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

141–150 of 318 posts

Re: Two envelopes problem

#141

Post by shannifin »

wccrawford wrote: post #101
shannifin wrote: post #98

Earlier quoted context omitted.

> Many people interpret the envelope problem as being the same situation. It just seems obvious to me that it's not the same situation. That is, I think the "paradox" is ultimately about the temptation to treat these as similar problems in the first place.

Right, the difference here is that the calculation that determines the expected payout is the same as the calculation that determines how much it costs to switch. With the coin, the cost to play is set and is lower than the expected outcome.

I was thinking more along the lines of there being no need to actually calculate anything at all because the only relevant information is the 50-50 probability of choosing the higher envelope. And since you gain no new information after choosing, there's no need to calculate anything afterward either. The "paradox" is in making one think there's something to be calculated beyond the 50-50 chance.

Re: Two envelopes problem

#142

Post by kloch »

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 situations.

This is in contrast to the three-door problem where information is gained by opening the first door. https://en.wikipedia.org/wiki/Monty_Hall_problem

Re: Two envelopes problem

#143

Post by bottled_poe »

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…

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

Re: Two envelopes problem

#144

Post by jonath_laurent »

Some excellent critiques have been provided in this thread already but I would like to offer a different perspective.

In programming terms, the switching argument (see original link) is incorrect because it does not typecheck. And it does not typecheck because variable A is used out of its scope when writing down the (5/4)*A expected value.

Indeed, variable A is tied to a specific random outcome and so it simply does not make sense to refer to it in an expected value ranging over this same outcome.

Trying to formalize the argument in a proof assistant makes the mistake clear and obvious. However, the ambiguity of the English language makes it possible for such typing errors to sneak in undetected.

Re: Two envelopes problem

#145

Post by t_mann »

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…

There's an amusing amount of probability puzzles with seemingly paradoxical solutions out there, to the point that it's almost more surprising to see one where the intuitive answer is correct:

- https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

- https://en.wikipedia.org/wiki/Penney%27s_game

- Waiting time paradox

- Honorary mention: https://en.wikipedia.org/wiki/St._Petersburg_paradox

(feel free to add more)

Re: Two envelopes problem

#146

Post by CaptainNegative »

xigoi wrote: post #105
CaptainNegative wrote: post #36

Earlier quoted context omitted.

The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number. Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and…

But couldn't you make the same argument for guessing that it's higher iff the number is greater than 42? Therefore, if the number is 9, both strategies tell you to do different things, yet in each case, your probability of winning is >½. There must be a hole in the logic somewhere…

> But couldn't you make the same argument for guessing that it's higher iff the number is greater than 42?

Yep.

> Therefore, if the number is 9, both strategies tell you to do different things, yet in each case, your probability of winning is >½. There must be a hole in the logic somewhere…

For any nonnegative value C, the probability of Gaussian draw x being larger than (iid) Gaussian draw y is strictly bigger than 1/2 when conditioned solely on x > C.

The particular choice of C might change which specific draws of x and y the strategy succeeds with, but as you noticed for every C a thresholding strategy of the above form does give you some pointwise nonzero advantage over random guessing.

Re: Two envelopes problem

#147

Post by cosmojg »

Comparing this to the Monty Hall problem:

Probability of choosing the correct door at time of choice:

Two Envelopes: 1/2 -> 1/2

Monty Hall: 1/3 -> 1/2

...because Monty opens one of the doors, providing information.

Re: Two envelopes problem

#148

Post by shp0ngle »

hutzlibu wrote: post #140
olalonde wrote: post #133

Earlier quoted context omitted.

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 informatio…

That means that no solution to the paradox is definitive, but also there is no argument.

Switching back and forth obviously doesn’t have any positive. What they mean by “is not definitive” is what explanation is the definitive one.

Re: Two envelopes problem

#149

Post by jonath_laurent »

mike_hock wrote: post #64
dllthomas wrote: post #17

Earlier quoted context omitted.

I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A . But with any actual distribution (... I think?) it does , at which point... no paradox. It's true that it's not clear (at least to me , but perhaps more generally) what distributi…

No, joe_the_user is exactly correct. The expectation they compute in step 7 requires that A is a constant. The formula seems to just be the regular formula for expectation: Value that B can take times probability that it takes that value, summed over all possible values. But the "possible values" they put in are A/2 and 2A, which makes no sense. A is a random variable. Don't forget that random variables are really fu…

This comment explains the fundamental reason why the reasoning is incorrect. I offer the same perspective in a different comment (https://news.ycombinator.com/item?id=31569991).

In my opinion, the Wikipedia article is making a disservice to readers by mentioning unnecessarily complex mathematical arguments involving bayesian reasoning and infinite distributions. I believe all of these are distractions from the more fundamental typing error that invalidates the switching argument.

Post reply on HN