Probability: Envelope Paradox

Factual Questions
1

You are given two envelopes, A and B, each containing some money. You are told that one contains exactly twice the amount of money that is in the other. You are then allowed to pick an envelope to keep.

Let’s say that you tentatively decide to choose A, then attempt to determine whether or not taking B instead would be a good idea. A is equally likely to be the envelope with the larger or smaller amount of money. Call the amount of money in A “x.” Now, B has a 50% chance of containing 2x and a 50% chance of containing .5x. The expected value of the money in B is the average of these two possibilities, 1.25x, which is greater than x. Therefore switching to B would be a good idea.

Of course, then one could call the amount of money in B “y” and show that the expected value of the money in A is 1.25y, so switching back to A would be a good idea. This reasoning can go on ad infinitum. It would seem that the grass is always greener in the other envelope, so to speak.

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?

2

The flaw in your reasoning is that you reasoned that your approach even slightly resembled an exercise in probability. Your statement that envelope B contains either .5X or 2X is mathematically correct, but without significance in terms of assessing the probability of the envelope containing more or less money than envelope A. The “expected value” of the money in envelope B is NOT 1.25, and I’m not sure what might have led you to believe that simply averaging the two possibilities, in terms relative to another factor, would have any significance whatsoever.

Think of it this way: envelope B contains either “J” amount of money or “K” amount of money. “J” could be the half or any other lesser amount, and “K” could be the double or any other greater amount. The odds are 50-50 that envelope B will contain one or the other.

Oh, and if you find a way to average J and K, let me know, will you?

3

  1. Since BOTH have the option of being double or half the amount of the other, they cancel each other out. Just picking up the envelope supplies you with no real information anyway. This much you knew.

  2. Now, go ahead and open it up. What happens if you get $7.39? Since there’s no way to divide that in two in real money, you know for certain that the other envelope is the one with double the money. If it’s an even amount, you’re back to square one. It still becomes more profitable to switch envelopes anyway. Math is sometimes funny that way.

  3. So yes, it is profitable to ALWAYS switch envelopes. But since you can’t determine any information from looking at no envelopes, and since you can’t pick after having seen both, just pick one, look, and then switch.

then again…

  1. You greedy capitalistic bastard. Just pick an envelope and be happy about free money.
4

You have two random variables floating around: the amount of money in the envelope, and whether the other envelope has more money or less. But there’s only one random variable. Once the question of whether you picked the envelope with the most money has been determenined, the issue of how much money you have is determined as well. And once the issue of how much money you have is determined, the issue of which envelope you have is determined. You’re confusing random variables with unknown variables. If the envelope you pick up has x dollars, the amount of money in the other envelope is unknown, but it isn’t random: it’s determined by your choice of the first envelope. The contents of the envelopes is set from the beginning, and is unaffected by your actions.

5

TBone: Your reply hasn’t really cleared things up for me. Isn’t the average of J and K just (J + K)/2 if J and K have an equal probability of being true? It’s true that I can’t assign a monetarty value to either one, but I know that K is four times J, so can’t I find the average relative to what I know?

The Ryan: I don’t really understand your response either. How do I treat an unknown variable if I can’t treat it like a random variable? Isn’t it essentially the same as a random variable if I repeat the experiment a large number of times? Why does it matter that the amount of money in both envelopes is determined ahead of time?

Let me ask a question to see if I’ve messed up at a basic level. Is it acceptable to treat the envelope that I have not chosen as though it has a 50% chance of having twice as much money and a 50% chance of having half as much money?

6

You pick up envelope A, which has X amount of money in it. Envelope B then has either 2X or .5X… But the Xs aren’t the same in the two cases.

Think of it this way: Let’s say that one envelope contains $1, and the other envelope contains $2. A priori, your expectation is a buck fifty. Now, you pick up one envelope. There’s two possibilities: First, you might have picked the big envelope, so the one on the table contains .5 * $2. On the other hand, you might have picked up the small envelope, so the one on the table contains 2 * $1. In other words, the expectation of the other envelope is $1.50, exactly the same as the expectation of the one you’re holding.

7

Close but no cigar.

For any value of X JasonFin is correct. Say A contains $10 then by swapping you receive either $5 or $20 at an average of $12.50 because nothing else is possible.

But since the combinations that include a $10 envelope are $5 and $10 or $10 and $20 then in order to pick up the $10 envelope you must sometimes pick up the $5 and sometimes the $20.

In four trials you will end up with -
$5 from 5/10
$10 from 5/10
$10 from 10/20
$20 from 10/20
for a total of $45 at an average of $11.25.

