2 envelopes problem revisited

Factual Questions
61

My answer is "Ignore technical formalizations of probability theory. Let’s look at all the possibilities, and how good it is to switch in each possibility. It looks like this:



YOUR ENVELOPE

| 1 | 1 | 1 |   |   |   |   | 
| - | - | - | 1 | 2 | 4 | 8 | 
| 8 | 4 | 2 |   |   |   |   | 
+---+---+---+---+---+---+---+

+---+---+---+---+---+---+---+ +--
|   |   |   |   |   |   |   | |
|   |   |   |   |   | 4 |   | | 8  T
|   |   |   |   |   |   |   | |    H
+---+---+---+---+---+---+---+ +--  E
|   |   |   |   |   |   |   | |   
|   |   |   |   | 2 |   |-4 | | 4  O
|   |   |   |   |   |   |   | |    T
+---+---+---+---+---+---+---+ +--  H
|   |   |   |   |   |   |   | |    E
|   |   |   | 1 |   |-2 |   | | 2  R
|   |   |   |   |   |   |   | |   
+---+---+---+---+---+---+---+ +--  E
|   |   | 1 |   |   |   |   | |    N
|   |   | - |   |-1 |   |   | | 1  V
|   |   | 2 |   |   |   |   | |    E
+---+---+---+---+---+---+---+ +--  L
|   | 1 |   |-1 |   |   |   | | 1  O
|   | - |   | - |   |   |   | | -  P
|   | 4 |   | 2 |   |   |   | | 2  E
+---+---+---+---+---+---+---+ +--
| 1 |   |-1 |   |   |   |   | | 1
| - |   | - |   |   |   |   | | -
| 8 |   | 4 |   |   |   |   | | 4
+---+---+---+---+---+---+---+ +--
|   |-1 |   |   |   |   |   | | 1
|   | - |   |   |   |   |   | | -
|   | 8 |   |   |   |   |   | | 8
+---+---+---+---+---+---+---+ +--

Is it good or bad, on average, to switch? Well, on the one hand, on each column, we can see that switching is good on average. On the other hand, on each row, we can see that switching is bad on average. On the other hand, along each -diagonal, we can see that switching is a wash, on average. That’s odd; how can the analysis change based on how we group things? Well, this square shows us how; it’s just a brute fact that regrouping can change things in this sort of way, in this sort of example. We know that regrouping won’t change things if there are only finitely many possibilities, but clearly, it can in cases like this. None of the paradoxical reasoning is fallacious; this is just the sort of case where there is no well-defined answer as to how good switching is on average, independently of how we group the calculation."

62

To sum up all of what Indistinguishable has been saying: Infinities are weird, and don’t behave the way your intuition says they would. In this case, the weirdness is that it’s not always unambiguous how to average over an infinite set.

63

and didn’t.

64

Perhaps it would help to consider a finite case, similar to Indistinguishable’s. Suppose you have eleven possible values in the envelopes: 1, 2, 4, 8, …, 2048. And suppose you have ten possible pairs of those envelopes you might be given: 1 & 2, 2 & 4, … up to 1024 & 2048, with each possibility equally likely.

You’re given a pair of envelopes, and open one of them, and it’s not one of the end cases (so not 1 or 2048). Should you switch? If your envelope had, say 128, then your expected value when switching is (64 + 256)/2 = 160. Yay, switch!

This is just like the paradox you see when the amount in your envelope is unspecified. But here it’s easier to see how this occurs. 5 percent of the time, you’d open the envelope with 2048 in it, and you’d know that you shouldn’t switch. If you double the number of possibilities, with 21 envelopes up to a couple million, the chance of picking the end-case envelope is cut in half. Increase the number of envelopes enough, and the chance of an end-case envelope can be made arbitrarily small.

It’s tempting to say “Well, the chance of getting the lowest or highest value is so small, we can ignore it.” Then it looks like it’s always beneficial to switch. But the benefit comes from knowing you haven’t selected the one envelope with the largest amount.

The OP’s analysis essentially does this, ignoring the end cases, when it assumes p always has the same value. If you’ve picked the highest-valued envelope, though, p = 0.

