Envelope Problem
- Thread starter Meistro
- Start date
This is a matter of CE vs EV.
Regarding CE, you can either keep the guaranteed $100 or switch the envelope for the risk of losing $100 for the potential of $50 or $200. As you can see, this is all subjective depending on someones lifestyle and how much money they have.
Regarding EV, you could switch and net your imaginary $25 in EV that you've just generated. You can receive a potential of $50 or $200 instead of the original guaranteed $100. $125 average from both - $100 = +$25.
The answer to the question is you take the 100 every time.
100 CE > 25 in EV
Regarding CE, you can either keep the guaranteed $100 or switch the envelope for the risk of losing $100 for the potential of $50 or $200. As you can see, this is all subjective depending on someones lifestyle and how much money they have.
Regarding EV, you could switch and net your imaginary $25 in EV that you've just generated. You can receive a potential of $50 or $200 instead of the original guaranteed $100. $125 average from both - $100 = +$25.
The answer to the question is you take the 100 every time.
100 CE > 25 in EV
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Alright, I think I have a solution. Your best strategy is to switch once, no matter the amount you start with. Your gain here depends on variance. The reason why switching once is so clearly +EV is because you are never going to end up at an even result. Half the time you will win and half the time you will lose. If you switch twice and get one win and one loss you are back to where you started, no gain. But let's say you switch twice and get two wins. $100, $200, $400. +300. But what happens if you get two losses? $100, $50, $25. -75
So if you ran the game twice and one time happened to get two wins and the other time happened to get two losses you would be way ahead.
Since the law of large numbers dictates that the more trials you run the closer you get to the expected result, your best strategy is to play the game once.
So if you ran the game twice and one time happened to get two wins and the other time happened to get two losses you would be way ahead.
Since the law of large numbers dictates that the more trials you run the closer you get to the expected result, your best strategy is to play the game once.
In case I am ...
With the question as it is phrased above, the situation is entirely symmetrical between the envelopes. There can be no possible reason to switch; there is no information on which to base a switching decision. (As Sonny intimated.)
This is hopefully the common-sense, obvious answer that is everyone's starting point. But it's hard to pin down the flaw in the logic that seems to support switching -
It seems like a paradox, and you can begin to convince yourself that common sense must be wrong. But, for once, common sense is entirely correct.
There's a wealth of reading material on this problem and variations of it -
https://en.wikipedia.org/wiki/Two_envelopes_problem
I found reference no. 10 from the Wikipedia page quite helpful - https://arxiv.org/abs/1411.2823 (sections 1.3 and 2.3, in particular)
Reading that paper, the only thing I am still struggling with is the rationale behind the stated need to -
Without a proper grasp of that, I feel like I'm still leaning more on common sense than on mathematical proof.
But the interesting thing I take away from that paper is that you can make switching favourable if you recast the problem slightly -
Two envelopes are labelled A and B. An amount is placed in envelope A, and a coin is tossed to determine whether to place half or twice that amount in envelope B.
Now you have an asymmetry. If you are offered envelope A, you should ask to switch to envelope B.
With the question as it is phrased above, the situation is entirely symmetrical between the envelopes. There can be no possible reason to switch; there is no information on which to base a switching decision. (As Sonny intimated.)
This is hopefully the common-sense, obvious answer that is everyone's starting point. But it's hard to pin down the flaw in the logic that seems to support switching -
It seems like a paradox, and you can begin to convince yourself that common sense must be wrong. But, for once, common sense is entirely correct.
There's a wealth of reading material on this problem and variations of it -
https://en.wikipedia.org/wiki/Two_envelopes_problem
I found reference no. 10 from the Wikipedia page quite helpful - https://arxiv.org/abs/1411.2823 (sections 1.3 and 2.3, in particular)
Reading that paper, the only thing I am still struggling with is the rationale behind the stated need to -
Without a proper grasp of that, I feel like I'm still leaning more on common sense than on mathematical proof.
But the interesting thing I take away from that paper is that you can make switching favourable if you recast the problem slightly -
Two envelopes are labelled A and B. An amount is placed in envelope A, and a coin is tossed to determine whether to place half or twice that amount in envelope B.
Now you have an asymmetry. If you are offered envelope A, you should ask to switch to envelope B.
just guessing here (I just skim read parts of it, & would likely never understand hardly any of it):
it seems (just like my 5th grade math teacher used to say, "correct answer, did you check your math?" me, "no teacher")
to where they are sort of getting the same answer using two rationales, their own and that of others (a sort of math check).
equating equation (1.1.1) especially the value X in the introduction going about it using their contrived rational where they state,
"In this variation the content of one of the two envelopes is revealed
to both players.
Existing literature does not appear to include a way to replace the variable X of variation 1.1 with the numerical value of the amount
that is revealed (see for example the paper of Graham Priest and Greg Restall [3], section “Opening the Envelope”).
This is exactly what we will try to achieve here. "
edit: I guess as well, their use of the weighted average shows a bit of what's going on (in an informational sense) that perhaps wasn't previously depicted. end edit
it seems (just like my 5th grade math teacher used to say, "correct answer, did you check your math?" me, "no teacher")
to where they are sort of getting the same answer using two rationales, their own and that of others (a sort of math check).
equating equation (1.1.1) especially the value X in the introduction going about it using their contrived rational where they state,
"In this variation the content of one of the two envelopes is revealed
to both players.
Existing literature does not appear to include a way to replace the variable X of variation 1.1 with the numerical value of the amount
that is revealed (see for example the paper of Graham Priest and Greg Restall [3], section “Opening the Envelope”).
This is exactly what we will try to achieve here. "
edit: I guess as well, their use of the weighted average shows a bit of what's going on (in an informational sense) that perhaps wasn't previously depicted. end edit
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Clearly switching once has a positive expectation.
Would you play a game where you pay $100 to play, and then flip a coin, where if it lands on heads up get $200 and if it lands on tails you get $50? If not, you can't be considered much of a gambler.
The paradox is that the long term expectation of constantly switching if you are given this choice with every resulting sum, is to break even, whereas with a single trial you have a positive expectation.
Optimal strategy is to switch some finite amount of times, that gives you a decent chance of an extreme result but not so many that you start to get bogged down by the law of large numbers.
Would you play a game where you pay $100 to play, and then flip a coin, where if it lands on heads up get $200 and if it lands on tails you get $50? If not, you can't be considered much of a gambler.
The paradox is that the long term expectation of constantly switching if you are given this choice with every resulting sum, is to break even, whereas with a single trial you have a positive expectation.
Optimal strategy is to switch some finite amount of times, that gives you a decent chance of an extreme result but not so many that you start to get bogged down by the law of large numbers.
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Meistro, I get the impression you haven't read any of the material I referenced. If you had, you ought at least to be beginning to doubt yourself by now!
Read the Wikipedia page. Read the paper I picked out from the references. And also take a look at another of the references (no. 33) - https://arxiv.org/abs/1202.4669 , which explains things a lot more clearly than anything else I've seen.
Read the Wikipedia page. Read the paper I picked out from the references. And also take a look at another of the references (no. 33) - https://arxiv.org/abs/1202.4669 , which explains things a lot more clearly than anything else I've seen.