Probability: Envelope Paradox

Factual Questions
21

Ugh. I don’t know if this is going to help or just muck things up more, but I’ll give it a shot.

There’s a parallel puzzle to this called “the wallet game,” which, if it actually worked, would cure world poverty.

Basically, every day, everybody puts a randomly determined amount of money in their wallet. You then compare wallets with someone else who has done this. The person who has the least money in his wallet wins the contents of the other person’s wallet. So, if you have $X on you, the most you could every lose playing this game is X. But, by definition, whenever you win, you will win more than X. Since your odds of winning are 50/50, if you play this game enough, you’ll end up richer, because every time you win, you’ll win more than you lose those times that you lose. Of course, this logic applies to everyone who plays this game, which is how it cures poverty.

The faulty logic is that your odds of winning aren’t 50-50. The more money you have in your wallet, the lower your odds are of winning. The same goes for the envelopes–higher sums means increased likelihood that the other one will have half as much; lower sums means a better chance that the other envelope has more money. The error comes in assuming a 50% probability simply because there are two outcomes.

22

Yes, $4.50, I mis-divided.

The point is, the ‘always switch’ strategy which is supposed to always win, doesn’t do any better than the ‘always stay’ strategy. If you know what the host’s choices were, you can do a lot better than that-but if you don’t, then looking at your first choice doesn’t give you any information about his strategy.

To put this another way-suppose the host doesn’t pick the pair of envelopes at the 50-50 rate, but picks the $2/$4 pair 90% of the time and the $4/$8 pair 10% of the time. Now your strategy of always switching when you get $2 or $4 is suboptimal … you don’t have any way of telling when you pick your envelope with $4 which pair he’s taken, and without any other information about his strategy you’re stuck. (You can’t even do a Bayesian analysis without some priori probability between the two possible envelope pairs.)

And in general, you won’t know even what the possibilities were for the envelopes, much less what kind of selection strategy the host used to pick the pair he presented to you. Zero information means zero potential for reasoning your way to an effective strategy.

23

Messiah (how can you expect to be addressed as this, I have to kneel while posting),

Your explanation is fine by me, I’m just trying to answer the original question without having to say:

“I hope you understand this?”.

But congrats for the elegance of your explanation of why the choice is moot.

24

SCSimmons

The problem with your last explanation is that you won’t randomly pick one set of envelopes 90% of the time. You’ll pick it 50% of the time. If you are picking one set of envelopes 90% of the time over the long run, you are either in the very rare subset of the statistically extreme, or some factor you haven’t told us about is guiding you to choose the 2/4 set over the 4/8 set. Given the second assumption, you should adjust your strategy to not switching and come out ahead.

25

Let’s say the maximum prize is Y.
If you don’t switch your expected (average) winnings
are Y*(1/2)+(Y/2)(1/2)=Y(3/4).
If you pick an envelope and declare “There is X
in this envelope”, then how do you express what is in
the other envelope? Well from the setup of the game
we know if X=Y then the other envelope has Y/2, and
if X=Y/2 then the other envelope has Y. Let us
define a function, f, which has argument the number of
dollars in the envelope you picked, X, and value the
number of dollars in the other envelope. Then as discussed
above: f(X)=Y/2 if X=Y and f(X)=Y if X=Y/2.
So what is the expected value (average value) you get
by switching? Well it is the expectation of f(X), denoted
E[f(X)]. Applying the definition of expectation, we
get
E[f(X)]=f(Y)P(X=Y)+f(Y/2)P(X=Y/2)=
=(Y/2)(1/2)+Y(1/2)=Y*(3/4).
This is the same as not switching. Thus it makes no
difference if you switch or not.

26

Once again I say, we want to answer the original question :

“It seems obvious that in reality, with only the information given, there is no reason to choose one envelope over the other. Switching should not produce any advantage or disadvantage. Where does the flaw come into my reasoning? I’ve been puzzling over this for months, unable to come up with an answer. Can anyone help?”

I’m telling you boy, he don’t wanna define no function, he isn’t a mathematician, he just wants an explanation of why it doesn’t make any bloody sense at all to him.

Since many posts ago we agreed that the question is a mere trick, I mut spit on your math.

27

JasonFin, I think I see the problem in your thinking. The term “expected value” does not refer to the value of B when you are holding A, but rather the value of your earnings by playing the game. That is, if you play this game n times you can expect to earn 1.5xn, ignoring Monty Haul scenarios where the person with the envelopes attempts to manipulate your choice.

