Logic Problem for GoT

[Ulysses]

Relatively simple explanation as to why GoT's approach is invalid here: Two-envelope paradox.

More detailed explanation

Just thinking logically about this problem before jumping straight into EV calculations should make it clear that one can't just apply GoT's basic EV calculations here when dealing w/ finite numbers. See Brocktoon's posts for more on that logic.

[Ulysses]

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Problems like this one is why people who know a decent amount of game theory get frustrated watching game shows.


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Problems like this one are why some people should step back and think about problems in a common-sense way before jumping straight into EV calcs.

[maurile]

This is known as the "two-envelope paradox." I used boxes instead of envelopes to make it harder for people to Google the answer.

[Ulysses]

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In your question, YOU SHOULD ALWAYS SWITCH.

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If that is the case, how does opening the envelope influence your decision? You will switch for any value of x, yes? If so, why do we need to open the envelope? But wait...

[Ulysses]

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If AFTER you picked the first box a guy flipped a coin and put either 2X or 1/2 the amount of the box you picked in another box, then you should ALWAYS switch because of the reasons that me and GoT thought it was a +$25 EV to switch.

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Exactly.

[GuyOnTilt]

No, they're exactly the same.

No, they're not. In the envelope paradox and variations thereof, there are situations and conditions where the EV of switching is zero. The problem you gave is not one of these.

If there is a series of pairs of envelopes (or boxes) as Caro presented then there are situations when switching is neutral in EV. In his example, there are 4 pairs: x and 2x, 2x and 4x, 4x and 8x, and 8x and 16x. He himself stated that if we chose an envelope and were allowed to look at the value inside and saw that it was either 2x, 4x, or 8x, we should always choose to switch with its paired envelope. If it were x, we should trade as well, and if it were 16x we should obviously not trade. Then he went on to say that if we didn't know the range of the values in the envelopes, switching has 0 EV. All of the above is true. Let's say out of the 4 unmarked pairs of boxes/envelopes, we choose pair number 1, and of the two boxes in pair 1, we should Box A. We open Box A and see that it contains $200. Should we switch? The answer is it doesn't matter. We have no clue as to the range. We weren't given the information of where $200 ranked in the different values of the boxes. If could just as easily be the lowest amount as a middle amount as the highest amount. Therefore, the EV of switching if it were each of those would cancel out to equal 0. If it the bottom of the lowest pair, we'd gain $200. If it were the highest of the lowest pair, we'd lose $100. If it were the lowest of the 2nd pair, we'd gain $200. If it were the highest of the second pair, we'd lose $100. And same thing with pairs 3 and 4. Therefore, it doesn't matter whether we choose to switch or not. That is the problem that Caro presented.

In the problem you presented, WE DO KNOW THE RANGE OF THE VALUES IN THE BOXES. We pick one of the two boxes and look inside and there's $100. Now, we know that there are 2 possibilities (just as there were 8 possibilities in Caro's problem): It's either the highest of the pair, or the lowest of the pair, against perfectly similar to the process used to calculate Caro's problem's EV. In the first possibility, we chose the lowest of the two boxes, and we'd gain $100 by switching. In the second possibility, we chose the higher of the two and would lose $50 by switching.

I still maintain that in the original post, switching is +EV.

GoT

[RocketManJames]

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In the problem you presented, WE DO KNOW THE RANGE OF THE VALUES IN THE BOXES. We pick one of the two boxes and look inside and there's $100. Now, we know that there are 2 possibilities

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GoT, you're wrong here... the range is known AFTER THE FACT. If you knew the range before you opened the box, you'd be in a bounded scenario and some EV calculations could be made. BUT, the range was unknown to you BEFORE you opened that first box. I believe this makes all the difference.

-RMJ

[maurile]

Maybe you should ignore Caro's explanation. It obviously didn't make the solution clear to you.

Focus instead on my question of whether two people who get different boxes should both be willing to pay me a penny to switch.

[GuyOnTilt]

If that is the case, how does opening the envelope influence your decision? You will switch for any value of x, yes? If so, why do we need to open the envelope? But wait...

I understand how choosing to switch makes no difference from a common sense and Bayesian perspective, but I don't understand how that overrules EV calcs. One must be flawed somewhere since they're in direct contradiction with each other. I'm at the point where I'm starting to think that it makes no difference, but can't find a flaw anywhere in the math to point to and say, "That's why."

GoT

[maurile]

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I understand how choosing to switch makes no difference from a common sense and Bayesian perspective, but I don't understand how that overrules EV calcs. One must be flawed somewhere since they're in direct contradiction with each other. I'm at the point where I'm starting to think that it makes no difference, but can't find a flaw anywhere in the math to point to and say, "That's why."

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Yes, that's exactly why this is such a great problem. You can get the correct answer by using some common sense, but to actually prove it mathematically is extremely difficult (some people think it's impossible). The link I provided before isn't the one I meant to give. The paper I had in mind is no longer available on the Net as far as I can tell. Try this one, though, for an overview of the Bayesian analysis.