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[Ulfarr]

ln(k*x) = lnk + lnx 

I'm not sure if this makes it more intuitive but by multiplying x with k the result is always whatever value coresponds to lnx plus the same number (lnk) so if you plot both ln(k*x) and lnx in the same graph they should be "parallel" to each other.

Say for x2, x1:

[ln(k*x2) - ln(k*x1)] / (x2-x1) = [lnx2 - lnx1] / (x2-x1)

Since k is a constant it's ultimately irrelevant in the rate of how the function changes in regards to x.

[Vector]

I really like that explanation, Ulfarr.

[Ulfarr]

Thanks Vector 

[JoshuaFH]

I started watching a certain Vtuber Tabletop Game and they use a slightly more complex method of determining success or failure when rolling for things than DnD. They roll Xd10's, depending on the chosen skill that's being rolled for, and count the number of rolls that are 8 or greater, with additional rolls given afterward for every 10 that is rolled. Each roll that is 8 or greater is a Success, and you need a certain number of successes in order to determine how well you do in a given action, and that's how success or failure is determined.

I'm enjoying the video but I'm mentally hung-up on how the hell would you calculate the average number of Successes for any given number of d10's being rolled? Intuitively, it feels like 3d10's gives an average of ROUGHLY 1 success, 5d10's will give roughly 2 successes, 7d10's giving 3, and do so on for every additional 2 additional dice; but something tells me my intuition is wrong but I'm not sure how I'd write it out to determine it mathematically.

[Arx]

Unless I miss my guess, every roll gives you a 0.3 chance of a success and a 0.1 chance of getting another go. That next go has a 0.3 chance of success and a 0.1 chance of another go, and so it goes infinite.

That means we can write a sum for it, which (since our numbers are all <1) should converge and give us a non-infinite number.

My brain is too fogged to go to first principles right now (>_<) so I will just present without comment that since we're multiplying by 0.1 every time we have an event, the sum of probabilities of the events is

Code: [Select]

sigma (n to infinity) 0.1^n. Which we can multiply by 0.3, since we're counting up successes not the chance to go infinite, and we get

Code: [Select]

0.3 * (sigma (n to infinity) 0.1^n), and then we can add a 0.3 for our initial event.

Because convergent series was a long time in my past, I plugged it into WolframAlpha which tells me that every roll has a 1/3 chance of yielding a success. Assuming I did the maths right. And that more-or-less lines up with your intuition.

[WealthyRadish]

fakedit:
bleh, ninja'd, but I'll post anyway

It's easier if you just look at one die roll. You'll get a 2/10 chance of a basic success, and a 1/10 chance of an exploding roll that provides one success, another 2/10 chance basic success, and a 1/10 chance of another roll, onward forever. It looks like this:

0.2 + 0.1(1 + 0.2 + 0.1(1 + 0.2 + 0.1(1 + 0.2 + 0.1(...) ) ) )

or

0.2 + 0.1(1.2 + 0.1(1.2 + 0.1(...) ) )

The key observation of this expression is that the '1.2' term in each sub-expression is the only one that is actually getting added to the sum, so think about how you could express each '1.2' term's actual value in the sequence (if you were to multiply out the '0.1's).

Doing this, the kth term in the sequence can be written as 1.2 * 0.1^k. The entire expression is now:

0.2 + 1.2 * sum[k=1 to infinity](0.1^k)

The next key observation is that this is a form of the geometric series, which converges to a very simple formula (1/(1-r) where r is a fraction between 0 and 1). It should be noticed that the geometric series starts at k=0 instead of 1, which can be easily fixed...

(-1) + (1 + 0.2) + 1.2 * sum[k=1 to infinity](0.1^k)

-1 + 1.2 * sum[k=0 to infinity](0.1^k)

-1 + 1.2/(1 - 0.1)

... which simplifies to 1/3.

For n dice rolls, they're all independent events, so you just multiply by n, so the final average number of successes in this case is n/3.

[JoshuaFH]

Thanks, I appreciate it. The level of math is a little out of my ken, I'm not sure how the sigma works in equations, and I couldn't think how to write an equation that technically involves the possibility of infinite rolls.

[prefuzek]

Unless I miss my guess, every roll gives you a 0.3 chance of a success and a 0.1 chance of getting another go. That next go has a 0.3 chance of success and a 0.1 chance of another go, and so it goes infinite.

That means we can write a sum for it, which (since our numbers are all <1) should converge and give us a non-infinite number.

My brain is too fogged to go to first principles right now (>_<) so I will just present without comment that since we're multiplying by 0.1 every time we have an event, the sum of probabilities of the events is

Code: [Select]

sigma (n to infinity) 0.1^n. Which we can multiply by 0.3, since we're counting up successes not the chance to go infinite, and we get

Code: [Select]

0.3 * (sigma (n to infinity) 0.1^n), and then we can add a 0.3 for our initial event.

Because convergent series was a long time in my past, I plugged it into WolframAlpha which tells me that every roll has a 1/3 chance of yielding a success. Assuming I did the maths right. And that more-or-less lines up with your intuition.

Intuitively (without reaching for series formulas) this is just 0.3+0.03+0.003+ ..., which is 0.3333... or 1/3.

[chaotic skies]

Anyone know a better way to at least somewhat understand linear algebra than my horrifically boring professor lecturing for 45 minutes to an hour?

[Rockeater]

Can you be more specific, what you know and what you need?

[chaotic skies]

I'm taking a college course and the course is online so it's primarily videos of my professor lecturing for a while and writing the notes, with little explanation of what's really going on beyond 'this is just how it works'. I'm trying to find a way to understand what's actually happening, rather than just regurgitating formulas and theorems. Any format is fine, or even just a place to start - I'm not asking for a replacement for the class, just something to supplement it

[Quarque]

Watch the youtube videos of 3Blue1Brown on youtube on the subject. Clear explanations, crisp presentation.

[NJW2000]

There is a best way to learn mathematics: prove it and use it.

Assemble a list of every theorem in the course, and the definitions and assumptions used in that proof. Make sure it is in order.

1. Take the first theorem. Starting with those definitions and assumptions, try to prove each theorem. You can look at the proof the professor does, just make sure you're able to reproduce it independently.

2. Get hold of some questions requiring the theorem, or a proof that uses it. Attempt the questions, or follow the proof. If the proof is part of your course, this also counts towards step 1 for that proof!

Repeat for each theorem.


The problem? This is quite difficult, very slow, and unless you're doing questions specifically made for your course, may give you depths of intuition and understanding where you don't need it. Also, some proofs may be completely outside your professor's remit, and more deep or complex than the actual course. Also, while it's probably ok-ish for linear algebra, there are some areas of mathematics where even your professors won't really want you to bother learning the proofs. So some caveats here. But I don't think there is a more solid foundation than this.

[Vector]

@Chaotic Skies: 1. Feel free to ask your questions here (eg "wtf is an eigenvector") 2. Try Linear Algebra Done Right by Axler

[chaotic skies]

Working on homework (that's do tonight, I'm very responsible I promise). What the fuck does this even mean?