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Or the "-1" simply addresses the fact a tied 'roll pair' has no variance between the results? That factor deals with the diagonal line of matching/tied rolls?
Yes. The important thing to understand is that for counting the number of different possible results, the dice are treated seperately, so e.g. a [4,1] and a [1,4] are two dice combinations. And as you correctly said, there's only one occurance of [4,4].
I think it's easiest to explain by example. Let's say we want to know the number of different combinations that result in a on a d6. We need at least one die rolling a , and the other not rolling higher. So the combinations are 1) die A rolling a and die B rolling any of the lower results (, , or ), 2) die B rolling a and die A rolling any of the lower results, and 3) both rolling a . The number of lower results are, of course, the target result minus one (in the example case 3), so either of the first two has a number of combinations that's x-1, with the third combinations being unique. To get the sum of the three possibilities how to get to the result, we add them together: (x-1) + (x-1) + (1). For the example, that's 3+3+1=7. If you look at MrCharisma's picture, and count the number of white dice that say , you'll see it's indeed 7!
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Ummmm . . . .
I used MrC's "exploded view" mathematics to run an analysis on 1d8 w/ advantage and got 5.81.
And then I plugged "8" into Derklord's equation . . . and wound up with negative fractions? WHUT?? Are there some parentheses/brackets that are left out of it?
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Ummmm . . . .
I used MrC's "exploded view" mathematics to run an analysis on 1d8 w/ advantage and got 5.81.
And then I plugged "8" into Derklord's equation . . . and wound up with negative fractions? WHUT?? Are there some parentheses/brackets that are left out of it?
The actual number should be 5.8125, but I'm assuming that's what you got and you were just rounding.
Now as to Derklord's equation, I assume you just got your order of operations incorrect.
(n+1)*(2/3*n-1/6)/n
So the inside/outside brackets I assume you know, but remember that multiplication and division happen before asdition and subtraction, so if we pretend you don't know the order of operations and fully add brackets everywhere it becomes: (n+1) × ([2/3 × n] - 1/6) ÷ n
The square brackets are probably the ones you missed, giving you (2/3) × (n- 1/6) for that part of the equation, rather than (2/3 × n) - 1/6. Now for mathematicians we know that those square brackets are implied by the order of operations, but as a self-professed "math-hating humanities student" it's reasonable to assume you forgot that.
So if we plug some numbers into that equation again:
D6: (6+1) × ([2/3 × 6] - 1/6) ÷ 6
= (6+1) × ([4] - 1/6) ÷ 6
= (7) × (3 5/6) ÷ 6
= 4.472222222...
D8: (8+1) × ([2/3 × 8] - 1/6) ÷ 8
= (8+1) × ([5 1/3] - 1/6) ÷ 8
= (9) × (5 1/6) ÷ 8
= 5.8125
D20: (20+1) × ([2/3 × 20] - 1/6) ÷ 20
...
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I used the incomplete equation, for one thing.
Y'know, math at 3 AM ain't really my thing, eh? lol
But, the square brackets did make a difference.
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RE: "math-hating humanities student"
I come from a strong religious/spiritual background, and the wife of one of our elders was my >ONLY< math instructor from 9th grade through 12th grade. My discipline in the "generally accepted standards of respect" (so to speak) was NOT sufficient to prevent me from wearing my
3. Find x.
There it is ------> x
. . . t-shirt in front of her on a regular basis.
Maybe that helps illustrate how keenly I meant it.
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Yes -- "implied" operations are of no use to me. I require explicit instructions.
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I'm reposting this one because I would like to get a tiny bit of feedback on it.
The best of two dice will be a bit over 70% of the max.
Obviously this is just an approximation, but it can save time and is pretty close.
(Because 2^0.5 / 2 = 0.71…)
Help me break down the steps, here. I'll give it a shot.
"Two raised to the power of .5" represents the two rolls, each averaged in value?
And then you divide that result by two . . . to represent the fact that you can only choose one of those two results?
So, by extension, if you were to roll three matching dice and take the best result of the three, would it look like this?
3^0.5 / 3 =
86.6% of the maximum result?
Application
So the equation to determine the average result of rolling 1d8 with advantage would be . . . .
8 * (2^0.5/2) ?
Resulting in . . . 5.66 on average?
1) Is there a reason why that root equation ( 2^0.5/2 ) is inaccurate in comparison to Derklord's . . . bulkier composition?
2) Is there any mathematical soundness in that equation such that it can be applied to the questions I'm asking, and that it be "responsible"?
3) IF we were to assume that (2^0.5/2) was applicable to my questions concerning 'advantageous averages', was my expansion to include three rolls correct?
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Disclaimer: The formula in my second post is wrong, it's supposed to say 2/3*n-1/(6*n)+1/2 (it had -1/2). Also, I should stick to lowercase n, to differentiate from the unit of force.
But, the square brackets did make a difference.
And here I wasn't sure if I should include the the redundant multiplication signs*... good thing I didn't! The multiplications sings are necessary for spreadsheets anyway, though.
