So you thought Monty Hall was tricky?

[MrsLime]

If you liked the 'Monty Hall' problem, you're going to hate this one!

Imagine I have a sum of money and two envelopes.  I put one-third of the money into one envelope, and two-thirds of the money into the other envelope (i.e. one envelope contains twice as much as the other).  I give one envelope to Snoopy and one to Danafish, and tell them that before they look inside, they can either keep their envelopes, or swap them with each other.

Now, Snoopy thinks to himself: "Imagine my envelope contains £20.  Then there is a 50% chance that Danafish's envelope contains £10, and 50% that Danafish's envelope contains £40.  So if I swap envelopes, 50% of the time I lose £10 and 50% I gain £20.  So, on average, I gain £5 by swapping.  Therefore I should swap!"

At the same time, Danafish is thinking to herself: "Imagine my envelope contains £20.  Then there is a 50% chance that Snoopy's envelope contains £10, and 50% that Snoopy's envelope contains £40.  So if I swap envelopes, 50% of the time I lose £10 and 50% I gain £20.  So, on average, I gain £5 by swapping.  Therefore I should swap!"

But of course there is only a fixed sum of money between the two envelopes, so how is it possible that they both gain by swapping?

[londonpokergirl]

Not this one again    let me remember.....

[MrsLime]

Ok, anyone from the Slough Palace can ignore this thread!

[AndrewT]

I thought it must be something to do with the phrasing of the question, but it isn't.

I had a think, then gave up and googled the answer.

Then my brain melted.

[londonpokergirl]

Ok, anyone from the Slough Palace can ignore this thread!

hehe me and patrick out then

[Mbuna]

Both players win as they have risked 0 to win a either 1/3 of a sum of money or 2/3 of a sum of money .

50% of the time a player will win 2/3 of the money regardless of wether they swap or not
and 50% he will win 1/3.

I think

[Mbuna]

I personally would swap if my envelope was the thinest.

[Colchester Kev]

If some mug was giving away free money in envelopes, i would cut a deal with the other envelope holder and do an even chop

[thetank]

The envelope with a third has 2/6 of the dosh
The envelope with two thirds has 4/6 of the dosh

Equal chance of having both, so your EV before the envelope opens is a half of 2/6 + 4/6. This is 3/6.

If you swich and lose, you drop 1/6 from your EV, if you switch and win, you're up 1/6 from your EV.

As 1/6 - 1/6 = 0 neither player will either gain or lose from swapping.

....but Snoopy is a jammy fish, so I'd probably swap if I was Dana. 

[thetank]

But that's all common sense of course. Everyone knows that neither player gets anything out of swapping.
So one can conclude, as AndrewT said,  there must be a logical fallacy in the wording.

Here is my theory.....

Imagine my envelope contains £20.

This assumes that there is a 100% chance of his envelope containing £20 (or a scalar multiple thereof)
In truth, there's a 50% chance it has £20, and a 50% chance it has another amount. (25% £10, and 25% £40)

So before you know what's in the envelope it is not only £20, it is also £10, and also £40. Something to do with a cat belonging to Schroedinger..blah blah blah, but it's at this point that the logical fallacy draws it's root, and skews the sum.

 

[MrsLime]

Here is my theory.....

Imagine my envelope contains £20.

This assumes that there is a 100% chance of his envelope containing £20 (or a scalar multiple thereof)
In truth, there's a 50% chance it has £20, and a 50% chance it has another amount. (25% £10, and 25% £40)

So before you know what's in the envelope it is not only £20, it is also £10, and also £40. Something to do with a cat belonging to Schroedinger..blah blah blah, but it's at this point that the logical fallacy draws it's root, and skews the sum.

