The Magic Café
Username:
Password:
[ Lost Password ]
  [ Forgot Username ]
The Magic Cafe Forum Index » » Puzzle me this... » » 2 envelopes puzzle/paradox (0 Likes) Printer Friendly Version

 Go to page 1~2 [Next]
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Hello all,
the monty hall problem posted not long ago reminded me of another interesting controversial puzzle/paradox.

Here is a very deep paradox—you can find it on the net quite easily though the solution is not always correct.

You have 2 envelopes, one has a sum of money and the other double the amount. You can't deduct from the outside anything about the contents.

Now say you choose envelope A and assuming it has X dollars in it (X unknown now - envelope was not opened yet), let's look at the other alternative:

You have a 50% chance to have 2X in the other envelope and a 50% chance to have 0.5X in the other envelope. Your expectation is therefore E = 0.5 * 2X + 0.5 * 0.5X = 1.25X !!!

Therefore we must switch, as on average it is always worthwhile to do so—as shown from the calculation.

Now you go to the second envelope and apply the same argument—again you must switch.

So where is the fallacy here???

Nir
Bong780
View Profile
Regular user
Toronto, Canada
172 Posts

Profile of Bong780
There's another proof:

Lets assume D is the difference between the two envelopes (ie. the lesser amount of the two). If you gain in doing the switch, your gain is D. If you happen to lose, your loss is also D.

Since the expected gain/loss is the same, it won't matter if you switch or not.
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Bong you didn't answer where the problem is in the paradox I presented. You just gave another answer.
N.
Ember
View Profile
Regular user
London
121 Posts

Profile of Ember
I'd say the fallacy is that there is any fallacy.

In both cases the math gives the best strategy you can use. In both cases you are faced with either doubling your money or only losing half. So your best strategy, given no further information, irrespective of which envelope you have, is to switch.

The fact that your best choice is the same in both cases is interesting but does not constitute a fallacy.
Jonathan Townsend
View Profile
Eternal Order
Ossining, NY
27122 Posts

Profile of Jonathan Townsend
Nir, you are asking them to write some equations and then do some math. You've seen how it goes when I ask them to call an EFFECT by a generic name.

Yes, the two sets of equations LOOK identical. It probably helps to use X for the value in one set, and Y for the other. That way, the abs(DIFFERENCE) becomes the invariant quantity.
...to all the coins I've dropped here
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Actually no need for math equations, though I would suspect a deeper knowledge of probability theory is required.

This is one of the most beautiful paradoxes I know in the sense that the calculation I have given is true (or is it?)—you really have to always switch ad infinitum... the problem is deep deep in the basis of the question...

Well, do some reasearch, read the online solutions, it is indeed fun to read...
N.
TomasB
View Profile
Inner circle
Sweden
1143 Posts

Profile of TomasB
My guess is that the formula seems to dictate that the contents of the envelopes may change regarding which you pick. In one case there is X and 2X and if you pick the other envelope they state that the contents are X and X/2 when of course it can't change.

It's a wonderful paradox as formulated. And I think you can even say that you may open the selected envelope to find out what X is and _still_ you have the paradox.

/Tomas
Top Hat
View Profile
Inner circle
We peed on you!
1077 Posts

Profile of Top Hat
Nir,

This is a great question. I discussed it with a friend about 10 years ago and can't remember the outcome of the discussions.

However, here are my re-worked thoughts:

First, lets' describe the true reasoning one should follow if asked to carry out this test:

If one envelope has twice as much as the other, then we are justified in assuming a particular case. So, let's say that one envelope contains $5 and the other contains $10.

Now - I pick up one of the envelopes.

The chance that the envelope I am holding contains $10 is 1 in 2 (or 0.5).

The chance that the OTHER envelope contains $10 is exactly the same - 1 in 2 (or 0.5). It is in NO WAY affected by the fact that I have already selected an envelope, because the contents of the envelopes always remain fixed.

Therefore, there is NO advantage in changing my choice, because if I did, I would still end up with the same chance (1 in 2) of holding the $10 envelope.

That is the correct reasoning to follow.

So why is the "paradox" reasoning false? Let's see...

The statement "you got 50% chance to have 2X in the other envelope and 50% chance to have 0.5X in the other envelope" is NOT correct.

Why?

Because you CANNOT have "50% chance to have 2X in the other envelope" AND "50% chance to have 0.5X in the other envelope" at the same time.

