The Envelope Paradox: What Does Maths Say?

[colin9876]

envelope paradox??

Maths can sometimes be nearly as interesting as physics...
There are two envelopes on a table. Both have cash in, you don't know how much but one has twice as much as the other
You can pick just one envelope, and you are allowed at most one swap.
You choose envelope A say, and see that has £100 in it. Are you better off on average to swap and choose envelope B?

Envelope B could have £50, or £200 in it - 50% chance of each
So expected value if you swap is 0.5*50 + 0.5*200 = £125
So on average you are better off swapping - how can that be?

 

[WarPhalange]

Exactly how you calculated it.

Look at it this way: 200 is 100 more than what you have, where as 50 is 50 less than what you have.

So basically, if you lose, you lose 50. If you win, you win 100. So on average you will win more money than you lose, even if you lose and win the same amount of times.

Say 4 wins and 4 losses, 4 wins = 800, 4 losses = 200, that's 1000, divided by 8 = 125 per win.

The only reason this works is because you already have a benchmark by looking at the first envelope.

 

[matt grime]

No, that is not the explanation. The analysis - that you're best to switch - in the paradox is independent of looking in the envelope. The real problem is that the paradox assumes that there is a uniform distribution on the infinitely many values that may be in the envelopes. This is not possible.

 

[quadraphonics]

Right, a different version of that paradox which might clarify the issue is to allow unlimited swaps, but without looking inside the envelopes. If swapping envelopes really yields a 25% expected increase, then swapping again should yield even more increase, and so on without bound.

I wonder if this paradox has any implications for improper priors that are sometimes used in Bayesian statistics? The same thing is often done there, trying to assign a uniform distribution on infinite support.

 

[DaveC426913]

No, that is not the explanation. The analysis - that you're best to switch - in the paradox is independent of looking in the envelope. The real problem is that the paradox assumes that there is a uniform distribution on the infinitely many values that may be in the envelopes. This is not possible.

I don't follow the logic here. Are you saying it is better to switch, or it is not?

 

[quadraphonics]

I don't follow the logic here. Are you saying it is better to switch, or it is not?

The whole point of a paradox is that there is a valid line of reasoning supporting either conclusion.

 

[matt grime]

I don't believe that is the case, quadraphonic. The only way that the reasoning is valid to swap is if there is such a thing as the uniform distribution on the natural numbers. There isn't.

Let's put it this way. Suppose that this is a game show, then the monies in the envelope must be less than the budget on the show. So if you open the envelope and see that the amount is more than half the show's budget, then you stick, otherwise you change. But here we have to observe the money in the envelope (and make a reasonable assumption about the budget of the game show) in order know what to do.So, to answer Dave's question: since you know neither the priors nor the posteriors, in the original case switching doesn't do you any good (or harm).

 

[DaveC426913]

So, to answer Dave's question: since you know neither the priors nor the posteriors, in the original case switching doesn't do you any good (or harm).

Ok, we know this intuitively. It seems that the key is to find the flaw in the logic of the OP's analysis. Why does his math seem to show that switching will on average yield a better result?

 

[WarPhalange]

Right, a different version of that paradox which might clarify the issue is to allow unlimited swaps, but without looking inside the envelopes. If swapping envelopes really yields a 25% expected increase, then swapping again should yield even more increase, and so on without bound.

Are you assuming you have an infinite amount of envelopes or what? Your rules aren't making any sense to me.

If you have $100 in your envelope and switch to one and see that it has $200, switching again you would switch back to the first one which had $100, right? You only have 2 envelopes to pick from.

 

[matt grime]

Why does his math seem to show that switching will on average yield a better result?

For the 4th time: because you have assumed a prior distribution that is uniform on the natural numbers. You have no idea what the correct distributions are, so you can't do any reasoning. If you had some idea you could decide what to do, as I explained above.

E.g. if you know there is at most 1000 dollars in there, and you open to find 800, don't swap! Dually, if you open to find 1 dollar, and know that the amount is an integer number of dollars, then do swap.

 

[matt grime]

Are you assuming you have an infinite amount of envelopes or what? Your rules aren't making any sense to me.

If you have $100 in your envelope and switch to one and see that it has $200, switching again you would switch back to the first one which had $100, right? You only have 2 envelopes to pick from.