Now back to JasonFin’s clever little puzzler, by swapping we net 1.25X by not swapping we get X. Therefore for any combination that includes X we get on average 1.25X+X/2=1.125X just as we have seen above.

Thanks JasonFin. Do you know the old TV show classic of 3 doors behind which are little prize, huge prize and nothing. You choose a door. The host opens one of the other doors to reveal either little prize or nothing and asks if you want to swap for the remaining door. What do you do?

8

http://www.straightdope.com/classics/a3_189.html

You keep quiet and back slowly, slowly away…

9

These are 2 different questions, and both have been done here before.

The Monty game show answer is pleasingly simple, once you know.

The other one was one of my first threads, and I found it hard to understand. But wise Dopers finally convinced me that there would probably be clues in the amount of money in the envelope you examine.

  • does the person offering you money have a budget? (if you’re on a show called ‘Win a Million’, and they offer you $500,002 - TAKE IT!)

  • is the sum of money indivisible by 2?

10

Look, if you open an envelope and find $40, the other envelope could contain $20 or $80. You don’t know which. But there is NOT a 50/50 chance that the other envelope contains $80. The envelope is fixed. There is a 100% chance that it contains [whatever it contains] and a 0% chance that it contains [the other possibility.]

One of the easiest errors to make in probabilities is to assume that “it either does or it doesn’t” amounts to 50% probability. I’ve got a state lottery ticket in my hand, it will either win or it won’t, so I have a 50% chance of winning? Bah. But that’s exactly what you’re saying when you assume a 50/50 chance that the other envelope contains $20 or $80.

In fact, you started with a 50% chance of picking the best envelope. You have no new information by having picked that envelope, and so you still have a 50% chance of having picked the best envelope.

If you open the envelope, then you get additional information and may need to rethink. That additional information may be helpful (like if it contains an odd number of dollars) or not (like if it contains $40, say).

However, in the absence of new information, it is not correct to say that you are better off switching. (In the related but very different Monty Hall three-door problem, you get new information when one door is opened, and thus need to rethink your probabilities.)

ASIDE PARADOX (IGNORE IF YOU ARE NOT MATHEMATICALLY/LOGICALLY INCLINED, IT WILL GIVE YOU A HEADACHE):
Assuming that the person setting up the problem is reasonably bright, they would never allow a set-up where one envelope contained an odd number of dollars, right? But that means they could never allow a set up where one envelope COULD contain an odd number of dollars. That is, if you open an envelope and find $10, you know that the other envelope wouldn’t have $5 (if it did, and you had chosen that envelope, you would have known to switch.) So if open an envelope with $10 in it, you know to switch.

Similarly, if you open an envelope with $20 in it, you know that the other envelope couldn’t contain $10. Why not? Because if you had picked a $10 envelope, you would know to switch, as reasoned above.

Continue the discussion ad infinitam. There can be no such envelopes. QED.

11

  1. What C K Dexter Haven said.
    (Although to his last point; if I were doing this, I’d use powers of 2 e.g. $16).

  2. Probabilities only change when you acquire new information.
    (e.g. until you open the envelope, you have no new information)

  3. The only reason that you might get new information (and thus want to swap envelopes) is that there are ‘limits’ on money. These include budgets and amounts that don’t halve.
    (Your envelope contains $0.01; Your envelope contains more than half of the game show budget)

  4. If you write a positive rational number in one envelope, and double that number in the other, then I don’t believe that knowing one number makes any difference to the probabilities.
    And if you don’t believe me, your envelope now contains:

0.04856331

What’s in the other one?

12

C K,

Can you further explain what you mean by the 100%, 0% scenario? Because as it stands right now, I disagree with you.
If you do not know the information, what the reality of the situation is is of no consequence when you’re figuring out the equations. True, if you pick up one envelope, the reality of the situation may be that the other envelope has a 100% chance of having half that amount, but you don’t KNOW which envelope is which. You’re not dealing with the reality, merely your perception of it. And your perception is that the other envelope could equally have half the amount you now hold or double it.

Take this example from poker. Ever play Texas Hold’em? You have two cards in the hole which are yours and yours alone. You start off after the first round of betting with three additional community cards. If your two hole cards are a pair of threes, then, on average, 2 out of every 15 times you will have trip 3s after the community cards have been dealt. This is taken into account based upon the fact that you have no clue what the opponents are holding. In reality, they may hold the other two 3s in the deck and you have a 0% chance of getting another 3. In reality, they may hold no 3’s and your odds go up to 20%. But since your reality is based upon the actual reality that you have no clue what they hold, you have to count their cards as completely random and just assume that you have a 1 out of 7.5 chance of hitting another 3.
Same thing with the envelopes.

