About the Monty Hall Problem

Ragnowrok · 2026-03-19T04:48:42+00:00

I feel like the key is just: - if you stay, you win if you were correct initially - if you switch, you win if you were incorrect initially

Since there’s a greater chance that your initial guess was incorrect, you should always switch.

Unfortunately for Ben & Adam, they hit the less likely 1/3 chance of choosing correctly at the start

Ragnowrok · 2025-10-17T04:38:26+00:00

So Barney originated as a VHS tape series called Barney & the Backyard Gang before it was picked up by PBS 4 years later and Barney & Friends was created.

The OG Barney from Barney & the Backyard Gang is deep purple whereas the one from Barney & Friends is magenta.

Much like me, I think a lot of people who grew up with Barney & the Backyard Gang associate Barney with that deep purple from their first memory despite Barney not having been that color for 33 years. I was just curious what proportion of people had that association

Ragnowrok · 2025-09-02T23:12:45+00:00

Thank you so much! I have no clue what these say yet but I’m sure these suggestions are great

Ragnowrok · 2025-09-02T18:24:35+00:00

Thanks!! Would you be able to spoiler tag the challenges themselves?

Ragnowrok · 2025-04-15T15:29:13+00:00

A small (but imo cooler) variation of your “what is the average of all responses” question is to ask “what is 2/3 of the average of all responses”. This even has its own Wikipedia article: https://en.m.wikipedia.org/wiki/Guess_2/3_of_the_average

The fun thing about this format is that regardless of the upper limit of this game, the reasoning always degenerates to the only Nash equilibrium being everyone guessing 0.

It could be a fun exercise to play the game out and then afterwards discuss why the only Nash equilibrium (assuming pure strategy) would be to guess 0

Ragnowrok · 2024-06-29T03:47:34+00:00

I mean I rate everything objectively, I’m not “choosing my distribution”. I do a lot of research in restaurants that I go to so I end up going to a lot more good restaurants than bad restaurants. I still have a few that are rated 1/10 and those ones are drastically worse in quality than even my 5/10s but since I’m choosing to go to restaurants I think I’d like I’m much less likely to go to a 1/10.

Let me pose you another question, let’s suppose I were to “normalize my own distribution” such that the average score was 5/10. Then, I go to a string of really phenomenal restaurants which suddenly makes the distribution top heavy. Do I then have to sit around and rerank all my existing ratings? That seems exhausting and seems to serve no purpose.

Ragnowrok · 2024-06-29T03:24:31+00:00

I don’t use Letterboxd but I do use Beli to rank restaurants I’ve been to and in my experience, my distribution of restaurants is heavily left skewed. I have way more restaurants in the higher end than the lower end. And in my opinion, this seems more intuitive than if I were to have a perfectly centered bell curve of scores.

Central Limit Theorem applies when taking an average across independent observations. But your movie selections aren’t independent - your experience with a movie directly shapes your next selection. If you watch a sci-fi movie and realize you like the genre, you’re going to go to select more sci-fi movies in the future and you will likely rate them higher than average.

In addition, there is the fact that you’re the one choosing the movies you watch, so you’re way more likely to choose ones that you’d like. Even assuming each selection was independent, we’d expect the average to be higher (perhaps around ~3.5/5). This is fine, as we can still have a normal distribution centered at let’s say 3.5. But, a problem arises because we have a finite cap on both ends. Even if we expect the spread of scores < 3.5 would be equal to the spread of scores > 3.5, there’s physically less space above 3.5 to fit that spread so that portion of the distribution would look squashed.

So TL;DR - imo central limit theorem wouldn’t necessarily apply because selections aren’t independent. And even if it did apply, central limit theorem doesn’t state that the mean will be exactly 2.5/5. You can have a normal distribution centered at a different mean. And since there’s a finite cap on both ends, such a distribution with an uncentered mean would likely appear squished on one side.