The point of the paradox is that the reasoning is independent of the envelope's contents, so you can assume that you didn't open the first envelope. So you swap, but now you're in the same situation, so swap back, do this arbitrarily many times and you have an infinite expected gain.

This is an utterly bog standard paradox explained in hundreds of places all over the place. Google for an explanation for Devlin, say from the AMA.

 

[DaveC426913]

For the 4th time: because you have assumed a prior distribution that is uniform on the natural numbers.

This is too abstract an answer to be meaningful (to me at least). You might as well simply say "The OP's argument is flawed because it contains an error."

Again, I know it's wrong but I can't say where the OP's argument contains a flaw.

I suspect this is an incorrect conclusion: "50% chance of each". Is it because the doubling/halving rule means that the distribution is geometric rather than linear? i.e a 50% chance of each would actually only be the case if the other envelope had L50 or L150 (evenly spaced, not multiplied)?

The more I ponder the more I think that's what you're saying, but my brain is hurting.

 

[matt grime]

I have explained it to you by example twice, though you may not have seen the second time as it was an edit. Sorry, that you don't get the explanation, but that isn't my fault.

Let me repeat the example. Suppose you know that there is *at most* 1000 dollars in the envelopes ('cos that is all that the offerer can afford, say). Then clearly opening and discovering 800 Dollars means you don't swap. The posterior (loosely the guess at what the monies might be) has changed 'cos of this information.

The only way you can say, in the original version, that both possibilities of there being more or less in the other envelope are the same is if you assume that *all* possible dollar amounts might be in the envelopes. Since we can assume that there really are only a finite number of dollars in the world, this is clearly nonsense (and mathematically impossible anyway since there is no such thing as a probability measure on the natural numbers that assigns equal probability to all of them).

 

[mrandersdk]

Never seen this paradox before, very interesting, As suggested a simple google search gave some good answers. This site explains what i guess matt grime is trying to say

http://homepages.inf.ed.ac.uk/amos/doubleswapsoln.html

 

[quadraphonics]

I don't believe that is the case, quadraphonic. The only way that the reasoning is valid to swap is if there is such a thing as the uniform distribution on the natural numbers. There isn't.

Right, of course my statement was in the context of the assumptions, which are, as you say, problematic. But I don't think it provides a very satisfying resolution to say that "since you don't know the priors, switching doesn't help you." Because, indeed, its our very ignorance of the priors that leads us to want a uniform prior on the amounts in the first place. On top of that, using an improper prior still leaves us with perfectly sensible posteriors, so we lack the usual easy way of removing them.

My understanding is that this paradox is still an open question. Perhaps that's not the case...

 

[Hurkyl]

Because, indeed, its our very ignorance of the priors that leads us to want a uniform prior on the amounts in the first place.

No, it doesn't! Lack of evidence for an alternate prior distribution does not constitute evidence for a uniform prior distribution! And even worse, what would ostensibly be "uniform" here isn't even well-defined! What would be uniform in one parametrization of the sample space is non-uniform in other parametrizations.

 

[matt grime]

But I don't think it provides a very satisfying resolution to say that "since you don't know the priors, switching doesn't help you."

Perhaps I could have chosen my words better, but under no prior distribution is it always preferable to switch irrespective of the amount in the envelope, and that is provable. With some priors switching is preferable if you know the amount in the first envelope, under others it is not. Under many the expected gain is zero.

Because, indeed, its our very ignorance of the priors that leads us to want a uniform prior on the amounts in the first place.

Such a thing can't exist, so this doesn't help.

My understanding is that this paradox is still an open question. Perhaps that's not the case...

Open in what sense? There is a perfectly good explanation for it.

 

[colin9876]

I find it the most intriguing paradox- many sites give poor and differing explanations.
For the record, the point about the upper budget is not relevant - I could have said a genie with infinite resources filled the envelope.

(i) So either its better to swap ...

or it makes no difference and therefore
(ii) the probability that envelope B only has £50 in is 2/3 (and not 1/2 as we assume)
...
well (ii) is unlikely to be correct so
IT IS better to swap if you find ur envelope has £100 in it!

 

[matt grime]

I find it the most intriguing paradox- many sites give poor and differing explanations.
For the record, the point about the upper budget is not relevant - I could have said a genie with infinite resources filled the envelope.

