My Math Forum Solution to the Two Envelope Paradox

 Number Theory Number Theory Math Forum

April 20th, 2017, 12:58 PM   #11
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 Originally Posted by romsek you're talking nonsense. If you know the value of your envelope then if it is $x$ you will swap, if it is $2x$ you will not. You will always end up with $2x$
You know the value, like 10 dollars, but you don't know if you have x = 10 or 2x = 10. In other words, you don't know which is which still.

April 20th, 2017, 01:06 PM   #12
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 Originally Posted by AplanisTophet You know the value, like 10 dollars, but you don't know if you have x = 10 or 2x = 10. In other words, you don't know which is which still.
the scale of the bet having an influence on the betting behavior may well be a characteristic of the human personality.

it has nothing to do with math though.

April 20th, 2017, 01:13 PM   #13
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 Originally Posted by romsek the scale of the bet having an influence on the betting behavior may well be a characteristic of the human personality. it has nothing to do with math though.
Assume that people will always do what will make them more money, independent of personality.

If you know your envelope has 10 dollars, then you'll swap because you'll have a 50% chance of getting 20 dollars and a 50% chance of getting 5 dollars.

That's where:

50%(0.5*10) + 50%(2*10) = 12.50

...comes from. You're making "25 cents on the dollar" on average for swapping.

If you don't know the value of your envelope, then independent of personality, you're indifferent.

Get it?

April 20th, 2017, 01:31 PM   #14
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 Originally Posted by AplanisTophet Assume that people will always do what will make them more money, independent of personality. If you know your envelope has 10 dollars, then you'll swap because you'll have a 50% chance of getting 20 dollars and a 50% chance of getting 5 dollars. That's where: 50%(0.5*10) + 50%(2*10) = 12.50 ...comes from. You're making "25 cents on the dollar" on average for swapping. If you don't know the value of your envelope, then independent of personality, you're indifferent. Get it?
you're also losing 25 cents on the dollar on average from swapping.

I'm done here.

April 20th, 2017, 01:37 PM   #15
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 Originally Posted by romsek you're also losing 25 cents on the dollar on average from swapping. I'm done here.
Ok, but now you're the one talking nonsense...

April 20th, 2017, 02:35 PM   #16
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 Originally Posted by AplanisTophet Ok, but now you're the one talking nonsense...
if i have two equiprobable outcomes, $x$ and $2x$

then the expected value is $\dfrac 3 2 x$

If I have the $2x$ envelope and swap for the $x$ envelope.

Then I come out $\dfrac 1 2 x$ below the expected value.

the way you are looking at this problem that is equivalent to losing $\dfrac 1 2 x$

April 20th, 2017, 03:02 PM   #17
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 Originally Posted by romsek if i have two equiprobable outcomes, $x$ and $2x$ then the expected value is $\dfrac 3 2 x$ If I have the $2x$ envelope and swap for the $x$ envelope. Then I come out $\dfrac 1 2 x$ below the expected value. the way you are looking at this problem that is equivalent to losing $\dfrac 1 2 x$
As I said in the OP, I only acknowledge an average value of 1.5x as you say.

When someone gets an envelope, they don't know if they have x or 2x. If you let them peek in the envelope and see 10 bucks, they 'correctly' assess that they have a 50% chance of getting 20 and a 50% chance of getting 5 if they swap. This is an illusion created by frame of reference in the sense that only one of those two outcomes is possible and one isn't.

Let me be clear now. Let's say X = 10 and 2X = 20. Based on what the person is given, they see 10 in the envelope, they'll get 20 every time but if they see 20 in the evelope, they'll get 10 every time. From their point of reference, they have to swap thinking it will earn them that 25 cents on the dollar, but it can't, because 5 is not an option when they see 10 in that envelope and 40 is not an option when they see 20 in that envelope. The true average value will be 15 (half the time the person will end up with 10 and the other half the person will end up with 20).

I'm saying that the notion there can be a 50%(0.5x) + 50%(2x) = 1.25x option where swapping earns you that 25 cents on the dollar is nonsense. I don't see a true paradox.

So then, by saying 1.5x is the only possible average outcome, you are agreeing with me, yes?

 April 20th, 2017, 03:05 PM #18 Senior Member     Joined: Sep 2015 From: USA Posts: 1,783 Thanks: 919 I'm not getting dragged into an extended discussion with you.
April 20th, 2017, 03:11 PM   #19
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 Originally Posted by romsek I'm not getting dragged into an extended discussion with you.
Can I get a yes or no at least? I don't think an extended discussion is necessary (maybe it is, I just don't think so).

If you can't give a yes or no, then I think you're saying that you think an extended discussion is necessary, that there is more to be discussed but that you just don't want to.

 April 20th, 2017, 03:49 PM #20 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 3,747 Thanks: 39 Since this was posted in the number theory section rather than the probability section, my first thought was that perhaps the amount of money in the envelopes must be an integer number of dollars. If that were the case, then there would certainly be some strategies available to maximize your gain. For example if the first envelope contains an odd amount, then obviously you should exchange it for the other envelope.

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