Sunday, April 04, 2004

The 2-envelopes paradox

A really interesting math paradox was discussed at our university's "philosophy retreat" (at a lodge up in the mountains, it was very cool) a month or so ago. Crooked Timber posted a variation of it late last year. Anyway, the basic formulation is as follows:

Imagine you are offered a choice of two envelopes, and all you know is that one of them has twice as much money in as the other. So you pick one at random, but just as you are about to open it, you are given the option to swap envelopes, if you want.

Now, it should be obvious that there is nothing to be gained by swapping to the other random envelope. But the more you think about it, the more you begin to doubt this. After all, however much money you have at the moment, the other envelope either has half of that, or else double. Each possibility is equally likely, so by averaging them [ (2 + 1/2) /2 = 1.25] you realise that by swapping envelopes you will, on average, make a 25% profit!
This answer is clearly incorrect... but what is wrong with the math used to work it out?

Here's the solution I came up with...

First, formalise the scenario:
We have two envelopes, with values x and 2x (for some unknown x).
You take one at random. Call whatever value is in this envelope "y", and whatever is in the other one "z".

There is a 50% chance that you have the lesser envelope (y = x, z = 2x), so swapping in this case would double your money (z = 2y).
The other 50% chance is that you have the greater envelope (y = 2x, z = x), so swapping will halve your money (z = y/2).

It seems that based on the variable y, the expected value of swapping (to the other random envelope z) would yield a net increase of y/4 [E(z) = .5 * 2y + .5 * y/2 = 5/4 * y]. Yet this is clearly ridiculous (especially since you can do the same calculations on z to reach the opposite conclusion).

What has gone wrong?
The problem is that, within the E(z) calculation, the y in "2y" is different from the y in "y/2". That is, the variable y is being used to simultaneously represent two different values (x in one place, and 2x in the other).

Of course there is no problem with using a variable to represent some (single) unknown value (such is their purpose). BUT in this case the expected profit is calculated using two different instances of y simultaneously, whilst (deceptively) treating it as though it were a single variable (with a single value).

This inconsistency causes the bizarre results. The y that gets doubled is different from the y that gets halved. It is easy to lose sight of this crucial fact.

Note that there are only 2 unique numerical values involved in the entire scenario: x and 2x (all other variables (y and z) are equal to one or the other value).
Contrast this to sort of scenario insinuated by talking about having one envelope containing y (or x, or whatever you want to call it), and another which contains either half or double this... here you have the base case x, but also two other possible values, x/2 AND 2x.

This demonstrates a crucial difference between the original scenario, and this other (deceptively similar) implied scenario. To highlight this difference, consider the range of possible values:

1) in the original scenario, possible values range from x to 2x (so the maximum is double the minimum).

2) in the implied scenario, possible values range from x/2 to 2x (so the maximum is QUADRUPLE the minimum).

So how should we solve it instead?
There are several alternatives, but I think the most illuminating are probably the following two methods:

1) Substitute in x instead of y. Notice that when calculating the expected value of z, the "y/2" part is using y = 2x (so "y/2" can be replaced with "x"), whereas the "2y" bit is using y = x (so "2y" is to be replaced with "2x"). Substituting this gives us [E(z) = .5 * 2x + .5 * x = 1.5x], which is a zero profit, as expected.

2) Use "y" consistently.
To do this, we must calculate the expected initial value of y, before you work out the profits. This is easy enough: E(y) = 1.5x. Notice that the other envelope z does NOT possess "either half or double" of the expected value of y. The wording of the problem sort of implies this lie, which is why the problem sounds so confusing at first.

By using y consistently (i.e. by calculating its expected value), we can expose this flaw in the so-called "paradox". Solving for z gives E(z) = 1.5x also, so there is no gain to be made in swapping.

The math works... one just needs to be careful to apply it properly ;)

Update: I've just noticed that Matt Weiner posted about this paradox several times in January. A quick skim shows his answers to be a lot more complicated than mine... I'm not sure whether that means my solution here is insufficient, or if he's just missed the easy answer. Any ideas?

2nd Update: Crooked Timber is back into it!

4 comments:

  1. If you look in the first envelope you pick and see what's in there, that amount can't be a variable anymore. The paradoxical argument can still be performed, but your solution breaks down. Sorry about that. ;-)

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  2. If you look in the first envelope you pick and see what's in there, that amount can't be a variable anymore. The paradoxical argument can still be performed, but your solution breaks down. Sorry about that. ;-)

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  3. Yeah, tricky. I guess if we fix the value of y (e.g. by looking inside the envelope), then we simply have to deny that there is a 50% chance that z = y/2 and a 50% chance that z = 2y.

    Though defending this denial will require further explanation than I've offered here. Oh well.

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  4. this problem ONLY becomes philosophical ONCE you do not know algebra. The correct maths are: half of x plus double of y over two is equal to 1:
    (1/2*2 + 2*1/2)/2=1
    so there is equal possibility for any case.

    the possibility of 1.25 only exists for the place, before recieving the envelopes there is a 125% possibility to increase his zero money by 75%.

    ReplyDelete

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