But could not have put a uniform distribution on the possible values which is what the paradox requires. If you believe the 'budget' is the important point then you've not grasped the theory of probability.

The main point of using a budget is to demonstrate that a non-uniform (and mathematically possible) prior shows that in some cases switching is bad, and in others is good.

 

[quadraphonics]

Perhaps I could have chosen my words better, but under no prior distribution is it always preferable to switch irrespective of the amount in the envelope, and that is provable. With some priors switching is preferable if you know the amount in the first envelope, under others it is not. Under many the expected gain is zero.

Sure, that's no problem. If I'm given a prior on the amounts in the envelope, the rest is easy to see. But without being given any further information (which is the way that the paradox is invariably stated), we have no basis for assuming any particular prior, and so no way of deciding whether to switch or not.

Such a thing can't exist, so this doesn't help.

And yet, people use improper priors all the time, and get perfectly sensible results. And, anyway, it's easy enough to rephrase the paradox so as to avoid referring to probability at all. The explanation about there being no uniform probability distribution on the natural numbers only tells us that probability theory, as it is formalized, doesn't allow us to know nothing about a distribution on the natural numbers. And yet, it is manifestly the case that we know nothing about the distribution in question. Rather than provide an explanation, this ends up begging the question of whether probability (or, really, Bayesian decision theory) is up to the task in the first place.

Open in what sense? There is a perfectly good explanation for it.

In the sense that, to my knowledge, there is no consensus on a satisfactory explanation. The explanation about uniform distributions on the natural numbers may satisfy you, but it leaves many others wanting.

 

[matt grime]

Sure, that's no problem. If I'm given a prior on the amounts in the envelope, the rest is easy to see.

Who is talking about knowing the prior amounts in the envelopes? I'm talking about knowing the prior distribution of the (smaller, wlog) amount in the envelopes.

But without being given any further information (which is the way that the paradox is invariably stated), we have no basis for assuming any particular prior, and so no way of deciding whether to switch or not.

Yes, that is what we've been saying.

And yet, people use improper priors all the time, and get perfectly sensible results

Where (not doubting, just curious)?

In the sense that, to my knowledge, there is no consensus on a satisfactory explanation. The explanation about uniform distributions on the natural numbers may satisfy you, but it leaves many others wanting.

That explanation tells me that the paradox as stated is nonsense. By all means formulate a better statement. Since the whole point is about calculating expectations, using standard probability theory, then you must accept the premises of standard probability theory, or the paradox (as with all paradoxes, really) falls apart by its own inherent inconsistency.

Or let's put it this way: there's no paradox to a mathematician - the question is based on faulty mathematics. That should not stop people attempting to make sense of it in a different way if they so wish. After all innovation is good. But don't be under the impression that there is a paradox here at all. So by all means invent a new system which generalizes probability, and expectation, and allows the situation of the question to occur. It would appear to be fraught with potential inconsistencies, though, and has no obvious link to the real world (should anyone think that important).

 

[quadraphonics]

Who is talking about knowing the prior amounts in the envelopes?

Nobody. Please read the text you quoted more closely.

Yes, that is what we've been saying.

Who is "we?" And how can you be satisfied by a resolution that still can't tell you whether to switch or not? Note that this is not the same as saying that switching or not are equivalent strategies.

Where (not doubting, just curious)?

The uninformative case of most of the classic conjugate priors for exponential family distributions are improper. For instance, the classic case of Bayesian estimation of a Guassian mean, using a conjugate (i.e., Guassian) prior. If you want to make the prior uninformative, you have to set the variance to infinity, but this is improper. Doing so isn't really a problem, however, as the resultant estimator is identical to the MLE (and MVUE). Which illustrates the issue: we know that such estimators are perfectly fine (indeed, optimal), and we see that the Basyesian estimators also reduce to this case in a natural, intuitive way, except that, technically, the uninformative prior "does not exist." It seems obtuse to resolve the situation by insisting that the uninformative Bayesian estimator be excluded; a better solution would be to re-write the rules to handle this properly (although I'm not aware of anyone having done that yet).

That explanation tells me that the paradox as stated is nonsense. By all means formulate a better statement. Since the whole point is about calculating expectations, using standard probability theory, then you must accept the premises of standard probability theory, or the paradox (as with all paradoxes, really) falls apart by its own inherent inconsistency.

