The envelope problem, and a possible solution

[NaobisDad]

THank you for your reply Jason. You make a couple of very interesting points. You also seem to argue that the size of the value and the sampling method are of critical importance. A number of others have argued this, but I sill can't see the point. I suspect that I lack specific knowledge to fully appreciate those arguments. if you would be so kind, I'd really appreciate it if anyone could make a bit more of a detailed post such that laymen like me can understand.

The point about the relation to poker and point of determination is exactly the point that led me to question this explanation myself. However, there are some differences between pokersituations and this situation that might be of influence. I will not yet go into those here. But I do want to discuss that, because initially I was going to relate this post to exactly that. Which is why I started a new post before, instead of replying in the original thread.

[BBB]

In the envelope problem, to come to any conclusion other than that of argument #2 requires you to make one or more assumptions which are not given in the original problem. If you cannot derive any useful information from viewing the content of one of the envelopes, then it doesn't matter if you switch or not. Argument #2 is flawed if and only if you somehow gain information beyond that presented in the original problem by viewing the contents of one of the envelopes.

The flaw in argument #1 is that it assumes that, regardless of the amount that we see in the envelope we open, there is a 50-50 chance that the amount in the other envelope is double this amount, as opposed to half this amount. This statement cannot possibly be true for all values that we may encounter in the envelope.

Put simply, when we see that our envelope contains $100, we know that the other envelope contains either $200 or $50. We do not know, however, the chances that it contains either amount.

[NaobisDad]

@BBB: thank you for your reply. You said: "The flaw in argument #1 is that it assumes that, regardless of the amount that we see in the envelope we open, there is a 50-50 chance that the amount in the other envelope is double this amount, as opposed to half this amount. This statement cannot possibly be true for all values that we may encounter in the envelope."

It is possible to make a distribution such that those chances were equal, according to your rational it would then always be +EV to switch. I think this is not right. The reason I think this can be illustrated by making explicit how I expect the game to look.

The way I understand it, you have two envelopes, one containing value N and the other the value 2N. One envelope is picked randomly, and then there is the option to switch or not. Let's say we always switch (for reasons I stated earlier). Most would probably agree that whether you switch is rather trivial.

In the second condition the only change we make is that we flip open one of the envelopes and reveal it's contents. Then we switch again.

Nowhere in the original problem is it stated that the value you're shown influences the probability that the other value will be higher or lower. It simply states that you had 50% chance to pick either N or 2N, and that now that you know one value, you have a 50% chance that you hold either N or 2N. Reasoning on from there they then state that your EV is 50% * value/2 + 50% * value*2.

If argument 1 is wrong, which I feel must be the case, then I think that it is because somewhere a logical step wasn't allowed.

For those who state that sometimes switching might affect EV, do you think that this would still be the case if you were switching blind. And do you think that seeing the value matters?

If not, then you probably agree that agrument 1 is probably flawed, where is it flawed?

[jason1990]

Okay, here's a little thought experiment. Imagine that I am the one placing the money in the envelopes and you are the one choosing. There are infinitely many methods that I may use to decide what money to place. I will present two of them for discussion.

The first I call Method 1. What I do is roll two dice. I multiply the sum by 100 and place the result in an envelope. I then flip a coin. If it's heads, I place twice that amount in the other envelope. If it's tails, I place half that amount in the other envelope.

The second is Method 2. It's the same as Method 1, except this time I roll only one die.

Now you open an envelope and it contains $800. What is the probability that the other envelope contains $1600?

First, suppose you somehow know that I used Method 1. Since you can see $800, you know that I either (a) rolled a 4 and flipped a head, (b) rolled an 8 and flipped a tail, or (c) rolled an 8 and flipped a head. Only in case (c) will the other envelope have $1600. The probabilities of these events are (a) 3/72, (b) 5/72, and (c) 5/72. Hence, the conditional probability of (c), given that you see $800, is 5/13. So the EV of switching is

(5/13)(1600) + (8/13)(400) = 861.5 > 800

and it is +EV to switch.

Next, suppose you somehow know I used Method 2. Then it is impossible for the other envelope to contain $1600, so the EV of switching is only 400, and it is -EV to switch.

Of the infinitely many other methods I might use, some will generate neutral EV, some +EV, and some -EV. So the EV depends on the method, which is not known to us, since it's not stated in the problem. This is why the only right answer is "it depends" (both on the method and on the amount of money you see in the first envelope).

Now, one might be tempted to argue that since we don't know the method, then it's EV neutral to switch. But to say this, one would have to assign probabilities to all of the different methods and do so in such a way so that the resulting overall EV was 0. Such an assignment of probabilities is clearly not given in the problem.

Finally, for BBB's sake, let me repeat what I said before about Argument 2. It is EV neutral to always switch. Argument 2 is correct about that. But the problem did not ask about the EV of always switching. It asks about the EV of switching.

P.S. Suppose I use Method 2 and you open an envelope and discover $200. Then you can compute that the probability of the other envelope containing $400 is 1/2. So it may sometimes happen that the chance of the other envelope being greater is 50-50. But (under Method 2), this will only happen when you see certain values. It will not always be the case that the chances are 50-50. One of the main points of this puzzle is that there is no selection method that will always generate a 50-50 chance in this situation.

[rufus]

Actually, (at least in part) the trick is that the EV for envelopes where it 'makes sense' to switch is undefined or infinite.

