Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

Two envelopes problem

en.wikipedia.org

Post reply on HN
Print view View original

121–130 of 318 posts

Re: Two envelopes problem

#121

Post by civilized »

I'm going to present a critique similar to this one ( https://news.ycombinator.com/item?id=31568047) but I'll try to slightly rephrase.

Suppose the amount in the envelope you choose is $100. If your amount is the smaller one, the other one has $200. If your amount is the larger one, the other one has $50. So the expectation supporting the switching argument is computed over two different scenarios, where the amounts in the envelopes are {50, 100} and {100, 200} respectively.

This calculation seems to violate a hidden logical constraint: the set of amounts in the two envelopes must be the same no matter what conditional branches we consider for the expectation.

Another related issue - if you switch, the argument for switching back assumes the amount in your original envelope is 4x what you originally said it was.

Re: Two envelopes problem

#122

Post by andrewclunn »

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 own up to the fact that empiricism and applied mathematics is where the rubber meets the road.

Re: Two envelopes problem

#123

Post by selimthegrim »

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…

I’m not sure how literally applying Bayes’ theorem and reducing the sample space accordingly is owning logicians but you do you.

Re: Two envelopes problem

#124

Post by andrewshawcare »

lisper wrote: post #108
andrewshawcare wrote: post #97

Once you pick an envelope, you no longer stand to only gain money. You can lose money and that has to be reflected in the potential value of each envelope. After the first selection, you must express the envelope value as the potential of what each envelope holds (the probabilities from the initial selection) which makes selecting again a wash. Let A = 50 Envelope 1 is 100 Envelope 2 is 25 First selection 1/2(100) +…

> Let A = 50 Envelope 1 is 100 Envelope 2 is 25 No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2. A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.

Impossible. Schrodinger’s envelope, then. You can’t reflect the probability to be 2A AND 1/2A if A represents the value of an envelope.

The value of the other envelope must include the probability that it is the original A (not some sleight of hand new A’ that, itself, is based on an expected value).

Re: Two envelopes problem

#125

Post by one-more-minute »

civilized wrote: post #121

I'm going to present a critique similar to this one ( https://news.ycombinator.com/item?id=31568047 ) but I'll try to slightly rephrase. Suppose the amount in the envelope you choose is $100. If your amount is the smaller one, the other one has $200. If your amount is the larger one, the other one has $50. So the expectation supporting the switching argument is computed over two different scenarios, where the amounts…

To expand on this, the switching argument is actually correct if you happen to know that the envelopes contain either {x, 2x} or {2x, 4x} dollars (where x = 50 or whatever), and you observe that A=100. These are effectively hidden assumptions of the argument.

If you wanted to, you could explicitly model all possible values of the envelopes, and all possible observations of A – and you'd get the right answer again, including in the case above. You just can't define A in a path-dependent way, and then treat it as having the same value in all paths.

Re: Two envelopes problem

#126

Post by GeorgeRichard »

cedilla wrote: post #120
GeorgeRichard wrote: post #117

> Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Unless the imaginary “giver” shows you what was in the other imaginary envelope you’ll never know. We don’t even know if th…

It's a thought experiment. It's not about envelopes of cash, it's about logic and game theory. Their usefullness can hardly be overstated. But for what it's worth, there are a ton of thought experiments inspired by or lifted from the real world. The Monty hall problem is taken from a game show - it even has a contestant who must choose one of three doors...

Why is it a thought experiment? After all envelopes, cash, experimenters etcetera exist. It could be a real experiment. On the other hand what theory or hypothesis would it be testing?

What’s the problem with Monty Hall? The solution (switch to the other door) is demonstrably true. I believe it’s called a veridical paradox.

Re: Two envelopes problem

#127

Post by t_mann »

stefs wrote: post #111
t_mann wrote: post #90

Earlier quoted context omitted.

hmm, I don't really see where the probability of 50% comes from? even if we assume that we don't have any contextual knowledge about what amounts would typically be carried in wallets, the probability of winning would surely have to depend in some way on the amount currently in my wallet? in the extreme case, if I had zero cash, my probability of winning (not just drawing) would be zero, since we can't carry negative…

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

Re: Two envelopes problem

#128

Post by hammock »

lisper wrote: post #89
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…

The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.

Right away I was taken aback by the wikipedia article’s quick dive into mathematical solutions to the problem, when it seemed to me that what would be missing is consideration/framing of the initial setup.

That initial setup being, the act of starting the game with someone is(?) confers value in and of itself.

So I’m glad the real solution is wonky.

Re: Two envelopes problem

#129

Post by pontus »

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.

I wrote a pretty detailed analysis about this problem here in case you're interested:

https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa...

It all depends on how the envelopes are prepared.

Re: Two envelopes problem

#130

Post by phkahler »

civilized wrote: post #121

I'm going to present a critique similar to this one ( https://news.ycombinator.com/item?id=31568047 ) but I'll try to slightly rephrase. Suppose the amount in the envelope you choose is $100. If your amount is the smaller one, the other one has $200. If your amount is the larger one, the other one has $50. So the expectation supporting the switching argument is computed over two different scenarios, where the amounts…

So the issue is assigning the value A to the selected envelope and then doing conditional calculations for only the other envelop.
Post reply on HN