Can somebody explain the two envelope paradox to me

Factual Questions
281

Game 1 and two can be approached the same way and are functionally equivalent. The only difference being that game one is approached algebraically and game two gives a numerical value.
So for game two we calculate for example E[Y given that X=$100]=$125.
Game One we calculate E[Y given X]=1.25X

The real eye opener for me in this whole scenario is the realisation that expected value alone is not always sufficient basis for intelligent decision making. (I realise that this is slightly at variance with what I wrote earlier, but I have had a few more days mulling over it.) In this case, the aparent paradox is that E[Y|X]=1.25X and E[X|Y]=1.25Y, which appears at first glance to be a contradiction. Careful inspection reveals that E[Y] is not the same as E[Y|X]. The latter has a condition included in the calculation and has a subtly different meaning. It is not too difficult to contrive a scenario where we are told the contents of one envelope are 100 times the other which would intensify the paradox.

So, sorry septimus. I think a statement like
" the erroneous E[Y] = 1.25 E[X] derives, as you say, from misusing expected value."
is actually unhelpful. You are right – it is erroneous, but only because it opens up confusion between E[Y] and E[Y|X]. This conditional expected value is actually a reasonably logical thing to calculate given the scenario.
(And while I am at it, septimus I don’t follow your reasoning concerning P(Y>X)>1/3. I am not sure how this probability arises in a two envelope case nor why 1/3 should be the critical point for decision-making.)

My first instinct was to doubt the validity of the calculation, but I am satisfied that it is correct. And there are some better mathematical minds than me who have supported this notion. My second instinct is to question exactly what information is given to us by the calcualtion E[Y|X]=1.25X Therein lies the whole topic of discussion. Indistinguishible’s comment somewhere in the first 50 posts was, in effect, “true but so what?”

Game three is interesting to me for two reasons:

  1. It is a generalisation which, as is the nature of mathematics, serves to illuminate the problem further.
  2. We don’t have any information about the probability distribution of X without making some assumptions. These are not even hinted at in the OP. Talk of an upper bound and “Even Bill Gates doesn’t have that kind of money” etc might have something of a pragmatic sense to it, but still doesn’t equip us any further to make an intelligent decision and nor does it illuminate the problem further. In my thinking of the problem I have considered X to be a positive real.

So, I am looking forward to Indistinguishible providing us with a connection to the problems of the summation of the integers – when he is done grading mid terms. :slight_smile:

282

Let me keep trying. :smiley:

(1) Before opening an envelope, p(Y > X) is 0.50 exactly. How about after opening an envelope, and knowing the value of X? The common-sense problem does not tell us what p(Y > X) is, once X is revealed!. Now some of you, including Indistinguishable, seem to assume that p(Y > X) = .5 for any X once X is revealed, but that’s an assumption which is not part of OP and which is in fact counterfactual for any real-world instance of the two-envelope scenario.

(2) E[Y | X] is a (“peculiar”) shorthand for EY | X=x. Please be careful of abusing this notation.

(3) That one should switch, once X is known, whenever p(Y > X) > 1/3 is easily seen. One’s expectation on switching is
E[Y] = p (2X) + (1-p) (X/2) = X (3p + 1) / 2
To switch you want E[Y] > X which becomes
X (3p + 1) / 2 > X
or
p > 1/3

283

indistinguishable, I’m still trying to see where the infinite sums come from.

It doesn’t appear to change anything when you look in the first envelope. So even though by doing so, you have specifically limited the values available to either $5, $10, or $20 (for instance), you still suggest there is an infinite sum occurring?

septimus, you seem to be only offering a distraction. Sure, there are cases that are self-limiting, that using extraneous information can tell more about the probabilities of the values in the envelope. But the paradox arises in a case where you don’t have that information available.

I’m left with j_sum1’s comment:

Someone else asked the question, “What do you think the expected value equation is telling you?”

I’m really new to this expected value stuff. From what I glean from this thread, it is supposed to be a way to statistically evaluate which choice gives the best rate of return. That seems to be saying the statistical probabilities of what is in the other envelope, vs what is in the current envelope.

In the first case, you don’t open the envelope, then the EV calculation says that Envelope 2 has an EV of 1.25 of Envelope 1. But if you recompute, you find Envelope 1 has an EV of 1.25 Envelope 2 - the paradox is how can each be 1.25 the other, but by comparing the EVs in that case, you can say that the EVs are equal, so there is no benefit to swapping.

In the case you open the first envelope, then you are calculating a rate of return of 1.25 across a large sample (any single incident may not turn in your favor, but repeats of the same problem will turn out so). You see $10. You know the other envelope cannot have the same $10 in it. So the EV calc suggests a positive return on the bet. Where is the flaw?