For example, let's say I keep flipping a coin, and count h, the number of consecutive times it comes up heads. Then I put $(3^h) in one envelope, and $(3^(h+1)) in another. What's the expected value of an envelope?

[jason1990]

The EV of the envelope is not relevant. What matters is the conditional EV of switching, which is always well-defined and finite. It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p>1/3. The value p is always a well-defined number between 0 and 1.

[BBB]

[ QUOTE ]

Nowhere in the original problem is it stated that the value you're shown influences the probability that the other value will be higher or lower. It simply states that you had 50% chance to pick either N or 2N, and that now that you know one value, you have a 50% chance that you hold either N or 2N. Reasoning on from there they then state that your EV is 50% * value/2 + 50% * value*2.


[/ QUOTE ]

Hi Naobisdad,

All we are told in the original problem is that one of the envelopes contains twice the other. This tells us that once we open one envelope, that the other contains either half as much or twice as much. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct.

As an example, suppose that the designer of the game opted to put 50 and 100 in the envelopes (but that we don't know this). Clearly, it is -EV to switch in this case if we open the envelop with 100, as there would be a zero percent chance that it would be 200, and a 100 percent chance that it would be 50. Basically, the two envelopes contain X and 2X. The EV of switching no matter what will be -X if we chose the 2X envelope, and X if we chose the 2X envelope. Since there's a 50% chance of chossing either envelope, the net EV of switching no matter what is zero.

As a more involved example, suppose that we're told that the numbers were chosen by randomly selecting some integer k between 0 and K (where K is some positive integer that we are not told), and then placing $2^k in one envelop and $2^(k+1) in the other. Then, clearly, we should swtich if our envelope contains $1. But, what about for all other values? Suppose our envelope contains $2 or more, and we've decided to switch no matter what. Then, if our envelope contained $2^K, we would automatically lose $2^(K-1) by switching. If our envelope contained less than $2^K but more than $1, chances would be 50-50 that the other envelope contained more, so we would expect to gain on average 1/2 the amount in our envelope by switching (i.e., argument #1 from the original post applies for those cases). So, the total EV of switching no matter what would be: Sum(i=1..K-1){1/2*2^i} - 2^(K-1) + 1 (the +1 at the end is because we always gain $1 when we see $1 in the envelope and switch). The first term is a geometric series (minus the i=0 term); the summation is: 1/2*(1-2^K)/(-1)-1/2, or 2^(K-1)-1. So the total EV of a switch = 2^(K-1)-1 + 2^(K-1) - 1, with equals zero. So, actually, in this setup, we should switch if the envelope we open contains $1 (since the EV for switching in this case is +$1), and we should keep our envelope otherwise (since the net EV for all other cases is -$1). Note that in this example, argument #1 (50% gain by switching) holds for most of the cases, but this is offset by the 50% loss which happens when switch the largest possible amount, which offsets the 50% gains in the cases of the smaller amounts.

To prove that, without inferring any other information about the problem, that switching is EV neutral, consider this: Suppose the deal is that we're splitting the money with another player, with whom we were not allowed to discuss a strategy beforehand. After we choose, the envelopes will be resealed and presented to the other player. He will be handed the envelope that we chose, will open it, and will decide which envelope to keep. If it were +EV to always switch, then we would of course have switched. But from his perspective, it would also be +EV to switch. Except that clearly it cannot be +EV to switch twice, since you end up back at the same envelope. So, absent any other information, switching must be EV neutral.

[rufus]

[ QUOTE ]
The EV of the envelope is not relevant. What matters is the conditional EV of switching, which is always well-defined and finite.

[/ QUOTE ]

Let's say I pull the envelope, and find that it contains 8 trillion dollars. Now, there is only about 1 trillion dollars in currency in circulation, so there's a problem. If the EV goes to infinity, then, for any scarce resource, there is a non-zero probability that the envelope contains more of the resource than exists, ergo, if the amount in the envelope is close to that (or larger) it makes hardly any sense to switch envelopes.

[ QUOTE ]
It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p>1/3. The value p is always a well-defined number between 0 and 1.

[/ QUOTE ]

Actually, that's incorrect. It makes sense to switch if the average expected value (EV) of the other envelope is larger than the value of the held envelope.

[RolloRolf]

Okay this is what i'm thinking. Let's say you pick an envelope and recieve 100 dollars. Now we know for a fact that we (before we began) had an EV of +75 (50 and 100)OR an EV of +150(100 and 200). From now on switching will not change our original EV (either +75 or +150) it will only tell us which one we had. So when/if we switch we don't actually change our EV, we only believe we do. So instead of thinking of it as gambling 50 to win 100, you're actually gambling an EV of +75 OR +150 to win an EV of +75 or +150. You win and lose nothing by switching. EV=0

This is still somewhat flawed and doesn't really answer the paradox comepletely, but hey i tried :P

[jason1990]

[ QUOTE ]
[ QUOTE ]
It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p>1/3. The value p is always a well-defined number between 0 and 1.

[/ QUOTE ]

Actually, that's incorrect. It makes sense to switch if the average expected value (EV) of the other envelope is larger than the value of the held envelope.

[/ QUOTE ]
If x is the value of the held envelope and p is as I defined it, then the EV of the other envelope is

2xp + 0.5x(1 - p) = x(0.5 + 1.5p),

which will be greater than x if and only if

1 < 0.5 + 1.5p,

which is true if and only if p>1/3.