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Two envelopes problem

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Re: Two envelopes problem

#161

Post by pontus »

bottled_poe wrote: post #143
pontus wrote: post #131

Earlier quoted context omitted.

It depends on how the envelopes are prepared. If the two envelopes have x and 2x in them and one is randomly handed to you, there's a symmetry between them that tells you there is no benefit in switching. However, suppose x is placed in an envelope and handed to you. Then a fair coin is flipped. If it comes up heads, 2x is placed in another envelope and if it comes up tail, x/2 is placed in that envelope. Then you sh…

This is new information which changes the definition of the problem though.

Yeah, I agree. This setup though is the one to which the original "naive" argument applies which is why I think it's interesting to ponder.

Re: Two envelopes problem

#163

Post by zeroonetwothree »

metacritic12 wrote: post #156
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…

Generate a random number T from the arctan distribution. If the number in the other envelope is greater than T, switch. For envelopes with two different numbers n, m, the above is guaranteed to give you the higher one strictly more than 50% of the time.

To expand on this, you can use any distribution that has nonzero density over every interval.

Re: Two envelopes problem

#164

Post by camgunz »

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…

Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch".

But this omits (loses) the information of "both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the previous envelope contains >= $MONEY.

So if you forget that your current envelope also contains >= $MONEY you're stuck in an infinite loop, which you break out of only because you can only do a single iteration.

Re: Two envelopes problem

#165

Post by zeroonetwothree »

robocat wrote: post #78

> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch b…

The error is that the probability of winning is not 0.5, it’s a function of how much they have in their wallet. They probably don’t know the true distribution so they can’t calculate the actual expected value. But clearly your chance of winning is better if you have $1 than if you have $100000

Re: Two envelopes problem

#166

Post by zeroonetwothree »

t_mann wrote: post #127
stefs wrote: post #111

Earlier quoted context omitted.

wait - if you had zero cash in your wallet, you'd _never lose_, because you couldn't receive less.

You'd never win, because you can't have more than the other guy. A draw is not winning

You win if you have less

Re: Two envelopes problem

#167

Post by pontus »

zmgsabst wrote: post #139
pontus wrote: post #131

Earlier quoted context omitted.

It depends on how the envelopes are prepared. If the two envelopes have x and 2x in them and one is randomly handed to you, there's a symmetry between them that tells you there is no benefit in switching. However, suppose x is placed in an envelope and handed to you. Then a fair coin is flipped. If it comes up heads, 2x is placed in another envelope and if it comes up tail, x/2 is placed in that envelope. Then you sh…

Oh, weird. I’m not sure I’d have naively realized that the fixed amount in your envelope makes so much difference. For people confused like me, think about the outcomes: When the two are prepared together, say $2 and $4, then when you exchange your options are +2 and -2 with 50:50 odds. When the second envelope is prepared based on $2 in your envelope , then when you exchange your options are +2 and -1 with 50:50 odd…

I did a pretty deep analysis of this problem a while back here in case you're interested: https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa...

Re: Two envelopes problem

#168

Post by vehementi »

platz wrote: post #159
metacritic12 wrote: post #157

Earlier quoted context omitted.

Exactly! The logic of "the average of +100% and -50% is positive" only applies if the denominator is the same. In this case it isn't. The relative percentage "rule of thumb" doesn't even feel like a primitive in math. It seems wrong to reason from that.

why isn't this listed in wikipedia as an answer.

It is, it's the one where you treat both as having the same expected value

Re: Two envelopes problem

#169

Post by praptak »

My solution is this:

In point 2. the solution starts talking about probabilities without defining the sample space.

And this is why the solution is nonsensical. "One envelope contains twice as much as the other" does not define a sample space, so it is too early to start talking about probabilities.

If you say "the sample space is all pairs of natural numbers where one is twice as big as the other", then what are the probabilities assigned to each pair? They cannot be pairwise equal because the space is infinite.

So even if probabilities for (X, 2*X) and (X, X/2) happen to be equal for a particular X, they definitely cannot be equal for every X, so the "infinite swap" doesn't hold anymore.

Re: Two envelopes problem

#170

Post by platz »

vehementi wrote: post #168
platz wrote: post #159

Earlier quoted context omitted.

why isn't this listed in wikipedia as an answer.

It is, it's the one where you treat both as having the same expected value

that the expected value is the same that is an outcome of all solutions, so just stating that is reductive.

What is at question is the reason for getting to the expected value is the same.

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