Well, Wikipedia already has the non-probabilistic formulation for everyone to see if they're interested. And, yes, the whole issue is one of inconsistency (the definition of a paradox is that there are inconsistent correct answers, after all). But to simply say "it falls apart" is not useful; it "fell apart" at the starting when we produced multiple contradictory answers. This paradox endures because people, even laypeople, have a strong intuition that probability should be able to handle this simple scenario. So to attempt to resolve it by saying "oh, probability theory doesn't allow improper priors" isn't satisfying. There at least needs to be a good reason supplied as to why probability theory can't be made to accommodate improper priors (which, I realize, is a very tough question).

 

[matt grime]

Nobody. Please read the text you quoted more closely.

Indeed I did misread you, sorry.

Who is "we?" And how can you be satisfied by a resolution that still can't tell you whether to switch or not? Note that this is not the same as saying that switching or not are equivalent strategies.

We appears in this thead to be at least me and hurkyl (or was it hallsofivy). The resolution tells me that the paradox is nonsensical, and the questions it asks are nonsensical. I can be perfectly happy with that.

Well, Wikipedia

I stopped trusting that a long time ago. Thanks for the other info though.

EDIT: glanced at it, and I think it just proves what I suspected long ago: I have zero interest in subjective probability, or the semantics involved in bayesian statistics. Which should be a warning to others, I guess.

Personally I find saying probability can't handle it perfectly satisfying, since I have no preconception that it ought to handle it. Just as I have no problem with the fact that mathematics can't cope with fields of composite characteristic, or non-abelian cyclic groups.

In some cases one can do apparently silly things and get sensible answers. Occasionally there are reasonable explanations for this. E.g. making a perfectly sensible attempt to work out the sum 1+2+3+4+... (analytic continuation of the zeta function, if someone's wondering). Experience tells me not to get too fixated on trying to make it so simply because I wish it.

 

[Hurkyl]

And how can you be satisfied by a resolution that still can't tell you whether to switch or not?

Why would I consider "you've given me insufficient information" to be an unsatisfactory response to a question to which I've been given insufficient information to answer?

 

[M Grandin]

I have just thought about it a minut and return with perhaps better answer later.

But preliminary, I would say the expected value is infinite if unlimited span of value.
And both half and double of infinity is still infinity. Therefore infinity in both swapped
or unswapped choice.

 

[M Grandin]

Simplest possible explanation

As I suspected, there are very simple "solutions" to this paradox not requiering much
mathematics (mentioned in Internet, but I made the same calculus too ...):

Assume the lower value envelope A contains S and higher value envelope B contains 2S.
Likelyhoods getting A resp B of course are 0.5 both.

If you got A the gain by changing is 2S - S = S
If you got B the gain by changing is S - 2S = -S

The overall probability gain is therefore 0.5 x S + 0.5 x (-S) = 0

So you don't statistically win anything by exchanging, as common sense also tell. So
a correct calculus of this must also regard what you actually got in your envelope in either case.

 

[DaveC426913]

Assume the lower value envelope A contains S and higher value envelope B contains 2S.
Likelyhoods getting A resp B of course are 0.5 both.

If you got A the gain by changing is 2S - S = S
If you got B the gain by changing is S - 2S = -S

That makes perfect sense, both mathematically and intuitively.
But I'm still trying to find the flaw in the OP's explanation of it.

You choose envelope A say, and see that has £100 in it.
Envelope B could have £50, or £200 in it - 50% chance of each
So expected value if you swap is 0.5*50 + 0.5*200 = £125

 

[matt grime]

The flaw when viewed as a question about standard probability, as has been explained many times, is that you can only say the probabilities for each of the two possible values in the envelopes are 50% if *all* possible dollar values are equally likely to appear in the envelope. This is what you get from working out the posteriors - P(X)=P(2X) for all X - so a uniform on an infinite set. This is not allowed in probability theory, so we can say that the paradox is based on a mistaken premise.

quadraphonics mentions a logical reformulation, which also has a logical (posited) explanation that is not about probability.

 

[uart]

"If you got A the gain by changing is 2S - S = S
If you got B the gain by changing is S - 2S = -S

That makes perfect sense, both mathematically and intuitively. But I'm still trying to find the flaw in the OP's explanation of it.

