Statistics: Two Envelopes Problem

In this video, I discuss the two envelopes problem, a famous paradox in statistics.

[lionelward]

Awesome job in explaining this!! Very intuitive and clear explanation of a difficult subject

[jameswhite]

Excellent, concise, and thorough explanation.

[ED-sbrs]

Best explanation I could find of this. Thank you 👍

[jameswhite]

You should try an explanation of "the sleeping beauty" paradox, which I still find baffling, but have hopes of resolving. Then try the most challenging and profound of paradoxes: Newcomb's Paradox. I almost despair of ever solving it but in the attempt it has lead me to consider some deep concepts.

[samreenfatima]

It was a very good clarification! Thankyou

[mastercontrol]

Suppose you had a finite range (say D = [0, 100] ∩ Z). Let a ∈ D. I put this in an envelope marked A. Then, I toss a fair coin, and generate some number b like this: if I get heads, b = 2a, and if I get tails, b = a/2. I put it in an envelope marked B.

Now, when I hand my subject an envelope, I am always sure to hand him the A envelope, and although we generated a from a finite domain, isn't it still equiprobable that b = 2a or b = a/2?

In this case, doesn't the paradox hold?

[piprod]

I paused it at 1:41 after the explanation of the problem. I've not come across the paradox before so I'll try to solve it then watch the explanation.

The first scenario is definitely true. We're trading a box worth £10 for a box with either £5 or £20 or on average a box worth £12.50. £2.50 profit on average for exchange.

On the second scenario say they are £10 and £5. On average they are worth £7.50 and you'd only be trading it for a box worth £7.50. So there's no point in exchanging them.

[granrey]

An amount "Z" has been distributed among the envelopes in the ratios of "1/2 X", "X" and "2X".   for instance, the amounts could be: 10, 20 and 40 (Z=70). Each envelope has a exterior different color.

The game is played with two players and a host. 
The host is the only person who knows what amount is on each envelope based on the colors.
The players know that the envelopes have amounts of 1/2X, X and 2X. They don't know what  the Z amount is, neither which envelope is which.  

The envelopes are put on a table and the host will randomly take one of these two envelopes out of the game: either "1/2X"or "2X".  In a way that one of these two sets are left on the table.

Set (1). "X", "2X"
Set (2) "1/2 X",  "X"

Rules:
-In Set (1). The winner of the game would be the player who has  "2X"
-In Set (2). The winner of the game would be the player who has "X"
-The players are allowed to switch with each other as long they have not opened any of their envelopes.

The two players are aware of this but they don't know which of the two sets are on the table. All they know is that either "1/2X" or "2X" was taken out of the game.

At this point the game is identical to "The two envelopes paradox".  Each of the two players will select one envelope.  

A) what is the expected envelope that any of the two players have?
Envelope "X" is on both sets.  Meaning, it has a 1/2 chances of being in the hand of one of the two players.  Meaning the other 1/2 chances is divided by the other two envelopes "1/2 X" and "2X". 

As result, the equation for the expected envelope that any of the two players have is : 1/2 ("X") + 1/4 ("1/2X") + 1/4 ("2X"). 

  B) what is the chance of any of the players to win the game based on the expected values of the expected  envelopes they have?

This is based on the winning chance of every envelope.

"X/2" only shows on set 2 and it will always lose to "X".  As result the winning chance is 0%.
"X" this envelope is on both sets, it wins on set 2 but loses on set 1.  As result the winning chance is 50%.
"2X" this envelope is on set 1 and it will always win. As result the winning chance is 100%

Substituting the winning chances of each envelope on the expected envelope equation we get this:

1/2 (50%)+ 1/4 (0%) + 1/4 (100%)=  50 %.    Meaning, there is no need for switching.

[granrey]

 
My take.

expected value=
This formulas has two conditions that must be met as well to be correct:

-all the values of probabilities used in the formula must add to EP= P1+P2+....Pn
-all the values of values used in the formula must add to EV=V1+V2+....+Vn

The amount of V1+V2 used on your expected value has to be equal to the total amount of money of both envelopes, meaning EV= V1+V2. Meaning the expected value for any of the two envelopes is 1/2 V1+ 1/2 V2= 1/2 (1/3 EV)+ 1/2 (2/3 EV) = 0.5 EV..