CK, I have to say that there is no paradox. Of course you can divide an integer recursively until you have an odd number, but so what? Consider the case where the first envelope contains either $80, $160, or $320. If the first envelope contains $80, the second could contain either $40 or $160. Speculations along the lines of, “If I had picked $40, the other might have been $20, but if I had picked $20 I would have known to switch…” are useless, because once you go far enough you remove what you hold from the realm of possibility, which is obviously wrong. The fact remains that the player holds $80, and that the other envelope contains either $40 or $160. Similarly, if the player holds $160, the other envelope contains $80 or $320, and if the player holds $320, the other envelope holds $160 or $640. So basically there is an infinite number of values to choose from that would not let the player determine anything, with a lower bound somewhere I haven’t figured out yet. I should also point out that since the OP said “some money” we are allowed to use cents.

28

Look guys, I think “messiah” has done the math as succintly as it can be done. What we are trying to answer is the original question in a way that satisfies the questioner.

Look, the poor quy says he’s tossed this around for months. I read his post…he is not stupid…,give him the answer he wants…don’t play games.

29

<< Before I open bag A there is 50% chance that A has the most monetary value. If I look in A there still is a 50% chance that A has the most monetary value. Therefore if I find 10 dollars in A **then B has 50% chance of $20 and 50% chance of $5. ** >>

I have boldfaced the logical flaw in the argument. NO. If you find $10 in A, then B has a 100% chance of either $20 or $5. But that is NOT the same as a 50% chance of one or the other. Just because two things could happen, does NOT mean that they have equal likelihood of happening. I roll a dice, it could come up a six or it could come up SOMETHING ELSE. That doesn’t mean there’s a 50% chance of it coming up a 6 and 50% chance of it coming up something else.

In the envelope case, the result is already determined. Someone has put money in the envelopes. If we knew the amounts of money, the result would be a certainty. We don’t know the amount of money, we only know that one envelope has more than the other. So, if we find $10, there is a 100% chance that the other envelope contains either $5 or $20, and that’s all we know. We do NOT know the odds of it being $5.

The statement of the problem is set up to make you THINK of 50%. You first pick between the two envelopes, and that’s (presumably) a 50% chance. (ASIDE: If the money were all in singles, then you would see the diff between an envelope with $100 and an envelope with $200, I betcha.) Then the statement of the problem has one envelope containing “twice” the other, again to lead you to thinking of 1/2 or 50%. So the problem statement has stuck “50%” into your head, so that you will (psychologically) apply it falsely to the switching question.

Try to rethink the problem: suppose we have two envelopes and one has MORE than the other (forget about the “twice” which is a distractor to make you think 50%.) This is the same problem, right? OK, so you pick an envelope and it has $10. The ONLY facts that you have are (1) you picked one of the envelopes, and had a 50% chance of getting the better one; and (2) the other envelope could contain more or less than your envelope. The chance of you getting a better result by switching is [dramatic pause] 50%.

Trying again: Suppose you are dealt a card, face down, and the dealer has a card, also face down. Higher card wins. What is the chance that you will win? Well, without seeing either card, your best guess is 50% (you’ll either win or you won’t, and one of you must win.) But that’s just a wild-ass guess. You have, in fact, no clue. as soon as you look at your card, you get more information, and if it’s a 3, your odds just got considerably worse.

In the stated problem, you are just guessing that the odds are 50%, and that’s what causing the paradox. You’re trying to estimate odds that are unknown (because you don’t know what is in the other envelope). Your estimate is wrong, the odds are not 50%. The odds are either 100% or 0%. Hence, your calculation of expected value is based on a faulty estimate of the odds, leading to the paradoxical result.

I hope one of those explanations help.

30

I think the confusion of the OP stems from not precisely defining what the “game” is that we’re playing. Or rather, confusing two similar games. Consider two scenarios:

SCENARIO ONE
I give you an envelope containing X dollars (say, X = $160, just for fun). I then prepare a second envelope that contains either $80 or $320, with a 50/50 probability (say I flip a coin to determine which). You, of course, do not know how much the second envelope contains (only that it’s either $80 or $160 with a 50/50 probability). I offer to trade envelopes. Do you trade?