That is the MAIN FLAW in the given explanation!

The truth is, you can only have "100% chance to have 2X in the other envelope" OR "100% chance to have 0.5X in the other envelope" - and that depends on which envelope you first pick!

To explain:

Let's assume you pick the $5 envelope. The chance that the other envelope contains $10 is 100%. The chance that the other envelope contains $2.50 is actually ZERO!

So if you swap (not knowing what you currently hold), your chances of doubling your money are exactly the same as your chances of first picking the $5 envelope (= 1 in 2).

Alternatively, let's assume you pick the $10 envelope. The chance that the other envelope contains $20 is actually ZERO! The chance that the other envelope contains $5 is 100%.

So if you swap (not knowing what you currently hold), your chances of halving your money are exactly the same as your chances of first picking the $10 envelope (= 1 in 2).

SO... because either scenario is equally likely (because you are just as likely to initially choose one envelope as you are to choose the other) let's assume (rightly) that we DON'T KNOW what your envelope contains. Not knowing what the envelope contains does not alter the fact that there are only two possible scenarios (as outlined above). And this is regardless of the actual amounts of money involved!

So if you swap, you are equally as likely to double your money (1 in 2 chance) as you are to halve your money (1 in 2 chance).

So there is no advantage in swapping.

It is actually a simple situation with no paradox. The apparent "paradox" arises from the faulty central statement, as explained.
TH Smile Smile Smile Smile Smile
Jonathan Townsend
View Profile
Eternal Order
Ossining, NY
27122 Posts

Profile of Jonathan Townsend
And so, the excluded middle resigns from the paradox and slinks away in shame...
...to all the coins I've dropped here
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Tomas,

as usual you are right - but I wanted to bring it in later Smile

Top hat, your explanation is not right. you look at the cases where they already happened - then of course there is no way there is 50% case A and 50% B then we have a deterministic case! but this is why we have the expectation for (a broader term for what is basically an averege)!!! I showed a calculation of what on AVEREGE you SHOULD do - in that case it showed you should always switch. This is how you calculate those things in probability, we don't have a deterministic view on everything. the problem is not there Top Hat - it is much deeper than that...

keep thinking,

Nir
Top Hat
View Profile
Inner circle
We peed on you!
1077 Posts

Profile of Top Hat
Nir,

My explanation is correct. It is, in actual fact, the same argument that Tomas has used - but just expanded to make it easier to "see".

So - I reapeat: you CANNOT have "50% chance to have 2X in the other envelope" AND "50% chance to have 0.5X in the other envelope" at the same time. This is because you cannot consider X, 2X and 0.5X within one scenario. As Tomas said, you either have (X and 2X) OR (X and 0.5X).

That is why the formula is incorrect - because it uses X, 2X and 0.5X to consider the outcome, when in fact they cannot all be considered within one scenario (i.e. one choice of envelope).

(Re-read my explanation!)
TH Smile Smile Smile Smile Smile
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Top Hat,

I reread the explanation - it is incorrect - the fault is elsewhere, we can continue that through PM, as there is no use for a message ping pong over that.
basically you state that the message is incorrect but not give the real reason.
I can give you some simple example when this approach DOES work (of considering many different possibilities in the same time and giving "weights" to each one). again I do not want to reveal the answer so fast as some poeple enjoy thinking about it, but would be happy to supply you the answer on a PM basis.

Nir

p.s. what Tomas stated is just another way to look at the same problem - that even if you know the contents of a single envelope you still have the same problem...
Platt
View Profile
Inner circle
New York
1910 Posts

Profile of Platt
Wow, I couldn't follow half of the attempted explanations. I'm guessing it's something like this:

If you have $10 and lose $5, you have a 50% loss.

If you have $5 and and gain $5, you have a 100% gain.

So the percentages seem to be in your favor even though the actual $$ amounts will even out.

It's like the stock market. If it goes down 50% in value. It will have to, at that point, gain 100% in value to get back to where it was before the 50% drop. In essence, the % changes because your dealing with a different total sum, but the change(50% drop vs. 100% gain) is actually the same value.

Am I on to something here?
Sugar Rush is here! Freakishly visual magic. http://www.plattmagic.com
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Platt
sorry not even close Smile
Top Hat
View Profile
Inner circle
We peed on you!
1077 Posts

Profile of Top Hat
Nir,

I stand by my explanation. To further clarify:

Given that one envelope contains X and the other contains 2X (which we agree on), then...