Meaning 50% chance of getting half the value of the total on both envelopes on any envelope (no need for switching).

Also, if you decide to define the envelopes based on each other as you are doing it, it should be as  any of these two formulas (but you cant mix the formulas):

(a) V1=2xV2   Meaning EV= 2X+X= 3X

So, if EV= 30 and EV= V1+ V2, then X= 10, V1=10 and V2= 20


or

(b) 1/2 V1= V2  Meaning EV= 1.5X
So, if EV=30, and EV= V1+V2, then X=20, V1=20 and V2=10



So the expected value should be for any envelope based on equation (a) is;
1/2 (2x)  + 1/2 (X)= 1.5X, since X= 10, then expected value= 1.5 (10) = 15


So the expected value should be for any envelope based on equation (b) is; 
1/2 (X)  + 1/2 (0.5X)= 3/4X, since X= 20, then expected value= 3/4 (20) = 15

both solutions give the same expected value which is 0.5EV = 15

Also EP= 1/2+1/2= 1 on both equations 
On both equations the total of money is 30

the apparent paradox happens cause you are combining half of equation (a) with half of equation (b).  This way EV= 2X+0.5X= 2.5X (changing the the total amount of money on both envelopes)

which is false, the amount of money on both envelopes is 2/3 in one and 1/3 in another and the two equations that explain that are EV=2X+1X= 3X or EV= X+0.5X= 1.5X

[deboogs]

Thank you so much. This was killing me.

[alfonsoesteves]

Thank you very much, it's the best video about this paradox I have found so far.

BUT YOU DIDN'T ANSWER THE PARADOX! at least not completely.

I will name A the envelope you randomly choose first, and B the other one.
I will write expected value of X like this: E(X)

The only mistake in the reasoning that says E(B)=5/4E(A) is the presumption that P(B=2A)=P(B=1/2A), but if we are sure that the expected values of these envelopes are the same, it is imminent that P(B=2A) < P(B=1/2A), to be more specific it is 1/3<2/3. How is it possible that the envelope we choose first always have more (double) chances of being the "big" one? We could have chosen anyone!

Let's talk about the distribution function (DF):

I know that with a specific DF and a specific A (for example 100$), then it might be possible that P(B=2A)>P(B=1/2A), but we don't have a clue about the actual DF. In fact, I'd say it is more reasonable to think about this function as a monotonic decreasing function, the closer X is to zero the higher f(X) is. So the chances of P(B=2A) are always less than P(B=1/2A).

The DF you show us has a maximum, but we don't know if it is at X=1/10^1000, X=1, or X=10^1000. So we can't conclude anything from it.

But even if we define a new DF as some kind of average of all the possible DF, one of the following statements has to be true:
1) E(B)=5/4E(A)
2) 1/3 = P(B=2A) < P(B=1/2A) = 2/3

And both are paradoxical since we choose A in a totally random way.

[dfhwze]

It's quite simple. One envelope contains X, the other contains Y. If you have X, you have an expected gain by switching to Y. If you have Y, you'll loose that value again by switching. Since you don't know whether you have X or Y, it doesn't mather what you do.

[granrey]

Let say, there are 3 envelopes. 

the envelope you pick has X amount (cause you opened it)
one of the other two envelopes has 2X and the other envelope has X/2 (you dont know which envelope has 2X or X/2 but you know they do exist with those amounts)
Should you switch?
the average value of the last two envelope (expected value of switching) = 1/2 (2X) + 1/2 (X/2)= 5/4 X.

Since you have X, the expected gain on the switching is 5/4X-X= 1/4X. meaning you should switch.


This is the equation that you are using, which as I have demonstrated is for another problem all together.

[itskelvinn]

It hurt me trying to understand what you were saying but good video

[Yizak]

I feel like its not a very good idea to divide by infinity...