A real poser this time

Originally posted by Palynka
What about what I wrote?

If you see the choice of swapping as betting 0.5x to have a 50% chance of gaining an extra x.


Edit: To prove it doesn't matter if you swap.

You should see the problem differently because you already have X, so you have 0.5x to lose and x to gain.

OK

Originally posted by PBE6
This one will probably (I hope) start a discussion, aka "pissing match", so here it is:

You are presented with 2 envelopes, each containing a non-zero amount of money. You are told that one envelope contains twice as much as the other one, and then you are asked to choose an envelope. Before opening your envelope, you are given the choice to switch with ...[text shortened]... e.

The questions is: which is more advantageous, to switch or to keep your original envelope?

because at no point in the scenario is a decision made because of the value of the envelopes the odds of either having picked the more valuable envelope 1st or switching to it 2nd, therefore it makes no odds at all.

think it through logically,

you pick an envelope, a or b, there is a 50/50 chance that the envelope you picked is the more valueable.

you are asked if you want to switch, there is no change in the odds, they remain the same. there is still a 50/50 chance of either envelope being the more valueable

Originally posted by Acolyte
As the exhaustive analysis correctly points out, it is possible for problems to arise. However, this won't happen if the distribution X of the total money is 'nice', ie with finite mean and variance, which it would be unless the person offering you the envelopes was someone of infinite means.

In your original description of the problem, you said that ...[text shortened]... en you can start speculating on the distribution of X, and the problem becomes more complicated.

I dont understand why opening the envelope before you decide to switch or not makes any difference on your chances of improving your situation.

Either way, you've got a 50% chance of getting 2X, and a 50% chance of getting 0.5X.

Originally posted by richjohnson
I dont understand why opening the envelope before you decide to switch or not makes any difference on your chances of improving your situation.

Either way, you've got a 50% chance of getting 2X, and a 50% chance of getting 0.5X.

If you pick the envelope without opening it, and then switch, it's exactly the same as just picking the other envelope first, so the chance of getting the bigger one is 50/50.

However, if you open the envelope first, it may influence your decision on whether to switch or not. For instance, if there is a maximum amount which can be in any envelope, say $1,000,000, and an equal probability distribution amongst all the integer amounts, then if you found $500,001 or more in the first envelope you wouldn't switch (since the other envelope could only contain half the original amount).

It gets tricky if you say there is no maximum, or if you have a strange probability distribution. There's a link on my earlier post to a paper by someone who's thought a lot about it (a much better source than me).

Originally posted by richjohnson
I dont understand why opening the envelope before you decide to switch or not makes any difference on your chances of improving your situation.

Either way, you've got a 50% chance of getting 2X, and a 50% chance of getting 0.5X.

What does X represent in your statement?

Originally posted by DoctorScribbles
What does X represent in your statement?

Sorry, that was sloppy. X represents the amount of money in the envelope you choose first.

My question relates to the case where there is no maximum amount of money, and no funny money distribution algorithms:

Does it matter if you know X?

I can see that it may have a psychological effect on some people (i.e., you might be comfortable in wagering $5 if the 1st choice envelope contained $10, but not comfortable wagering $500,000 if the 1st choice envelope contained $1,000,000).

Originally posted by richjohnson
Sorry, that was sloppy. X represents the amount of money in the envelope you choose first.

My question relates to the case where there is no maximum amount of money, and no funny money distribution algorithms:

Does it matter if y ...[text shortened]... agering $500,000 if the 1st choice envelope contained $1,000,000).

EDIT: Lest I confuse you, I should point out that in your first post here, your first sentence is correct, but your second sentence is not. It is your second sentence that I am addressing here.

-----------------------------------

My question was somewhat rhetorical in nature.

X is not well-defined, even after you say that X represents the amount of money in the envelope you choose first, and even after you know the value of the money in that envelope.

Allow me to be more precise. You'd like to use X in this way, to claim two things:
1. X is the amount of money I see when I open an envelope.
2. The other envelope contains either 2X or .5X, with equal likelihood.

The problem is that in (2), X necessarily represents something other than what you claim it represents in (1). This is not obvious or intuitive to everybody, but nonetheless, it is so.

Let us be more concrete to illustrate that this is true. Suppose you find 20 in the First envelope. Is it really true that the Other envelope is just as likely to contain 40 as 10, and that by switching you expect to end up with 25 on average? Well, let us consider two cases to determine this.

Case 1: Suppose that the Other envelope contains 10. By your reasoning, if the Other envelope had been opened first, you would then say that the First envelope is just as likely to contain 5 as 20. Therefore, again by your reasoning, opening the Other envelope first, you expect the First envelope to contain (20+5)/2 = 12.5, and thus you would switch to choosing the First envelope, since 12.5 > 10.

Case 2: Suppose that the Other envelope contains 40. By your reasoning, if the Other envelope had been opened first, you would then say that the First envelope is just as likely to contain 20 as 80. Therefore, again by your reasoning, opening the Other envelope first, you expect the First envelope to contain (40+10)/2 = 25, and thus you would switch to choosing the First envelope, since 25 > 20.

