201–210 of 318 posts
The Monty Hall problem has a simple resolution. At the beginning you had a 1/3 chance of picking the good door, and 2/3 chance of picking a bad door, meaning you probably picked a bad door at the start, so you should switch.
Earlier quoted context omitted.
Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch". But this omits (loses) the information of " both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the…
The right resolution is: suppose the two amounts are $50 and $100. If the envelope you have contains $50 and you switch, your gain is $50. If it contains $100 and you switch, your gain is -$50. So your expected gain from switching is (0.5)(50) + (0.5)(-50) which is zero. The way that they fool you is by saying that you either double your money or you halve it, leading you to think that your expected ratio is 2.5/2. B…
Earlier quoted context omitted.
Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)
Who said there is a uniform distribution on the natural numbers?
Imagine we have a distribution that satisfies that assumption, and then someone tells you they've sampled from that distribution and found that the result is of the form (say) 7*2^k, for some k > 0. That conditional distribution for k would seem to have to be uniform, right?
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 actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it . Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This ha…
Earlier quoted context omitted.
This is incorrect. Bullet point 7 in the written formula does not refer to a corresponding pair of envelopes, as you imply. Instead, it refers to "the other" envelope only: 1/2 chance that "the other" envelope contains 2A, and 1/2 chance that "the other" envelope contains A/2. Notice that both references reference the same object.
That precisely what's wrong, because A is not the same quantity in both "A" part of the equations. The right formula is changing either the 0.5A by A or the 2A by A. Try with A being a real like 100 usd numbers to see it. Basically you can't have "two realities" in the same equation. U havé to pick one.
No it's not, that's not how you make EV calculations. As I already said in my previous post, the parts in the EV calculation do NOT refer to the pair of envelopes. Both parts refer to the same object, not to different objects. You are trying to change the EV calculation such that the different parts refer to different envelopes. This does not make any sense.
> Try with A being a real like 100 usd numbers to see it.
...and that's not a way to prove / demonstrate the correctness of math. Yes, if you change these numbers in this way, you get EV 0, which is the correct answer. Just because the answer is correct does not mean the calculation is correct. In this case it isn't.
I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained math…
The claim is that no solution is accepted as definitive, which is an empirically true fact about the academic literature. To respond to this with "no, my favored solution is definitive" is uninteresting.
There is no such thing in mathematics such as an unresolved dispute over whether or not a five line proof with elementary concepts is flawed or not. The only possible situations are: 1) the proof can be checked to be correct at the level of axioms, 2) an incorrect reasoning step can be pointed to or 3) the proof is ambiguous and/or people cannot agree on what is being proved.
There isn't really a mathematical literature about the two envelopes paradox because the paradox is not really interesting from a mathematical standpoint. Or at least, making it interesting from a mathematical standpoint would require presenting a different argument than the one presented in the Wikipedia article. There may be a scholarly debate about different philosophical or linguistic aspects of this paradox. However, there is certainly no debate about where the flaw is in the presented switching argument once you make it precise enough.
(And the answer should not involve infinite series unless you are looking at a different, strong-arm version of the argument.)
I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained math…
Earlier quoted context omitted.
There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it . Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This ha…
Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.
That being said, even if you let k go negative I think it's still a paradox.
If you're curious the resolution to this is that before you look in the envelope the expected value of both envelopes is infinite. Only after you look inside one of them does that collapse to two finite values that you can subtract.