If you have a 50% chance of having 2X in the other envelope (i.e the one NOT currently chosen), then it follows logically that you have a 0% chance (NOT a 50% chance) of having 0.5X in that envelope!

The fact that there is a 50% chance of the other envelope containing 2X is by virtue of there being a 50% chance that you are currently holding the envelope containing X. It is NOT by virtue of the probability that the other envelope contains one of two alternatives (2X or 0.5X) - because if you are holding the X envelope, there is no such thing as a 0.5X envelope!

And therein lies the flaw in the "paradox"... because you say that you have a 50% chance of having 2X in the other envelope, but then go on to say that you also have a 50% chance of having 0.5X in that envelope! That statement (on which you base your formula) is fundamentally wrong - because at that point you are introducing an amount (0.5X) into the equation which simply does not exist in the problem!

I am going to add something, to defend against a possible response to what I've said.

You may say - "OK, I look in the envelope I've chosen and find it contains X. I now know that the other envelope may contain 2X and it may contain 0.5X".

That is perfectly true. I do not deny that. However, it MUST BE ONE OR THE OTHER (i.e. 2X OR 0.5X).

If the other envelope contains 2X, we have simply first chosen the "smaller" envelope. If the other envelope contains 0.5X, we have simply first chosen the "larger" envelope.

But my explanation holds regardless of whether you first choose the "small" envelope or the "large" envelope. In the first case, I hold X (and the other contains 2X). In the second case, I hold Y (and the other contains Y/2).

Because the actual amounts involved are irrelevant, then Y is the equivalent of 2X, and Y/2 is the equivalent of X - so it amounts to the same problem.
TH Smile Smile Smile Smile Smile
Nir Dahan
View Profile
Inner circle
Munich, Germany
1390 Posts

Profile of Nir Dahan
Top Hat,

I am sorry this is not true. try to ask yourself this - how would you calculate your AVERAGE gain? The answer is exactly as I stated (this is called in probability theory the expectation, marked by E usually)- the problem is there is something in the basic conditions of the paradox that doesn't hold - even before the calculations!!! that is the heart of the problem. I will send you a PM and will maybe find a few internet links about the problem.

see you on the PM board...

Nir
Top Hat
View Profile
Inner circle
We peed on you!
1077 Posts

Profile of Top Hat
Smile

No - my explanation is true. The answer given by the formula is wrong simply because it assumes (before the calculations even begin) the possibility of two simultaneous scenarios which are, in fact, mutually exclusive! That IS the basic condition that does not hold! That IS the heart of the problem!

(Nir and I have swapped PMs. I understand that this is a mathematically deep puzzle. I think my explanation is rational, but realise there is far more to the debate!)
TH Smile Smile Smile Smile Smile
eggshell
View Profile
Regular user
Chorley, England
146 Posts

Profile of eggshell
The formula Nir states is perefectly correct, it would always be worth switching because of that value. However the missing piece here is that fact that until you do make a final choice you have 0 !!!!.
Jonathan Townsend
View Profile
Eternal Order
Ossining, NY
27122 Posts

Profile of Jonathan Townsend
There is something deep about this problem. If X were the amount in the envelope in hand, and since BOTH contents are unknown, at any time you could trade for 2x or x/2. I feel like a monkey with his hand on coconut (you know the trap) and can't let the thing go, yet can't get the nut out.

My instinct/intuition: pretending to be in this situation is to treat do the switch twice, then keep what I have, as the expectation values for the contents are equal.
...to all the coins I've dropped here
Platt
View Profile
Inner circle
New York
1910 Posts

Profile of Platt
The equation seems wrong.

If one envelope has X amount. And another has twice the amount or 2X. You'd have "X and 2X." Not "2x and .5X."

If X were, say, 10, that would be the equivalent of $20 in one envelope and $5 in the other.
Sugar Rush is here! Freakishly visual magic. http://www.plattmagic.com
The Magic Cafe Forum Index » » Puzzle me this... » » 2 envelopes puzzle/paradox (0 Likes)
 Go to page 1~2 [Next]
[ Top of Page ]
All content & postings Copyright © 2001-2021 Steve Brooks. All Rights Reserved.
This page was created in 0.2 seconds requiring 5 database queries.
The views and comments expressed on The Magic Café
are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic.
> Privacy Statement <

ROTFL Billions and billions served! ROTFL