65

Perhaps thinking about it this way will help. Suppose there are n possible amounts 2, 4, 8, …, 2[sup]n[/sup] each of which is equally likely. You are given an envelope and told you may open it then decide if you wish to switch. If you see 2 you’ll want to switch and will gain 2. If you see 2[sup]n[/sup], you would gain 2[sup]n-1[/sup] - 2[sup]n[/sup] = -2[sup]n-1[/sup] and would not switch. If you see 2[sup]m[/sup], your expected gain from switching would be 0.52[sup]m+1[/sup] + 0.52[sup]m-1[/sup] - 2[sup]m[/sup] = 2[sup]m-2[/sup] so you should switch.

What is the probability before seeing the contents that you would wish to switch? The answer is (n-1)/n since you want to switch unless you have the largest amount.

What is the expected gain from switching before you see the contents. Each possibility has probability 1/n so the expected gain is (2 - 2[sup]n-1[/sup] + 1 + 2 + … + 2[sup]n-3[/sup])/n. The last terms inside the parentheses starting with 1 sum to 2[sup]n-2[/sup]-1, so the expected gain from switching is (1 - 2[sup]n-2[/sup])/n. So the unconditional expected value of switching is negative.

As n tends to infinity, the probability that you would wish to switch if you knew the contents tends to one, but the expected benefit of switching tends to minus infinity.

Thus the fact that you with probability one would like to have switched does not guarantee that the expected value of switching is positive.

66

Yup.

Oh, hey… It took me a year and a half, but I’m finally done grading midterms.

67

Great memory Indistinguishable. I had forgotten I had even participated in that discussion.
I just started reading this one and thought – this paradox seems familiar.
The great thing for me is that I got your explanation on summation without nary a blink. I must have growed up in my math in the last couple of years.

68

To those who were worried about my non-Kolmogorov “blistributions”, let me propose the following unobjectionable Kolmogorov distribution:

Suppose the envelopes are both filled with money and labelled on the back, like so: First, one envelope is filled with $1 and the other envelope is filled with either $2 or $0.50, according as to a coin flip. Both envelopes are labelled on the back with the quantity in them.

Then, a coin is flipped repeatedly till it comes up tails; each time it comes up heads before that, the envelope contents are quadrupled and the labels are doubled (if the second envelope was originally filled with $2) or the envelope contents are quartered and the labels are halved (if the second envelope was originally filled with $0.50).

Finally, a coin is flipped to determine which of the two envelopes to give to you and which to leave as the other envelope. Both envelopes are label-side-down.

Now, what will happen if you turn over your envelope to read (and more importantly, condition on) its label?

No matter what label you read and condition on, you’ll find the expected contents of the other envelope to be greater than those of your envelope! (And symmetrically, if you had instead read the other envelope’s label, no matter what it was, you’d find the expected contents of your own envelope to be greater than those of the other). Two-envelope paradox!

In fact, if you draw out the diagram to calculate the expected value of switching in this game, with the axes tracking the envelope labels, you’ll find it to be exactly the same as the diagram I drew before.

This should make clear that the paradox has nothing to do with any technical difficulties with a uniform distribution on a countable set, and really does have everything to do with conditionally convergent summation.

69

That’s still rather insulated from the real world, though, in that the a priori (before you start flipping any coins) expected value of that game is infinite. For a real-world game, you’d have to include some escape clause for what to do if the value you’re supposed to put in an envelope exceeds your bankroll, and just what that escape clause is would determine unambiguously whether you should switch or not.

70

Well, I don’t think it’s meant to be about “the real world”, but there’s nothing preventing us from taking it as about “the real world”; just consider the prize to be points in a game of our devising rather than some limited resource (of course, money is just points in some human game, but most of us aren’t invested with the authority to award it at will…). The only constraint is time to flip the coins…

71

Even arbitrary “points” will still have a limited budget in the real world, though. If nothing else, you have a limit in your budget of how many zeros you can fit onto the page and still be readable.

And the fact that folks refer to it as a “paradox” is itself a sign that the mathematics is being related to the real world. All of the mathematical “paradoces” ultimately come from some bit of mathematics that isn’t actually modeling something in our world, but which looks almost like it is, so that the differences between the mathematical results and what actually happens are called into contrast.

