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Two envelopes problem

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Re: Two envelopes problem

#211

Post by judahmeek »

pontus wrote: post #199
kloch wrote: post #142

A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…

There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it . Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This ha…

So the real predictor is the probability.

Once the probability gets around 7/9*.5X + 2/9*4X, then switching is no longer attractive.

Re: Two envelopes problem

#212

Post by vivegi »

Reminds me of the "splitting the dinner bill" problem we had fun with in middle school. It goes like this.

Three friends go out for dinner and decide they will split the bill threeways. The bill comes to $25. Each friend gives a $10 note with $30 pooled together to pay the bill. The waitress returns with five $1 bills in change. They leave a $3 tip for the waitress with $2 in change remaining.

The question is: Of the original $30 paid if we deduct $3 in tips paid, we get $27. Adding the $2 in change remaining we get $29. Where is the other $1?

Re: Two envelopes problem

#213

Post by amalcon »

ncmncm wrote: post #176
kloch wrote: post #142

A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…

The goalposts here are different from a normal "paradox". The actual solution is explained right there on the page, the expectation for both envelopes (one has x, one has 2x) is x*3/2, so switching changes nothing. The "solution" demanded is supposed to be pointing out the specific logical error in the erroneous calculation that says switching should win. Citing the correct logic is not considered sufficient.

The paradox is supposed to be in the contradiction between this (incorrect) reasoning and the correct reasoning, not in the counterintuitive conclusion of the incorrect reasoning.

Re: Two envelopes problem

#214

Post by kmod »

pontus wrote: post #210
kmod wrote: post #206

Earlier quoted context omitted.

Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.

No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the exp…

If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope.

I assume you meant to use a exponential base <0.5 so that the mean is finite

Re: Two envelopes problem

#215

Post by jonath_laurent »

amalcon wrote: post #209
jonath_laurent wrote: post #183

I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained math…

The second simple answer from the article is basically this: pointing out that A is conditionally defined in step 1, but both definitions are used in later steps as if they are the same value. It then goes on to over-explain that a bit, but this answer is in there.

I agree with you. What I am taking issue with is the article presenting a multitude of (unnecessarily) complex refutations (as if one wasn't enough) and suggesting that there is no consensus on which one should be believed. This is not how people do mathematics!

A better article would first introduce a more precise version of the switching argument and then precisely identify the flaw in it. The more advanced mathematical and philosophical discussion that explores variations of the basic argument should be separated and contextualized clearly.

Re: Two envelopes problem

#216

Post by baobabKoodaa »

kmod wrote: post #203

If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory, and further false conclusions (such as "P and not-P" for P="it is better to switch") are not surprisi…

> If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory...

What? I can literally take 2 envelopes right now - physical envelopes - and I can literally put $10 in one envelope and $20 in the other envelope. Then I can shuffle the envelopes and hand one of them to you. Now the other envelope has equal probability of having half or double the amount of money as the first. How is this premise contradictory in your opinion? I can physically arrange objects in this fashion, so where's the contradiction?

Re: Two envelopes problem

#217

Post by [deleted] »

pontus wrote: post #210
kmod wrote: post #206

Earlier quoted context omitted.

Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.

No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the exp…

[deleted]

Re: Two envelopes problem

#218

Post by pontus »

kmod wrote: post #214
pontus wrote: post #210

Earlier quoted context omitted.

No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the exp…

If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite

The point I'm making is that if you open the envelope you will always conclude that it's advantageous to switch regardless of what you see in that envelope. In the case where you see $1, you're guaranteed that switching is a good idea. If you see anything other than $1 you come to the same conclusion but you have to rely on a probabilistic argument using expected values. Now, if you will decide to switch regardless of what you see in the envelope, then why even bother opening it? But, if you don't open the envelope, then you're back to the original problem where the symmetry argument is valid.

Re: Two envelopes problem

#219

Post by ChrisLomont »

jstanley wrote: post #181
ChrisLomont wrote: post #151

This reminds me of a fascinating problem (not sure if there's a name for it): You play the following game (say for many rounds). Someone puts different values in two envelopes (they can pick the values arbitrarily, even maliciously, each round). You want to pick the higher valued envelope. You pick an envelope, they show you the value of the other. You can hold your value or you can switch. It also works if the game…

FWIW, I think all 3 versions of your game are completely equivalent. In all 3 versions you first get to decide which envelope you look inside, and then you get to decide which envelope you open.

Yes, I agree they are equivalent, which is why I was able to see the same solution allows all three games. The last one gives the puzzle solver more things to fiddle with, even though it does not matter, perhaps making the puzzle more difficult or interesting.

Re: Two envelopes problem

#220

Post by kmod »

baobabKoodaa wrote: post #216
kmod wrote: post #203

If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory, and further false conclusions (such as "P and not-P" for P="it is better to switch") are not surprisi…

> If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory... What? I can literally take 2 envelopes right now - physical envelopes - and I can literally put $…

If I open an envelope with $10 then I know that there is $20 in the other envelope.

As soon as you assign a well-formed probability distribution to the money in the envelopes you will find that opening the first envelope is informative.

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