God uses some kind of St Petersburg process to work out the value of x (say roll a fair 100-sided die until it comes up 42, let r be the number of rolls that takes, let x = 10 * 2^r) we get just the same kind of puzzle without any sloppiness about infinity. (That variant is due to John Broome.)

Mats 11.03.03 at 10:40 pm

“we get just the same kind of puzzle without any sloppiness about infinity.” No Brian, we need to be sloppy here, because otherwise the distribution from which x is drawn has an infinite mean, so it’s no good swapping.

Otherwise, the distribution from which x is drawn has a finite mean, which means (when talking about positive numbers) that drawing a higher (or a high enough at least) number is less likely than drawing a lower number. Hence it is more likely that we’ve been handed the 2*x envelope when we see a large number and the 1*x envelope when we see a small number. If we know the mean, then its a good game and we should do a calculation to decide if we should swap or not.

If we don’t know the mean of what is in an envelope, we have to estimate it to be what we observe in the first envelope, so there is no use swapping.

The paradox lies exactly in saying that there is equal probability for having the smaller and the larger envelope, given that we observe the amount in the first envelope. This is true only for some distribution with infinite mean.

Boronx 11.03.03 at 10:47 pm

Turnbull has it right. The equation given has no relation to the problem except that it provides the answer that the angel wants.

Since you are dealing in ratios, the expected value is determined by the geometric mean, not the arithmetic mean: sqrt( 2y * y/2) == y

Mats 11.03.03 at 10:50 pm

To avoid sloppiness altogehter, try to do this with numbers that have to be finite! Like x is a number between 10 and 50. I get 100 in my envelope, so I don’t swap because I know there is a 100% chance *not 50%* that I’ve got the 2*x envelope.

Mats 11.03.03 at 10:54 pm

Now, the way you formulated it, if I get 10 days, I should of course swap since there are no 5 days leaves!

Mats 11.03.03 at 11:02 pm

Or taken the other way round:

Let’s say the chances really are 50/50 for getting a better deal when swapping. Given that first get the number n, chances for getting n/2 and 2*n are equal. But if the chances for getting a number and another that is four times as big are *always* equal, regardless of the number, then all numbers without limit has to be equally likely.

So you really have an infinite distribution. And you are being sloppy with infinites.

Boronx 11.03.03 at 11:04 pm

Seriously, though, you can make this equation work by asking what the referrent for ‘y’ is. Well, it’s got an expected value of it’s own: 1.5X. So if we take that into account, the expected value of the switch is: becomes: (2y/6)+(4/y/6) = y

boronx 11.03.03 at 11:05 pm

that’s (4y/6)

Mats 11.03.03 at 11:12 pm

How does the sloppiness then cause the paradox – it’s the same as my stripped pick a number game above. The distribution is infinite. So with any number of days we get in the first envelope, it is infinitely unlikely that we actually should have gotten such a low number, a finite one, from an infinite distribution. So we should actually swap. But this conclusion is only valid for infinitely unlikely cases, and hence irrelevant.

Brian Weatherson 11.03.03 at 11:39 pm

Mats, you do need to hand-wave a little to get the case that the expected gain from swapping is always y/4 for all y. But you don’t need to do so to get the case that the expected gain from swapping is positive whatever value y takes. (Which is why I presented the Broome case.)

In the case I’ve given there, if y = 10 then the (known) gain from swapping is 9, if y > 10 then the expected gain from swapping is (roughly) 0.23y. Now of course the expected value of y is still infinite (and that’s relevant to the puzzle) but the angel doesn’t want to use that. Rather, he argues that if you opened your envelope, you’d take the deal no matter what was there, so you should swap now.

Maybe to make the deal sweeter, he says that once you open the envelope, the cost to taking the deal will be 3 days in hell rather than 1 – you know that you’ll take this deal once you open the envelope, and you know it’s worse than the deal he’s offering now, so why not go for it?

cks 11.03.03 at 11:43 pm

There’s another variation on this problem that is interesting. In another case, the angel may tell you that the envelopes contain the values x and 2x in them. You get a chance to choose one of the envelopes and open it. After seeing the value in the envelope, y, you can decide whether to switch. If you randomly choose the first envelope, and then flip a fair coin to decide whether to switch, you’re equally likely to get x or 2x as your final value. The expected payoff for this random strategy is 1.5x. Is there a way to do better than this by using the value for y?

It turns out that there is. You can make the decision to switch envelopes random, but dependent on the value of y that you pick. If the probability that you switch envelopes a decreasing function of y, one that goes to zero asymptotically, such as e^{-y}, the conditional probability of switching given that you opened the envelope with the larger value is smaller that the conditional probability of switching given that you opened the envelope with the smaller value. The expected payoff is strictly larger than 1.5x.

Tom T. 11.04.03 at 1:06 am

Isn’t this just a variation on the “Let’s Make A Deal” probability problem that provoked a small firestorm among Marilyn vos Savant’s readers?

(Apologies to the non-US readers for the doubly incomprehensible pop-culture reference).

Matt Weiner 11.04.03 at 1:19 am

Tom T.–

I think so, more or less.

Note that the order of writing down the numbers and giving them to you matters. In Brian’s problem, God puts y and 2y in an envelope, and then the angel randomly decides to give you one of them. You shouldn’t switch.

Suppose it went like this: God writes a number y in an envelope and has the angel give it to you. He then flips a coin to decide whether the other envelope should contain 2y or y/2. You now have the option to switch.

Then you should switch–no matter what y originally was, you’ll gain an expected y/4 by switching.