72

Hm, I wouldn’t necessarily say that myself, but I have no reason to argue with it either.

Re: the limited resources issue, let me say this:

We might perfectly well imagine another world with unbounded ability to produce paper. It’s not our world, but it’s a perfectly plausible alternative world to think about. The inhabitants of such a world might ponder our two-envelope paradox themselves, and could not wave it away by saying “Well, clearly, in any real-world situation, there’s an implicit bound on the paper which can be made available…”. How are they to react to the paradox if enacted as a real-world game?

I would say they should react to the paradox by abandoning the idea that there is a clear Correct Move to take in the game which is determined by an unambiguous expected value calculation. It is manifestly, unarguably the case that there is not an unambiguous mean value (for the profit from switching envelopes), and, more arguably, there’s no reason to find this troubling. As I see it, the only reason to find it troubling is because one conjoins a bridge principle unifying mean values and decision-making with the silly idea that in every situation, there is always a particular decision to be singled out as the Correct One, or at least that there is always Platonically granted to any decision a status as either Good, Bad, or Neutral. Abandon this last superstition, and all is well.

Indeed, I would say we should react to the paradox the same way regardless of how bounded our paper-producing capabilities are.

73

Here’s one reason I don’t like the response “Of course, what really solves the paradox is the fact that in the real world, everything is finite.” It’s like the common response to Zeno’s paradox “Of course, what really solves the paradox is the fact that in the real world, everything is discrete”, as though Zeno’s paradox did logically disprove the coherence of continuous motion.

The actual resolution of Zeno’s paradox is to note that “Yup, you can indeed have infinitely many points in a finite span, just as you’ve noted with your very example. There’s nothing incoherent in that. Your intuitions are mistakenly upset at this, from training with cases of a different sort, but just gaze upon your example for long enough, and that sense should dissipate.” And I would say something similar to this.

74

For the past few days I’ve been keeping this problem in the back of my mind trying to boil it down to the simplest ideas I can. Here’s the problem as I saw it:

The “paradox” comes up when you consider yourself to be holding envelope A and being offered envelope B, which as far as you can tell, contains either twice or half the value you currently hold. Thinking this way leads to a conflict. I believe that the problem with this method of thinking is that envelope B does not hold twice or half the value of A. It’s value is already determined when it was sealed. You can’t randomly get a value in envelope B, you can only randomly choose envelopes A or B. In this way, what you get when you switch is just a result of which envelope you chose in the first place.

I mapped out these scenarios in probability trees in a video on my blog: http://davidkann.blogspot.com.au/2012/06/two-envelope-paradox.html

75

Simple answer:

Yes, the expected value of the other envelope is 5/4X, so you should switch.

NO!

Why? Because to determine the value of switching, you must subtract what you are giving up by switching which is the expected value in your envelope: 5/4X. So 5/4x-5/4x=0.

0 is the expected benefit of switching.

76

jtgain, what do you mean by X? You only get that the expected value of the other envelope is 5/4 X if you define X to be the amount in your envelope, so you can’t then turn around and say that the expected value of your envelope is 5/4 X.

77

I’ve spent the last 7 years developing an immunity to iocane powder.

78

Yeah, dammit, I know. I had a 2 1/2 hour drive today and I spent most of it trying to figure out this fucking problem. :slight_smile:

Yes, you have X. Switching gives you 5/4X in expected value. You should therefore switch forever, but intuitively you KNOW that either envelope gives you a 50/50 chance of the larger payoff. No point in switching at all.

No offense to other posters, but many of the responses make me go cross-eyed. Is there a simple explanation as to why the expected value calculation is wrong?

79

If you are referring to the expected value of the profit from switching, the simple answer is: It’s not well-defined. None of the conflicting calculations are wrong, as such. There just isn’t a well-defined mean profit to switching.

This is a phenomenon which can sometimes happen when trying to take a mean of something over infinitely many possibilities (just like the ambiguity in trying to figure out the mean of the integers).

80

Clarification added in bold