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Two envelopes problem

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Re: Two envelopes problem

#131

Post by pontus »

one-more-minute wrote: post #87

This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow. `B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where th…

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 should switch.

It's a subtle difference in how the envelopes are prepared, but it makes all the difference.

Re: Two envelopes problem

#132

Post by rtkwe »

andrewclunn wrote: post #122
the_sleaze9 wrote: post #118

I have to admit I remain unconvinced. Has anyone ever run a real world experiment to verify the math? For whatever reason I can't get past my intuitive feeling that this problem is just a simple 50/50 and these proofs are just fancy window dressing.

It's much like the Monte Hall problem, in that people often forget to include the portion about the revealed door ALWAYS being a loser. If it were a randomly opened door then the option to switch would actually just be 50/50. This can be shown by just running the simulation over thousands of iterations. The same is true here. It's not actually going to yield better outcomes to switch. Sometimes logicians just need to…

The difference between this and the Monte Hall problem is there's no new information revealed or pool reduction here.

Re: Two envelopes problem

#133

Post by olalonde »

the_sleaze9 wrote: post #118

I have to admit I remain unconvinced. Has anyone ever run a real world experiment to verify the math? For whatever reason I can't get past my intuitive feeling that this problem is just a simple 50/50 and these proofs are just fancy window dressing.

You remain unconvinced that switching is pointless? Unlike most paradoxes, this is one where the intuitive answer is the correct one. It's a paradox because of the flawed, but convincing, reasoning that is given in favor of switching.

Re: Two envelopes problem

#134

Post by davidguetta »

xmprt wrote: post #58
davidguetta wrote: post #53

The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

That's not what it's saying. It's saying one envelope contains A and the other contains either 2*A or A/2.

yes but the formula at bullet point 7 says it. That's the point. The formula is wrong and doesn't correspond to reality

Re: Two envelopes problem

#135

Post by hennell »

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 latter problem is more interesting, although as written they're meeting to compare wallets with this game, so both wallets should logically be empty.

But assuming a more surprise situation in which the game is suggested* the reasoning is pretty good for both if they're happy with a 50% chance at doubling their money + X. But I think maybe its the hidden prize that makes the logic work (or not work)? For one of them the prize is less then double their money (a prize they'd never win but still) so the deal isnt a good deal. Not sure how the math works for that, but a contest where you might win 2a or less then 2a is presumably a different reasoning.

*Where similar to the first point we should assume the suggesters wallet is empty.

Re: Two envelopes problem

#136

Post by davidguetta »

wccrawford wrote: post #100
davidguetta wrote: post #52

Earlier quoted context omitted.

Yeah, that's obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

It doesn't actually say that. It says the other envelope will either contain 2A or A/2. The envelope you're holding is A. The other envelop is either twice that, or half that.

the "writing" doesn't say it but the written formula at bullet point 7 says it. That's the mistake, the formula is wrong and does not describe reality.

Re: Two envelopes problem

#137

Post by davidguetta »

dotancohen wrote: post #80
davidguetta wrote: post #52

Earlier quoted context omitted.

Yeah, that's obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.

I don't think that this is so obvious to the layman, when A has already been defined as "the amount of money in the envelope I am holding".

Of course it's not obvious, that's why it's a "problem".

But if only look at the formulat at bullet point 7 there "2A" and "0.5A" as amounts but not "A". It just doesn't describe the current state of reality

not trying to brag, just reformulating the simplest way for everyone to see

Re: Two envelopes problem

#139

Post by zmgsabst »

pontus wrote: post #131
one-more-minute wrote: post #87

This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow. `B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where th…

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 odds.

(I had to diagram this all out; unintuitive to me!)

Re: Two envelopes problem

#140

Post by hutzlibu »

olalonde wrote: post #133
the_sleaze9 wrote: post #118

I have to admit I remain unconvinced. Has anyone ever run a real world experiment to verify the math? For whatever reason I can't get past my intuitive feeling that this problem is just a simple 50/50 and these proofs are just fancy window dressing.

You remain unconvinced that switching is pointless? Unlike most paradoxes, this is one where the intuitive answer is the correct one. It's a paradox because of the flawed, but convincing, reasoning that is given in favor of switching.

Is it?

"No proposed solution is widely accepted as definitive."

I mean for me this is totally a "cannot see the forest because of all the trees" situation and the intuitive solution is the right one.

But I thought the same of the monty hall problem (back in school) and only was convinved, after I wrote a small program to simulate it, which confirmed it.

But unlike in the monty hall problem, there is no new information here, when you are given the opportunity to switch. So the discussion seems weird to me.

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