Well the "paradox" is still there in that with the example you are given that the amount is 100. Therefore if you got A then S must be 100 whereas if you got B then S must be only 50.

Let me just offer a concrete example where you have a known prior distribution and show how the "paradox" evaporates when you have more information.

For convenience I'm going to scale the amounts down in this example so I can use a nice U(0,1) (uniform 0-1) distribution. It can easily be scaled back to any desired (finite) maximum envelope amount.

Say the values in the envelopes were chosen as follows. The lower amount is chosen from a U(0,1) distribution, the other of course being double this is and so has a U(0,2) distribution.

Since we are choosing an envelope at random then the sample monetary value that we find comes from a distribution function f = 1/2( U(0,1) + U(0,2) ). It is easy to see that the density of available monetary values is 0.75 from 0..1 but then it decreases to a density of 0.25 over the range 1..2.

Lets integrate to find the expected value (mean) that we will get if we don't swap.

$$E(x) = \int_{x=0}^1 \, 0.75 x \, dx + \int_{x=1}^2 \, 0.25 x \, dx = 3/8 + 3/8 = 3/4$$

Now let's look at what happens if we always swap. If x is in 0..1 then it's somewhat more likely that we have the low value envelope and therefore advantageous to swap. If we swap in this x_range we can expect that 2/3 of the time we will get doubled and only 1/3 of the time will we get halved (I haven’t shown this but it's easy). So the expected value is now 4/3 + 1/6 = 1.5x after the swap (for x in 0..1).

If however x is in the range 1..2 then our swap will always yield half as so we get 0.5x in this case.

So now we can calculate the expected amount we receive if we always swap as,

$$E_s(x) = \int_{x=0}^1 \, 0.75 * 1.5 x \, dx + \int_{x=1}^2 \, 0.25 * 0.5 \,x \, dx = 9/16 + 3/16 = 3/4$$

Exactly the same as before!

I don't know if that helped Dave but I often find it good to look at a concrete example.

 

[gel]

Perhaps I could have chosen my words better, but under no prior distribution is it always preferable to switch irrespective of the amount in the envelope, and that is provable. With some priors switching is preferable if you know the amount in the first envelope, under others it is not.

ok, try this.

Suppose 4ⁿ pounds is randomly placed in one envelope and 4ⁿ⁺¹ in the other, with probability 7ⁿ/8ⁿ⁺¹ (n=0,1,...).

Open any envelope at random, and find X pounds. Then, the amount in the other envelope, say Y, must satisfy E(Y|X)>=2X.
So it's always preferable to switch.

I think the real answer to the paradox is that you should use a utility function. And the answer will depend on how risk averse you are.

 

[matt grime]

Wow, you managed to make the probabilities add up to more than 1 in a really trivial way (n=0...).

Now, I seem to have read recently that it is provable that there is no prior probability distribution where it is *always* preferable to switch. I think you should work out those posteriors again, after making it a probability distribution in the first place, though.

 

[gel]

Wow, you managed to make the probabilities add up to more than 1 in a really trivial way (n=0...).

Sorry, that was just a typo. I fixed my post - try re-reading it now.

 

[Count Iblis]

Ok, we know this intuitively. It seems that the key is to find the flaw in the logic of the OP's analysis. Why does his math seem to show that switching will on average yield a better result?

Because one assumes a uniform prior (which is not really possible as pointed out earlier). But if it were then the expected amount of money in the envelope would have to be infinite. So, if you only find a finite amount of money in the envelope, you should switch.

 

[quadraphonics]

Because one assumes a uniform prior (which is not really possible as pointed out earlier). But if it were then the expected amount of money in the envelope would have to be infinite. So, if you only find a finite amount of money in the envelope, you should switch.

I think that this is a more satisfying way of stating the resolution than to simply point out that a uniform distribution with infinite support is not technically allowed. I.e., if the expected return under not switching is infinity, then the fact that switching provides 1.25 times the expected return does not actually recommend switching, since 1.25*infinity is still just infinity, so either strategy has the same expected return.

 

[Hurkyl]

If I open an envelope and see 100$, my expected return from swapping is between 50$ and 200$ -- i.e. not infinite.