And that paradox…That did give me a headache. But you’re right. There can’t be any envelopes. Weird.

13

I still don’t think that the conundrum has been fully explained, because the paradox still remains even without changing the probabilities.

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
If I had been offered the choice of $10 or a toss up between $20 and $5, I would take the toss, as on average I would make $12.50.

This is ignoring the paradox put forward by CK, which reminds me of the condemned prisoner paradox, which I believe has been brought up on the board before.

A prisoner condemned to death is told that he will be hung (my English upbringing) one morning of the next week (mon-sun) but is assured that he will not know the day of his death.
The prisoner reasons that it cannot be Sunday, as he would know for certain that it was Sunday, if he was not hauled away by Saturday morning.
Similarly it could not be Saturday, because if Friday morning were to pass and knowing from the rule above that it could not be Sunday, then it must be Saturday and again he would know the day of his death.
The same reasoning could be applied to each day; therefore the prisoner would not be hung.

14

its Hanged not hung
apologies (esp to moggy)

15

Enderw24,

it’s not quite the same as poker, where you know there are 52 cards in all, and what values they are.

The only new information you get when you open the envelope is that the sum of money may not be divided by 2 (swap!), or that you are near the limit of the real-life ‘budget’ (when you don’t swap).

messiah,

it is a good paradox, isn’t it.
But your strategy must be wrong. You say if you find e.g. $10, you would swap. OK, but you don’t need to look in the envelope for that.
You’re going to swap envelopes whatever money you find!
So there must be a fault in your logic.

16

**glee ** I can see that there is a fault in the logic, I just can’t spot where it is.

17

I thought JasonFin would see this from my earlier post.

The logical error is that X does not represent a number, it represents a physical choice.

Say you pick up A and it contains $10, then B contains either $5 or $20. Put down A pick up B. Whatever B contains changes the possible values of A. If B has $20 then A is either $10 or $40. If B has $5 A is now either $2.50 or $10. Plainly this is wrong.

For X to represent a number it must be, for instance, the lesser amount before an envelope is selected. Then the other envelope contains 2X. So you cannot pick BOTH envelopes and designate the contents as X.

18

OK, my turn to try. Ain’t easy to wrap your brain around this one, however …

Let’s take a simple example, where the host has two pairs of envelopes to show you: one with $2 and one with $4, and one with $4 and one with $8. He’s picking a pair of envelopes randomly, 50-50, and you’re randomly picking one of the two to open. You have two choices of strategy, keep or switch. It’s trivial to calculate that your expected win from this game is $3.50, regardless of your strategy. But wait-we proved that the switch strategy is better, didn’t we? Oh-but the proof assumed that there is a 50% chance that the envelope you open (whatever dollar value is in it) is the smaller envelope, and this is not the case. For instance, if you open the $8 envelope, the ‘switch’ strategy screws you, because it assumes that there is a 50% probability that the other envelope contains $16, when in fact there is a 0% probability.

Why does this happen? It’s because there are not just two possibilities in the absence of information. To calculate the real value of this game and determine ideal strategy, you need to know the host’s strategy. And you have no information about this strategy either before or after you draw the card-there are infinite possible envelope-pair-stuffing-and-choosing strategies for the host which have the value you drew being the higher, and the same number for it’s being the lower, of the pair you were given.

Any questions?

19

Not how I see it. Assuming I know what the possible combinations are.

If the only choices are 2/4 and 4/8 then over enough trials I will pick $2 25% of the time $4 50% of the time and $8 25% of the time for an average return of $4.50 per trial.

However, under the host’s rules, if I can look at what I’ve picked and change my mind -

every time I pick $2 I’ll swap it for the $4
everytime I pick $8 I’ll keep it
everytime I pick $4 I’ll swap it, half of the time for $2 and half of the time for $8

So now I get $2 25% OTT $4 25% OTT and $8 50% OTT for an average return of $5.50 per trial.

20

Don’t ask – I can see how your argument works for the 2nd iteration and onwards of the puzzle, but not for the first. I think the conundrum is solved as soon as we put limits on the values in the envelopes (min, max).

If we pick an envelope which has value x, then the second envelope (y) can either have 2x (if x<=max/2) or 1/2x(if x>=2min).

The first possibility (2x) happens therefore only when min<=x<=1/2max, resulting a y between (min and max inclusive) and so y on average = (min+max)/2

The second possibility (1/2x) happens only when 2min<=x<=max therefore results in a y between (min and max inclusive) with An average (min+max)/2.

So x will on average be (min+max)/2, but so will y, therefore the choice is mute.

I hope you understand this?