SCENARIO TWO
I place two envelopes in front of you. One of them contains twice as much money as the other. You open one, and find that it contains X dollars (say, X = $160 again). I offer to trade your envelope for the unopened one. Do you trade?

The important point here is that these scenarios are not equivalent. In scenario one, it should be clear that the best strategy is to trade. If you went through the same game 100 times, trading each time, you’d expect to get $80 half the time and $320 half the time, for an average of $200 every time you played. Better than the $160 every time if you stay.

In scenario two, it’s what you don’t know that hurts you. You do know that you’re playing either a $160/$320 game or an $80/$160 game, but you don’t know the probability split (as don’t ask showed in his first post). All you know is that the probability of playing a $160/$320 game plus the probability of playing a $80/$160 game is greater than zero. If you went theough this scenario numerous times, you would expect the first envelope to contain $80 or $320 (combined) at least as often as it contained $160. Factoring this in (again, as demonstrated by don’t ask), you find that it doesn’t matter if you switch or not.

31

Ender-this has nothing to do with being a ‘rare subset’ of a population. The 90% strategy is that of the host, who sets the values of the money in the envelopes and decides his own strategy for figuring out which envelope set to present to the player. If it’s his money :), it’s not surprising that he might pick the pair with less money in it more often, is it?

The point is, that without knowing the host’s strategy (including how many possible envelope sets with what amounts, and what method he’s using to decide which pair to present for the player’s choice), then there’s no way do determine a good strategy for the player (whether always stay, always switch, or something more complicated). Therefore, the original ‘proof’ that’s it’s always better to switch is invalid.

32

If you present the problem as “you have two sets of envelopes, one containing $2/$4 and one containing $4/$8 and he chooses which one he presents you with” then of course he can manipulate the frequency with which you get either. But that’s not the problem originally.
And you know what? I wrote quite a bit more and I’m erasing it all. I finally figured it out. It came as an epiphany and I can’t believe I didn’t see it sooner, it’s so simple. C K Dexter Haven I’m sorry, but I really don’t like your explanation. But now I finally understand what you were trying to say!

So let me see if I can help explain it further.
OK, you have two envelopes and you have no clue what either of them contain. I do though, because I created the game and I put the money into the envelopes. The envelopes have $4 and $8 in them.

If you pick up the envelope with $4 in it, you think the other envelope has a 50% probability of having $2 and a 50% probability of having $8. In reality, that’s not true. It has a 100% chance of having $8 in it because I put $8 in it. So, if you keep the envelope you chose, you’ll have missed out on $4.
Same thing goes for the other way. You think the other envelope has a 50% probability of having $4 and a 50% probability of having $16. Again, in reality, that’s not true. It has a 100% probability of having $4 in it.

But what happens is your perception of reality doesn’t change reality! No matter if you think the other envelope has $2, $4, $8, or $16, the reality of the situation is that it will ONLY have $4 and $8 every single time.

So now, if we play this experiment out hundreds of thousands of times (and you completely forget what the envelopes contain between each try), 50% of the time, you will choose the $4 envelope first. If you switch, you’ll gain $4. 50% of the time you’ll pick the envelope with $8 first. If you switch, you’ll lose $4.

End result, no matter how you play it, you’ll average $6 a pick.
I feel so much better now…

33

Well put, Ender.

There’s a useful general lesson in this example for dabblers in game theory (which is the main reason I was so bloody-mindedly continuing to post). When you’re trying to work out the values of potential strategies for one player in a two-player game, you need to remember that you need to evaluate the strategy against possible strategies of the other player, not just possible choices. This is sometimes hard to remember, probably because it’s too hard to do. In real-world games (rather than the toy games that theorists prefer to work on), there is usually an effectively unlimited selection of strategies. This is especially the case in limited-information games like this one, where the host’s choice of option is not revealed to the player. In general, the valid strategies for these types of games involves setting degrees of probabilities for selected available moves; and since probabilites are real numbers, that means aleph-one (the second degree of infinity) possible choices of strategy to account for. Ugly.

34

This is probably the best paradox I’ve ever seen, and I’ve devoted some time to thinking about it. I don’t agree with any of the solutions above, unless I’ve misunderstood them. The real answer IMHO is as follows:

It seems to me that the basis of the paradox is derived from the use of infinity in the initial distribution. By using an infinite upper bound for a uniform distribution (a true random number) we are assuming an expected value of infinity/2 (which is impossible, as a practical matter) and also makes any comparisons between it and results obtained after the first envelope is opened, meaningless.