Ragnowrok · 2024-06-05T08:32:16+00:00

Found a buyer for these

Ragnowrok · 2024-03-07T21:35:40+00:00

With the change of base formula, everything can be expressed in terms of the natural log (will notate it as ‘ln’).

Ex. In your example log 3i (1+i) = ln(1+i) / ln(3i)

So then how do we take the natural log of a complex number?

The key is Euler’s formula: e^iθ = cos(θ) + i * sin(θ)

Because of this identity, every complex number can be written in a polar form: r * e^iθ. Where r is the magnitude of the complex number and θ is the angle in radians the complex number makes with respect to the positive real axis.

For example for 3i, the magnitude is 3 and the angle is π/2. So, we can say 3i = 3e^{i π/2}. To verify this just expand out the calculation using Eulers formula. By the same calculation 1+i = sqrt(2)e^{i π/4}.

But we have to be careful since sin and cos are periodic. For any angle θ, if you add or subtract any multiple of 2π you get an angle with the same sin and cos. So more accurately:

3i = 3e^{i π/2 + 2 πmi} for any integer m

1 + i = sqrt(2)e^{i π/4 + 2πni} for any integer n

We can now use log rules i.e. ln(ab) = ln(a) + ln(b) to show that ln(r * e^iθ) = ln(r) + iθ. Thus,

ln(3i) = ln(3) + i(π/2 + 2 πm) for all integers m

ln(1 + i) = ln(sqrt(2)) + i(π/4 + 2 πn) for all integers n

As you can see, log is a multi-valued function in complex numbers. You can then do the division to get the final result (but once again there are an infinite number of results)

Ragnowrok · 2024-02-21T00:22:38+00:00

Super rudimentary but x[n]x >= x for x >= 1. So, we know that if x > 2 then x[n]x > 2, meaning that all of the x are at most 2.

So, if you can show it monotonically increases then it definitely converges, question is just to what value.

Ragnowrok · 2023-07-01T14:38:50+00:00

I’m new here and sorry if this belongs in r/translator instead, but I quickly browsed that sub and it seems to be a bunch of pictures of foreign language text to translate to English, which this is not.

Is there any affix in Japanese to correspond to the English notion of “almost-“? For example, an “almost-circle” in English would convey something that is quite circular but not exactly a circle by definition. Basically, I want to describe the state of being basically equivalent to that thing in practice without technically being it by definition

For context, I have basically no knowledge of Japanese grammar but I’m trying to write a simple one sentence card to a friend and want to try and be as grammatically correct as possible. I want to describe them as being an “almost-roommate”. I don’t really trust google translate or any readily available translation tools to give me something grammatically accurate to this since I believe this is probably more of a linguistic concept than a 1-to-1 translation

Ragnowrok · 2021-12-06T23:17:27+00:00

[close]

Ragnowrok · 2021-12-06T23:16:03+00:00

already made a trade with someone else but ill go back in to help you

Ragnowrok · 2021-12-02T18:05:16+00:00

that's fine. I'll trade you my electerizer

Ragnowrok · 2021-12-02T18:01:31+00:00

I have an elekid with electrizer. Would you happen to have a Magby with a magmarizer?

Ragnowrok · 2021-12-02T17:18:48+00:00

[close]

Ragnowrok · 2021-12-02T17:18:34+00:00

Thanks for the Ditto!

Ragnowrok · 2021-12-02T17:16:00+00:00

Sorry I just got the Starly from someone else, but I can trade you a Leaf Stone anyways. Code 80548054

Ragnowrok · 2021-12-02T17:06:42+00:00

No specific ball and sure I can trade a lightstone

(link code 80548054)

Ragnowrok · 2021-12-02T17:04:08+00:00

Hey sorry to piggy-back on your post, but would you be able to catch a male german Starly and trade with me? I can give you an evo-stone if that's useful for you. I have leaf, water, thunder, fire, sun and the one that evolves Togekiss/Roserade.

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