I think perhaps you've misunderstood what 'imagine' means and have been diverted by thoughts of kittens barking up the wrong tree.  Let me slightly reword the third paragraph:

Now, Snoopy thinks to himself: "I define X to be the amount in my envelope.  [So now, by definition, there is 100% chance that the envelope contains X.]  Then there is a 50% chance that Danafish's envelope contains X/2, and 50% that Danafish's envelope contains 2X.  So if I swap envelopes, 50% of the time I lose X/2 and 50% I gain X.  So, on average, I gain X/4 by swapping.  Therefore I should swap!"

Of course we agree that there can be no value in swapping!  But where is the logical flaw in Snoopy's reasoning?

[AndrewT]

Here is my theory.....

Imagine my envelope contains £20.

This assumes that there is a 100% chance of his envelope containing £20 (or a scalar multiple thereof)
In truth, there's a 50% chance it has £20, and a 50% chance it has another amount. (25% £10, and 25% £40)

So before you know what's in the envelope it is not only £20, it is also £10, and also £40. Something to do with a cat belonging to Schroedinger..blah blah blah, but it's at this point that the logical fallacy draws it's root, and skews the sum.

I think perhaps you've misunderstood what 'imagine' means and have been diverted by thoughts of kittens barking up the wrong tree.  Let me slightly reword the third paragraph:

Now, Snoopy thinks to himself: "I define X to be the amount in my envelope.  [So now, by definition, there is 100% chance that the envelope contains X.]  Then there is a 50% chance that Danafish's envelope contains X/2, and 50% that Danafish's envelope contains 2X.  So if I swap envelopes, 50% of the time I lose X/2 and 50% I gain X.  So, on average, I gain X/4 by swapping.  Therefore I should swap!"

Of course we agree that there can be no value in swapping!  But where is the logical flaw in Snoopy's reasoning?

The problem is that you have defined X to be the amount in Snoopy's envelope. Then you take two separate cases, one where Snoopy has the smaller amount and one where he has the larger amount. You then equate the two values of X, which you can't do because they're different.

The X in 2X is a different X than the X in X/2.

[Mbuna]

Here is my theory.....

Imagine my envelope contains £20.

This assumes that there is a 100% chance of his envelope containing £20 (or a scalar multiple thereof)
In truth, there's a 50% chance it has £20, and a 50% chance it has another amount. (25% £10, and 25% £40)

So before you know what's in the envelope it is not only £20, it is also £10, and also £40. Something to do with a cat belonging to Schroedinger..blah blah blah, but it's at this point that the logical fallacy draws it's root, and skews the sum.

I think perhaps you've misunderstood what 'imagine' means and have been diverted by thoughts of kittens barking up the wrong tree.  Let me slightly reword the third paragraph:

Now, Snoopy thinks to himself: "I define X to be the amount in my envelope.  [So now, by definition, there is 100% chance that the envelope contains X.]  Then there is a 50% chance that Danafish's envelope contains X/2, and 50% that Danafish's envelope contains 2X.  So if I swap envelopes, 50% of the time I lose X/2 and 50% I gain X.  So, on average, I gain X/4 by swapping.  Therefore I should swap!"

Of course we agree that there can be no value in swapping!  But where is the logical flaw in Snoopy's reasoning?

The problem is that you have defined X to be the amount in Snoopy's envelope. Then you take two separate cases, one where Snoopy has the smaller amount and one where he has the larger amount. You then equate the two values of X, which you can't do because they're different.

The X in 2X is a different X than the X in X/2.

To put it another way the envelope contains either the larger amount in (say 2X) if we swap we lose X

OR

The envelope contains the smaller amount ie X in which case if we swap we gain X


50% of the time we gain X 50% we lose X

 EV+ FROM THE SWAP = 0

[ItsMrAlex2u]

My brain is hurting now !!

[thetank]

The problem is that you have defined X to be the amount in Snoopy's envelope. Then you take two separate cases, one where Snoopy has the smaller amount and one where he has the larger amount. You then equate the two values of X, which you can't do because they're different.

The X in 2X is a different X than the X in X/2.

That sounds more likely.