284

P(X>Y) will of course depend on the probability distribution of X (which is the same as the probability distribution of Y). However, no such information is supplied – hence Little Nemo’s initial posing of the problem using X rather than any specified value.
Indistinguishible has suggested a dice to simulate the probability distribution of X on more than one occasion. I have considered a uniform distribution with an unspecified upper bound but greater than double the value of any opened envelopes.

The only way P(Y>X) is going to differ significantly from 0.5 is if you have some rather detailed prior information about the distribution of X and Y – for example, conforming to monetary denomenations or less than $27. Then you can reason sensibly using a different probability value.
Of course this will make the problem more complex than the one specified in the OP.

I am not interested in doing this. I am intrigued as to why a seemingly straightforward calculation on a realitively straightforward scenario produces a counterintuitive result and how to interpret this (correct) result for the purposes of making decisions. I am also interested in the underlying mathematical principles that have been exposed. Making guesses about the generosity of a hypothetical game host doesn’t interest me.

So. For the purposes of discussion
P(X>Y)=0.5
E=E[Y]
E[Y|X]=1.25X
This is a question about expectation and averaging and not about probability distributions. (Quoting Indistinguishible from I think about page 2.)

285

septimus, thank you for your reply. With all due respect (and I mean that sincerely), you haven’t really answered the question. Whether I hazard an unopened envelope or an opened one, the analysis should be the same. I posited an opened one just because it’s easier to see what’s going on. Thinking it over the past few days, I think the answer to my question (and, by extension, the OP) is psychological. This point was mentioned earlier in the thread. Indeed, I seem to recall that it was made by you but don’t have the time to hunt for it. Stated simply, if I trade (in either the opened or unopened case) and end up with the lesser envelope, I’ve only lost half of what I wagered. Whereas if I trade and end up with the greater envelope, I gain by twice. But, and this is crucial, it’s still a 50/50 chance either way. It’s just that the upside “feels” better than the downside.

As for why the expected value should be 1.5 rather than 1.25, several people have explained. Indeed, I think you mentioned this earlier. (M + 2M)/2 = 1.5M. To use an example, suppose you are offered two envelopes, one of which has $100 and the other $200. What would you reasonably bid for one of those envelopes, the one actually given being decided by the toss of a fair coin? You could reasonably bid anything up to $150 dollars. If you repeat the trial a hundred times bidding $149, you would expect (on average) to gain $50 net. In any event, as you (and others) have argued, where the OP goes sideways is neglecting to notice that the expected values of the envelopes are equal, so there can be no advantage to trading.

286

Correction. In the thought experiment in paragraph two, I intuitively divided the expected gain in half, since one would expect to win only half the time. Actually, that was already incorporated into the expected value calculation. To see this, run the numbers. (One of the reasons I prefer to work with examples.) Over 100 trials, one would expect to “win” fifty times for $10,000 (50 x 200) and fifty for $5,000 (50 x 100), for a total of $15,000, whereas the total wager would be $14,900 (100 x 149). Thus, the expected average gain would be $100. IOW, the expected value is 1.5M.

287

bump – if you are still up for it indistinguishible

288

p(x>y) =/= p(x>y|x)

289

I still don’t see where the infinity comes into play.

290

I’ve decided I need more background on this subject. I’m currently reading The Drunkard’s Walk: How Randomness Rules Our Lives by Leonard Mlodinow and then maybe I’ll try some Gerd Gigerenzer or Joel Best.

Although judging from this thread, maybe I should be reading Probability for Dummies.

291

Ok… I’m kind of interested in this problem but not interested enough to read 6 pages of detailed back and forth on it. Can someone PM me when they figure it out?

292

I think this is an instructive example: Suppose there are 3 equally likely cases

100, 200

200, 400

400, 800
Now, given X=400, you should switch. 400 is equally likely to be high or low.

and given X=200, you should switch. 200 is equally likely to be high or low.

So in the 200, 400 case, your calculation is correct. The expected value of the other envelope, given that your envelope has X, is .5 (2X) + .5 (X/2) = 1.25X no matter which one you call X.

**Every time you choose an envelope, it would be accurate for someone to say “given that an envelope has the value X of your current envelope, the expected value of the other envelope is 1.25X”. **You could keep switching, and they could keep saying it! Argh!

This may seem odd, but there is no flaw. It isn’t a paradox, it is just correct.

I think the above is the “paradox” you were having trouble with. However, the other flaw in your first post is that the person choosing the envelope would not have been able to perform the calculation. It could only be performed by someone who knew the distribution of values and the value of X.

Notice that your calculation would not have been correct in either of the other two cases. 800 is always the high, and 100 is always the low.

I claim that there will ALWAYS be cases such that your calculation is not correct. It is not possible that, no matter what value you get, it is equally likely to be high or low.

The link in the first post explains this very clearly. You could probably reread it and understand it with just a little online research on Bayes’ Theorem and uniform distributions. It is simpler than any of the explanations I have read in this thread.