In both cases, the very logic that would lead you to switch in the first place would be employed to show that it is more profitable to switch back. Thus, we have a contradiction, so your logic that leads you to switch --- (2X + .5X)/2 > X --- is necessarily flawed.

So, that is a proof that X can't have a consistent meaning across (1) and (2) above.

I can't easily explain why this is so, because it is so obvious that it is. It reminds me of the missing dollar problem, if you're familiar with that. There, if somebody believes that there really is a missing dollar, it's incredibly difficult to put into words why there isn't, except to say that there is simply no reason to arrive at that assertion.

I'll think it over some more to see if I can provide a better explanation. But just as addition, subtraction, and variables are simply abused in the missing dollar problem, so do many people abuse random variables and expected value in this problem, which we have seen.

Dr. S

Originally posted by DoctorScribbles
EDIT: Lest I confuse you, I should point out that in your first post here, your first sentence is correct, but your second sentence is not. It is your second sentence that I am addressing here.

--------------------------------- ...[text shortened]... s and expected value in this problem, which we have seen.

Dr. S

Lets state:

Total ammount in both envelopes = a

a is a constant and doesn't depend on probabilities. E(a) = a

E(first choice) = E(X) = (a)/2

You open one envelope. The ammount inside is X.

Ammount in other envelope = a - X
(Here is the key difference. The ammount in the other envelope should be seen as a-X, not 2X or X/2)

E(other envelope) = E(a) - E(x) = a - a/2 = a/2

Originally posted by Palynka
(Here is the key difference. The ammount in the other envelope should be seen as a-X, not 2X or X/2)

E(other envelope) = E(a) - E(x) = a - a/2 = a/2

Very good. Quite correct and quite clear.

Originally posted by DoctorScribbles
EDIT: Lest I confuse you, I should point out that in your first post here, your first sentence is correct, but your second sentence is not. It is your second sentence that I am addressing here.

--------------------------------- ...[text shortened]... s and expected value in this problem, which we have seen.

Dr. S

It is clear to me that switching does not give you an advantage. My question is whether or not it matters if you look in the originally chosen envelope before deciding to switch.

Originally posted by DoctorScribbles
Suppose you find 20 in the First envelope. Is it really true that the Other envelope is just as likely to contain 40 as 10

I don't know, you tell me:

Originally posted by PBE6
You are told that one envelope contains twice as much as the other one, and then you are asked to choose an envelope.

What are the chances that you picked the small envelope?

Originally posted by richjohnson
My question is whether or not it matters if you look in the originally chosen envelope before deciding to switch.

It does not matter whether you look in the envelope before deciding to switch or stay.

Originally posted by richjohnson
Originally posted by DoctorScribbles
[b]Suppose you find 20 in the First envelope. Is it really true that the Other envelope is just as likely to contain 40 as 10

I don't know, you tell me:[/b]

No, it's not.

Originally posted by DoctorScribbles
No, it's not.

Does that mean that the other envelope is more likely to contain 10 than 40?

Originally posted by richjohnson
Does that mean that the other envelope is more likely to contain 10 than 40?

No. Nor does it mean that the other envelope is more likely to contain 40 than 10. It means that the expected value of the contents of the Other envelope cannot be expressed solely in terms of the revealed contents of the First envelope.

If it could be expressed as you want it to be, you'd end up with a contradiction, as I showed before. I'll try again. Name the envelopes A and B.

I open A and find 10; you open B and find 20, but we don't share our information with each other. I conclude that B has an "expected value" of 12.5; you conclude that A has an "expected value" of 25. So, we both want to switch based on the same logic - a contradiction. That is, one strategy says that both A is better and B is better.

Now, if you're willing to live with this contradiciton for one iteration, let's take it a step further and actually perform the switch. Now I hold B, and you hold A, without opening them. I know that you had found either 5 or 20, so I know that I "expect" to find 12.5 when I open B; similarly, you "expect" to find 25 when you open A. But, this is another contradiction, for I always find 20 and you always find 10 in this scenario; the 12.5 and 25 EV numbers are completely bogus.

Now, at the beginnning of this scenario, it could have been that A found 20 and B found 10, in which case a similar analysis would show that I always end up with 10 and you always end up with 20 after we switch. Nobody ever averages a 12.5 or 25 finding, as your proposed method predicts one must, and thus your proposed method must be flawed.

Dr. S

Originally posted by DoctorScribbles
No. Nor does it mean that the other envelope is more likely to contain 40 than 10. It means that the expected value of the contents of the Other envelope cannot be expressed solely in terms of the revealed contents of the First en ...[text shortened]... one must, and thus your proposed method must be flawed.

Dr. S

I never proposed a method, only asked questions. (I retract my above statement regarding 0.5X and 2X, since you have shown this is inaccurate).

I don't want it to be expressed in any particular way, but I am curious as to how one would properly express the probability of the envelope not chosen being the one with more money.

ps- please bear with me, it's been a long time since I've done any real math.