I think this is what matters, not the infinite distribution. Unfortunately I am not capable of formulating exactly what I mean by “this” right now.

Keith M Ellis 11.04.03 at 7:02 am

As something of an amateur expert on the MHP, I have to say that I don’t think this problem is _at all_ similar to the MHP, except in the sense that the words “stay” and “switch” appear in the problem annunciation.

dsquared 11.04.03 at 7:03 am

No, it’s not the same as the Let’s Make a Deal paradox. That was basically about a popular confusion over Bayesian reasoning and had a definite solution. This one is about a genuine mathematical point; that there are distributions which don’t have means, while the normal rules of decision theory rather rely on their having them.

Keith M Ellis 11.04.03 at 7:16 am

By “MHP” I mean the

“Monty Hall Problem”, which is equivalent to the “Let’s Make a Deal Problem”; also known as the “Car and the Goats Problem” and the “Three Doors Problem”.

Matt Weiner, I’m pretty sure your reasoning and conclusion above is false.

Dsquared, calling it the “Let’s Make a Deal paradox” is misleading, since it’s not a paradox.

Keith M Ellis 11.04.03 at 7:29 am

Matt Weiner, I retract my claim about your comment. I think you’ve got it right: at least one major error is in thinking that the two scenarios you give are equivalent. They’re not, and your conclusions about each of them are correct, I believe.

Keith M Ellis 11.04.03 at 7:41 am

Note that in Matt Weiner’s two statements of the problem, _y_ is a random integer. In the first problem, _2y_ is also necessarily an integer. However, in the second problem, _y/2_ is not necessarily an integer. This should serve as an indication to our intuitions that the two problems are not identical.

Alas, I’m not bright enough to comprehend in just these few minutes what is going on in this problem. However, I strongly intuit that we’re unwittingly confused by an assumption that two “terms” (I mean this in the most abstract, conceptual sense) of the problem are equivalent when they are not. Language in some way is leading us astray.

Thomas Dent 11.04.03 at 2:23 pm

Your expectation is 3x/2 whichever you pick, if the angel has truly given you an envelope at random and doesn’t know which is which. Therefore your expected gain is zero.

To invert the angel’s (fallacious) argument and expose it as such, without loss of generality denote the envelope you *don’t* choose by z. Then the envelope you do choose has an expectation value of (z/2 + 2z)/2 = 5z/4. Therefore you expect to *lose* z/4 by switching.

The fallacy is in comparing the two equally-likely cases by setting y to a fixed value in both. *Without any further information*, and seeing just a single envelope containing y, we would be justified in considering that y was drawn from the same statistical distribution in both cases, therefore we would take the expectation of y to be the same in both.

However, we have the information that God has fixed the contents of both envelopes ahead of time, and since they’re both sitting in front of you they won’t change in value depending on which you pick, so we must be consistent between the two cases and set y = 2x in case A (where the other contains y/2) and y = x in case B (where the other contains 2y).

The angel’s argument works only if the contents of the envelopes could magically change to twice (or half) their values when you switched so as to conserve the value of y. Or for example if he tosses a fair coin *after* you look inside one envelope and *then* gives you either half or twice the number on the paper.

To explain once more, say God has put 20 in one and 40 in the other. If you picked 20, then y = 20 and your expectation is 30 (not 25!). If you picked 40, then y = 40 and your expectation is 30 (not 50!). In both cases the angel’s reasoning is fallacious.

This has nothing to do with finite or infinite probability distributions, it’s just about the consistent use of mathematical variables in comparing two cases resulting from the same fixed but unknown initial condition.

Thomas Dent 11.04.03 at 2:27 pm

Of course if x>10 and you sneak a look at the paper and it says 16 then you should switch, but this is cheating! (a.k.a. improper use of prior information).

Bill 11.06.03 at 4:30 am

Let x be the amount in the higher envelope. Assess your prior probabilities on what x is (you can always do this).

Let y be the amount you find in the envelope you pick. You now know that either x = y or x = 2y.

So, the probabilities of those two possibilities will be proportional to your prior probabilities that x = y and x = 2y before you saw what was in your envelope.

So (assuming you want to maximize your expected winnings), if before you saw anything, your probability that x=y was more than twice as high than your probability that x=2y, then you should not switch; otherwise you should switch.

Jonathan 11.06.03 at 9:35 pm

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.

Michael 12.03.03 at 1:40 am

Wow. The first comment and the last comment I read were both spot on. Doug said it right at the beginning: when considering whether you’ll get 2y or y/2, the values of y are different. If x = 200 (why not?), then you are either holding the 200 envelope (and the alternative is 2y = 400), or you’re holding the 400 envelope (and y/2 = 200). Thus, 0.5 * 2y + 0.5 * y/2 = 0.5 * 400 + 0.5 * 200 = 300, which is precisely what your expected value was when choosing originally. So Jonathan is correct: Stick with your original choice and expect 1.5x (300, in my example), or switch and expect 1.5x plus a day in Hell. Easy choice.

BTW, two more observations. (1) The way I read the problem is that the choice has to be made before seeing the contents of the selected envelope, so there’s no worries about whether the value is an even or odd number, or whether it’s less than 20. (2) The problem may be easily stated in a way that has nothing to do with distributions with infinite means or how God selected x at all. Just say that x = 20, and require that the choice of whether to switch must be made without opening the envelope. Then, presumably, it’s safe to assume the probability of selecting the 40 envelope is 0.5, but the angel’s “reasoning” still sounds persuasive… or does it?

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