Two Envelope problem

[TomCowley]

http://en.wikipedia.org/wiki/Two_envelopes_problem for those not familiar.

Let's modify the game a lot, defining a finite set of envelope pairs (2^X-1,2^X) with X as an integer on [1,Q]. This game is well defined, and switching both before and after opening is obviously 0EV over the sum of all possibilities. Now we can probabilistically model this game without dealing with any infinity-related messes.

If we choose an envelope pair at random, randomly open one of the envelopes, and then decide to switch, this is +EV in ALL cases except the case where 2^Q is opened. The case where 2^Q is opened is not in the spirit of the problem, because the original problem assumes that 2^Q+1 is possible. So, taking this finite snapshot, and calculating the EV of the switching, under the condition that with the chosen envelope, 2x and 0.5x envelopes are possible, switching is always +EV by the same fraction.

Let's modify the game closer to the original formulation, but still with 0EV. Let X be an integer from (-inf,Q] with envelopes as before. You choose an envelope pair at random, open one at random, and find 2^X. Then you can either switch or not.

This game cannot be modeled by a uniform distribution P=constant of probabilities of envelope pairs (because there are an infinite number of discrete values). If we take the standard assumption that all envelope pairs are equally likely, then even though an initial probability distribution cannot be defined, the expected value of the game over all switches is well defined (unlike the case where Q=inf) because it can be expressed by a convergent series and the expression of EV converges nicely to 0 (geometric series, I figure anybody who's interested in this can verify it), as it logically has to (the 2^Q forced downswitch wipes out the gain from all other switches). So in this infinite game, it is always +EV to switch when 2x and .5x envelopes are possible, AND THE MATH BEARS THIS OUT WITH NO PARADOX because it is ALSO 0EV to switch envelopes before opening.

Extended to the original problem, where Q->Inf, for any Q>X (which is a base assumption of the problem, that double the amount is possible), it is mathematically well-defined that switching is a constant +EV fraction, even though a uniform distribution can't be used for standard probabilistic methods.

It can be proved by induction that for all Q on the (-inf,Q] distribution, if X<Q, the EV of switching is positive. It can also be proved by induction that for all Q, switching with no knowledge of X is 0EV. The "paradox" appears to me to simply be that switching before opening and switching after opening do not sample exactly the same set of envelopes, even though the difference is "infinity" compared to "infinity -1".

Edit: technically I guess I should be using a limit to -inf too, but [Z,Q] limQ->inf of limZ->-inf behaves the same way.

[AaronBrown]

I'm a bit reluctant to answer, because there are some strong feelings about this problem among posters to this board.

I will content myself by saying that you have given additional mathematical detail to one side of the argument that switching must be zero EV. With a little more work you could prove the result that a Bayesian will only consider switching if the expected amount in the envelope under the prior distribution is infinite (with an infinite expected value, any realization is a disappointment, so it's always positive EV to switch). So if you insist on a consistent prior distribution, that is, if you know how the amount in the envelope was determined, there is no paradox.

However, there are plausible situations, like the initial statement of the envelope paradox, in which we don't know how the amount was determined. In real life, these are situations in which what we are offered affects what we think the range of offers is. For example, you are looking for your first job, with no idea of what salaries are. It could make sense to turn down your first offer, however big, because the bigger it is, the bigger your expectation of competing offers.

There are situations in which it seems to make sense to assume two things are equally likely, like the second envelope containing twice or half the amount of the first. The paradox shows this simple assumption can lead to trouble. Unfortunately the same assumption is embedded in many standard statistical methods. Rooting it out requires hairsplitting analyses that seem to have little to do with practical decision-making.

Some people become Bayesians and live with the hairsplitting and refuse to use common sense practical tools. Most people ignore the potential problems and live with the occasional contradiction. The point of the paradox is to point out the choice, not to make it for you.

[TomCowley]

Thanks for responding. What's really bugging me (or intriguing me) about this formulation of the problem is that I can't figure out where I'm handwaving at all (no summing alternating increasing series, no direct assumption of a uniform distribution over infinite space, etc), yet I come up with the two distinct results simultaneously: Switching, in general, is 0EV. Switching any opened envelope is +EV.

I'll put my formulation a little more formally. There are envelope pairs (2^X-1, 2^X) for each X on [Z,Q] (X,Z,Q all integers). Selecting an envelope at random means selecting a pair at random, which has odds 1/(Q-Z), and then selecting an envelope at random, for final odds 1/2(Q-Z).

The EV of switching before opening is the sum over all envelopes of P(envelope)(=1/2(Q-Z)) * EV of switching to the other envelope in the pair. It's trivial to write the series and prove by induction that it is always 0 (as Z->-inf and Q->inf).

It's also true that the conditional EV of switching, given a particular envelope, is well-defined as the sum over all envelopes of P(that envelope being the matching envelope given the first envelope opened)*value of the envelope - value of opened envelope, or in simple terms, 0.5*0.5X + 0.5*2X - X = +0.25*original value.

I achieve both results simultaneously, using what to me looks like sound mathematical reasoning (taking limits of functions/sums that provably converge or are constant). This is really odd to me. If you can tell me where I'm going wrong in getting both of these results for my version of the problem, I'd love to know. The EV of opening an envelope increases without bound as Q->inf, but I never use that number, so I don't see why that misbehavior would be a problem.

[HP]

edit: nevermind

[HP]

[ QUOTE ]
It can be proved by induction that for all Q on the (-inf,Q] distribution, if X<Q, the EV of switching is positive.

[/ QUOTE ]
Well, I guess, but it's infinitesimal

You realize that if you played this game, the average amount of money in the first envelope you open is 0

So you get to switch, increasing your EV of 0 by 25%

This infinitesimal gain in EV equals exactly the infinitesimal chance your envelope contains 2^Q, multiplied by 2^(Q-1)

Does this resolve the paradox?

[TomCowley]

Not really- %increase from switching an opened envelope is positive and constant.

Also, the limits for Z and Q going to + and - inf show unbounded EV for opening an envelope.

[HP]

[ QUOTE ]
Not really- %increase from switching an opened envelope is positive and constant.

[/ QUOTE ]
positive and constant, what do you mean by this?

If it's what I think it means, it's what I said also, and there's no disagreement. If you are implying the increase in EV is finite, I would disagree

Haven't checked out the Z and Q stuff yet, will look now

[HP]

[ QUOTE ]
It's also true that the conditional EV of switching, given a particular envelope, is well-defined as the sum over all envelopes of P(that envelope being the matching envelope given the first envelope opened)*value of the envelope - value of opened envelope, or in simple terms, 0.5*0.5X + 0.5*2X - X = +0.25*original value.

[/ QUOTE ]
so you are not counting the times we open an envelope with 2^Q inside, correct? Wouldn't that make up the overall difference?

To clarify my question, are you saying it's +EV overall even if you take a -EV hit by always switching when you receive 2^Q in your first envelope?

[TomCowley]

That's what's weird. For any value of X, once you open an envelope, the limit as Z/Q->-/+inf is always 0.25* original value. But over the sum of all switches, the EV of switching is provably 0 (because of the forced downswitch at the end). It's like as soon as you open the envelope, you know you're not the forced downswitch (since Lim Q->Inf means you're dealing with Q > X that you opened), so the EV of switching becomes positive.

[HP]

[ QUOTE ]
That's what's weird. For any value of X, once you open an envelope, the limit as Z/Q->-/+inf is always 0.25* original value.

[/ QUOTE ]
Really? I guess I'll work it out for myself, but I'm pretty sure it's 0