I tried to rework the problem without using infinites, but using an upper bound would reveal the answer in many instances (any number in the top half of the distribution must be the larger envelope). The best I could come up with is a system in which each envelope has a dollar amount between 0 and 100 that is completely unrelated to the amount in the other envelope, and also a number either “1/2” or “2”. If an envelope is chosen first the dollar amount is won, if the envelope is chosen second then the dollar amount is ignored and the number is multiplied by the amount in the first envelope to give the winnings. In this manner you would have a bounded random number but also no clue which envelope it was no matter what amount was in the first envelope.

But what would then happen is that any number greater then 100 could only be achieved by picking a “2” with the second envelope. As a result, the second envelopes actually would total more than the first, in aggregate.

In comparing this situation to the previously posed paradox, it seems to me that where they differ is in that the first scenario uses an infinite distribution. In truth, any second envelope must always have a higher expected value than the first, 25% specifically. The only reason that this is not so in the hypothetical scenario is solely because in utilizing an infinite distribution it removed the possibility of having another distribution with a larger expected amount. In other words, it should not be possible for two distributions with one producing a number either twice or half of the other to be identical - these are indeed contradictions - it is only being made possible by the use of an infinite distribution. So the inequality is a function of the infinite nature of the distribution - in effect by comparing the first and second envelope you are comparing two infinites. Thus no paradox.

35

IzzyR: The only problem with your analysis is that there is no distribution–the dollar amounts in both envelopes are fixed but unknown. That’s different from their being random.

The paradox arises when you introduce a random variable for the amount gained by switching, which is not random.

36

No, it’s not.

What does this mean?

37

OK, different example. I put a certain amount of money in an envelope and hand it to you. You have no information about the amount of money therein, so it is unknown. But to say that it’s random is incorrect, because I know how much is in there. In order for something to be random, it has to be unknown to everyone. Same thing with the envelope paradox–someone knows how much money is in each envelope.

My earlier comment (“the paradox arises…”) was meant to say that random variables can’t be used to model fixed but unknown quantities.

38

Well, beyond the fact that you’ve not provided any rationale for your assertion, I am quite sure that you are wrong. As an example, the unknown value of a parameter of a given distribution is in reality fixed, but it is treated as a variable, and an expected value can be computed for it (if the distribution of the parameter is known). In fact, the whole concept of Bayesian Probability is based on assigning an expected value to the probability of a fixed value being one or the other.

39

OK, here’s my rationale…If you know that something is fixed but unknown, you have more information about it than if you know nothing. Probabilities are very sensitive to what information is already (e.g., the “Monty Hall” problem), so unless we have a reason why these two cases are exactly the same, they should be assumed different.

Admittedly, Bayesian statistics are outside of my experience, but it seems to me that there is a difference between a random value and a fixed value that is treated as if it were random, as reasoned out above. Granted, there may be special cases where either approach is valid, but I think in general the distinction has to be made clear.

I’m not going to rule out the possibility that I’m wrong. But in order to be convinced, I need a reason as to why “fixed but unknown” and “random” are the same thing. If you can provide that, I will gladly eat every word I’ve said in here–after all, the fight against ignorance begins at home, right?

40

Ah, another thread staggers from its grave to eat the brains of the living …

I’m not sure what I can add to what I wrote before. Let me try to distill it, I guess:

When you are determining the value of a strategy in game theory, you do not test it against possible moves of the opponent, but against possible strategies. (n.b. this does not, for the most part, apply to complete-information games like chess, but it definitely applies to the game described by the OP.) The assumption after opening an envelope with $4 in it that the other envelope has a 50/50 chance of containg $2, and the same chance of containing $8, implicitly assumes that the host was limited in his possible envelope-stuffing strategies to ones that gave equal probabilites for any allowed dollar value to be either the low or the high value envelope in the pair selected to be presented to the player. This assumption is false; there are plenty of envelope-stuffing strategies for the host which do not meet this criteria. In fact, the restriction above is logically incompatible with the (much more likely) restriction that the host has a finite budget … Against any actual envelope-stuffing strategy, define it how you will, you will find that the expected gain of the ‘always switch’ envelope-picking strategy is exactly the same as the expected gain of the ‘always stay’ strategy. (Try it!) There may be a better strategy than these two, but these two are always equivalent.

Any questions?

-Christian