# Logic question

There are two envelopes containing money. You get handed one of them and it says on the back that the other envelope has an equal change of containing either half or double the amount of the current envelope.

You calculate the expected return to be bigger than 1 and change your envelope. It has the same text on the back. Do you then change back?

What’s interesting is that the first scenario is in the game boy game Wario Land. I always switched then, because expected return was higher.

The paradox only arises if you assume that all values are equally likely. But this is impossible: Infinite distributions don’t work that way. In any real-world distribution, the higher the value in your envelope, the higher the probability that the other one is half, and the lower the probability that the other one is double.

This isn’t so much a problem in logic as it is in evidence. What’s the evidence either written statement is even true, much less that the value can be estimated? Apart from opening both envelopes (which I gather by the rules of the game, you can’t do) there isn’t any. So there really is no expected value, just an assertion without evidence of probability or value (both of which you’d need, and neither of which you have).

So you might as well flip a coin.

Why? Give me a hypothetical where one dollar amount is in one envelope and the other is in the other; don’t, of course, tell me which amount is in which envelope — the one I’m now holding, or the one I could swap for — but do tell me what the dollar amounts would be in that scenario.

Wikipedia has a long article about this (and its predecessor problem, the necktie paradox). It’s actually not a trivial paradox at all, and there is no agreement on how to resolve it, but most proposed solutions rely on questioning the premise that the probabilities of the other envelope containing 2x or 0.5x (where x is the amount in my current envelope) are equal.

I would add that, even if it’s a metaphysical certainty, it’s still only a coin toss.

Even just granting that one envelope contains X, and the other 2X, you have no way of knowing which holds which. So the proper response is to flip a coin. And, without being able to examine the contents of both envelopes, you’ll never actually know if you “won.”

Again, it’s a problem in evidence, not in logic. No amount of logic will guide you to 2X. It will only cost you time. With metaphysical certainty as to the truth of each assertion, the “expected value” of each envelope is in fact the same: 1.5X. Without certainty or even knowledge as to the probability of truth, you have no expected value, only unknowns.

Either way, it’s a coin toss.

I haven’t read the article yet, but if I first got to know the amount in the first envelope, before I could choose whether to switch, the answer would be obvious.

And I did this over and over again in Wario Land, and made a lot of money. In that case it was a known amount.

I remember seeing a YouTube video not too long ago that gives what I consider a good explanation of what is going on. (Of course I’ve checked the likely suspects: Numberphile, StandUp Maths, etc. and no joy.)

Hidden underneath this if you look into the big picture is an infinite sum (the Wikipedia article discusses this, poorly, here and there). And if you are sloppily dealing with infinite sums you can do all sorts of stuff like “prove” 0=1.

People’s intuition on “randomly” selecting stuff from a “uniform” infinite set sucks.

Once you chuck out the idea of the amounts being from a distribution that doesn’t exist then the early point in the Wikipedia article applies. The envelopes contain fixed amounts of x and 2x dollars. Your expected win is 1/2 (x + 2x) = 1.5 x. Pick one and go home.

My take on it, after thinking some more.

Imagine you know the sums. They are 5 and 10.

Then it won’t matter, because either the first envelope contains a 10 or the second one does. Expected return is always 7.5.

Noe imagine you only know the sum of the first envelope. It is 10.

Then the other one can be 5 or 20. Expected return after switching is 12.5.

Now, imagine you get to know that the first envelope is a fixed, unknown sum. Then the second envelope was chosen at random to have either half or double that amount.

Then the expected return from switching would be 1.25X.

I think the paradox is, if someone handed you the other envelope, then the expected return from switching would still be 1.25x in your example. So, switch and your expected return goes up to 1.25x, then, what? Switch again? Because now you have the other envelope, and the first one is either half or double.

If someone hands you an envelope and tells you this, the information changes both sums, not just that of the other envelope.

I don’t think I know what “this” is in your reply. Someone hands you an envelope and the expected return from switching is 1.25x. So you switch. Now you have the other envelope – what’s the expected return from switching? Seems like it’s still 1.25x.

This is the Major Point of what goes wrong with people’s poor intuition. There isn’t a suitable way of achieving this “random” thing. Remember that it’s not just two values. It’s also all values that are half or twice those. And all values that are half or twice those, ad infinitum. You may not see infinity lurking behind this but it is when phrased the above way.

Again, people’s presumptions about randomness and distributions over infinite sets are almost always wrong.

So just ditch the whole concept of randomness in this context. It doesn’t work the way you want. So you are given two envelopes that another human being has chosen non-randomly to put x and 2x dollars in.

It’s still 1.25x. Since “x” isn’t going to change, the value isn’t going up every time you swap. It doesn’t matter how many times you swap, the value is still 1.25x. Choosing to not swap is no different than swapping twice. The expected value is still 1.25x. The more you run this experiment, the closer to the expected value you will get. If you just do it once, then you might end up with x, .5x or 2x. But if you run it a bunch, you will end up with 1.25x no matter how many swaps you do, provided the x-value of the initial envelope is random.

If you know what is in both envelopes, one is 10 and one is 5, but you haven’t opened either, then just pick one, it doesn’t matter.

If you pick one, it’s a 10, the other one will be a 20 or a 5, you switch because the 20 helps you more than the 5 hurts you.

It really comes down to the nature of the knowledge. Given the nature, if you’re allowed to open one, you don’t know more about the other one than it’s twice or half, you always switch. If you are just told it’s 10 and 5 or 2x and x, the numbers don’t matter, it doesn’t matter which one you start with. You open one and then switch. On average the switch will not help you, it’s the doubling. If you’re right ten times in a row it helps you more than if you’re wrong ten times in a row hurts you. If you’re right five times and wrong five times you wind up even.

I’m a bit nervous about posting stuff, since there has been a lot of analysis done already.
But here goes!

It seems to me it doesn’t matter which envelope you pick.

And to show that clearly, suppose you are shown the two envelopes, both with the same text.
Now it’s a purely random choice.

Does that make sense?

This. Unless we’re dealing with Schroedinger’s envelope, there’s no way both statements are true. Let’s say the envelope I select has a \$10 bill in it. That means the other one has either \$5 or \$20. I swap the envelopes. The first one still has \$10. Let’s say the second one has \$20. Now the statement on the second one means that the first one has either \$10 or \$40. But unless someone opened the first envelope and changed its contents while I wasn’t looking, or the money is in some sort of quantum indeterminate state, the statement on the second envelope is logically false.

What if they both contain \$0?

Then the other part of the premise, that the envelope contains money, is the part that is false .