Jesse

But remember, the envelope-stuffer is free to choose the probability function that determines how much money is in a randomly-generated pair of envelopes. In particular, there's no reason he can't make sure that the probability of getting a 100/200 pair will be equal to the probability of getting a 50/100 pair (although he could also choose the function so that you have twice the chance of getting a 50/100 pair as you do of getting a 100/200 pair).

Silent Acorns

I realize this. What troubles me about this problem is that the only way I can think of to determine the correct answer (that, in the long run, on average, it makes no difference if you switch or not) is by using the symmetry argument. If we assume that the other envelope has a 50-50 chance of being the higher or lower envelope, we conclude (incorrectly) that it is better to switch. Conversely, if we assume the correct answer and calculate the odds of the second envelope being the higher total we find it to be 1/3. Either way something is wrong.

One problem with these approaches is that the other envelope is NOT in some kind of quantum state, fluctuating between $50 and $200, just waiting for me to open it and force a decision. If it was, you should switch.

Lets consider it in more detail. Suppose I always put $100 in #1 and then flip a coin to decide weather I put $50 or $200 in #2. You then select an envelope:

0.50 chance of picking $100
0.25 chance of picking $50
0.25 chance of picking $200
average money in an envelope = $112.50

Now, if you find $50 or $100 in your envelope you should switch for a gain of $50 and $25, respectively. However, you might have selected the $200, in which case you are guaranteed to lose $100 if you switch. So an "always switch" strategy gives you:

0.25 chance of going from $100 to $200 (+$100) = +$25.00
0.25 chance of going from $100 to $50 (-$50) = -$12.50
0.25 chance of going from $50 to $100 (+$50) = +$12.50
0.25 chance of going from $200 to $100 (-$100) = -$25.00

Which sums up to an average gain of $0.00. In other words, no difference. Also, it doesn't matter what the odds of finding $50, $100, or $200 are. If you substitute x, y, and 1-x-y for the above odds (0.50, 0.25, and 0.25) you'll find that the switching stratagy always comes out even with the don't switch stratagy.

It's the focus on a single dollar total (like 100, or even X) that makes this problem tricky.

Jesse

Silent Acorns, you're definitely on the right track. My response below will give some of the rest of the answer away, so don't read on if you want to figure it out completely on your own...

The key is to think in terms of the rule the envelope-stuffer is using, and to realize he's always trying to outsmart you. For example, he could use the rule you mentioned, where he always puts $100 in one envelope and then has a 50% chance of putting $50 vs. $200 in the second one. But in this case, the person picking the envelope will know that if he gets $200, there is not a 50% chance the other will contain twice that and a 50% chance it'll contain half. But the envelope-stuffer can try to fix this. For example, he can use a probability distribution where he is equally likely to put any amount of money from $0 to $1,000,000 in the first envelope, and then a 50% chance the other envelope will contain half that and a 50% chance it'll contain double. So if your first envelope contains any amount less than a million, you won't be able to know if the second envelope contains half this or double this. However, if you get between 1 million and 2 million, you'll know not to switch, based on this rule, because the only way to get this amount is for the amount in the first envelope to have been more than 500,000 and for the second envelope to contain double that.

The envelope-stuffer can try to find even more clever ways to stuff the envelopes. For example, he could flip a coin a bunch of times, and if he gets tails on the first flip, the first envelope will contain between 0 and 1 million and the second will contain either double or half that; but if he gets HT, then the first envelope will contain between 1 and 2 million and the second will contain either double or half that; if he gets HHT, the first will contain between 2 and 3 million; HHHT, 3-4 million, etc. In this way there will always be some nonzero probability the envelope will contain any possible number, so there's no way you can automatically know if the second envelope you open will contain half or double what you found in the first envelope, as you could in some cases if he used the earlier schemes. Still, this isn't a perfect stuffing scheme, since depending on what you find in the first envelope you may be able to figure out that it's more likely that the second envelope contains half than that it contains double.

So the question is, is there any "perfect" scheme from the envelope-stuffer's point of view? If so, what is it?

Bumble Bee Tuna

Yes. Always put either 1 cent, 2 cents, or 4 cents in the envelope. He'd be insane to give out millions! Or even better, stop giving away his money to random strangers for no reason! That's the perfect scheme, IMO.

-B

Darth Dane

According to some physics, quantum physics, every occurence has 50 % of happening.

Take a deck of cards. You would say that there is 1/52 to pick ace of hearts. But according to quantum physics, the probability of picking ace of hearts is 50 %! Either it is ace of hearts or it is not ace of hearts.

Since the envelopes are identical, we will never know what is in the other one, until we actually open it, we may speculate all we want, but either the second envelope contains more money or it doesn't. Only by checking Reality can we find out what it Is




DD - Love Spliff

Lobstrosity

No, Dane, this is not a consequence of quantum physics. Where did you ever come up with that idea?

Darth Dane

No, Dane, this is not a consequence of quantum physics. Where did you ever come up with that idea?

Maybe it is not quantum physics, but the principle explained should still be valid.






DD - Love Spliff

SULPHUR

Do we toss out probability entirely

Darth Dane

Do we toss out probability entirely?

It wouldn't be practical. The earth may start to go fluid in 5 minutes, but if we live our lives as if it would, what would be the point of living? Probability tells us, that since it hasn't starting flowing away before now, and has afa we know, been around for billions of years, I will take it as probable that it will stay pretty much as it is´for a good time yet. I don't usually think about the possibility of earth floating away, but it is possible.

I live as if though I know that earth will stay put for a good while yet. Otherwise I would have a million things to fear and worry about, instead of living the fullest while we have physical life.






DD - Love Spliff

Lobstrosity

It's not quantum physics or any other kind of physics. The principle is not valid. You can argue that there are only two possible outcomes you can measure for a given process, but this says nothing about the probability of those outcomes.