*) Whenever there's no mathematical sign between two different parts, e.g. "2x" or "(x+1)(x-1)", a multiplication sign is implied, so that's the same as "2*x" and "(x+1)*(x-1)", respectively.
But yeah, brackets before exponents before multiplication and division before addition and subtraction. There were multiple posts about order of operation in 2015, so I assumed that to be existing knowledge. And to put it bluntly, that's a rule that every one should learn and memorize, as it's one of the most basic principles of (writing) math.
1) Is there a reason why that root equation ( 2^0.5/2 ) is inaccurate in comparison to Derklord's . . . bulkier composition?
Er, there's no root there. Also, the way i see it it's just an approximation that happens to be close, and isn't based on the actual mechanics.
2) Is there any mathematical soundness in that equation such that it can be applied to the questions I'm asking, and that it be "responsible"?
It's an approximation, not a calculation. For dice from 2-20, the divergence from the actual value is pretty small, between -0.34 and +0.32. For an advantaged d100 you'd be off by over 3.5, though.
My above posts do include the necessary tools to expand it to 'best out of three dice', although finding the equation to express the number of different combinations (to replace the 2*x-1) is non-trivial, and you'd need Wolframalpha (or learn that part seperately) to transform the the summation into a single formula.
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There is no actual roots, but they are probably talking about Square Roots. Which any exponent that is a fraction by definition is the nth root.
Oh, right, forgot about that. It's been a while.
Weird for the OP to know that but not the order of operations...
Wow. Lots of action. I'll try to get caught back up.
1) Concerning the use of the word "root" in "root equation":
I wasn't referring to square roots, or any mathematical function called by the designator "root". I was using it in the literary sense, as in "source" or "beginning".
2) Order of Operations:
My refresher on PEMDAS was supplied nearly six years ago in this very thread. I may not do much mathematical work, but I do recall that content.
But, given the fact that I'm a humanities student, the difficulty I experienced earlier with the equation you supplied (and later corrected), derklord, was that even though it was in bold font, it was incomplete. I focused on what "leapt off the page", rather than the whole of what you stated on the point.
You emphasized the part of the equation that needed the most working through, and I focused on only a fragment of what was pertinent to the question(s) I am asking. Hence my errant efforts.
3) An additional note on implied operations:
I do remember enough mathematics that I know where the * is implied, but since the spreadsheets need the explication, it certainly helps to keep me focused on what I'll need to plug into the cells once I've assimilated this process.
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2/3*n-1/(6*n)+1/2
Or have I missed something again?
Because if I'm examining the rolls of a d8, and am using 8 as n, then . . .
2/24 - 1/48 + 1/2 --> 4/48 - 1/48 + 24/48 --> 19/48 --> .3958333....
How is that number relevant again?
Wait. Nevermind. Same thing as before -- that's not the whole equation. Right?
(n+1) × ([2/3 × n] - 1/6) ÷ n
9 x (5 1/3 - 1/6) / 8 --> 9 x (5 1/6) / 8 --> 9 x 5.17 / 8 --> 46.53 / 8 --> 5.81
NOTE: 5.81625, but that is due to rounding the conversion of 5 1/6 to a decimal value to the hundredths.
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And I've got the handwritten "exploded d8" work in front of me that led me to the same result of 5.81. I know I can work through any simple advantage roll equation with that version.
I hope you're prepared to beat a dead horse.
I don't see how the "exploded view" process correlates to the equation that Derklord has spelled out. I don't even know if I can get to that level of comprehension, but I'd like to try. Can either of you walk me through it again?
Again, and as always: many thanks for your efforts.
This is the third daggum time the boards have eaten my reply . . . .
I'm getting more than a little upset.
I don't see how the "exploded view" process correlates to the equation that Derklord has spelled out. I don't even know if I can get to that level of comprehension, but I'd like to try. Can either of you walk me through it again?
If I'm completely honest I didn't follow exactly how Derklord's equation got related to the numbers either, but I checked it and it works, and I could at least help you work out why you were having trouble with it.
Having said that: If all you need is an equation to find the average roll then Derklord's equation does that perfectly
I've never had any of those PEMDAS-type mnemonics, but generally anything inside brackets is 1 number, so they get resolved before they interact with anything outside the brackets. Then ×÷ are resolved before +-
The only other thing to do if you're less practiced is to go through and manually insert × symbols betwern any 2 numbers that don't have a symbol between them . If it helps you can also go through and manually draw in brackets around multiplied numbers to keep everything in track . Drawing those extra symbols in can be really helpful if you don't do this on a regular basis.
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So as to how my "expanded view" relates to Deklord's equation, again I'm not sure. The point of my expanded view is to help you actually see the numbers so that you can manipulate them if you need to.
Once again I want you to go look at THIS PICTURE .
This picture shows us all the possible results of rolling 2d6 and picking the highest roll. But you can see patterns if you look. The easiest pattern I already told you:
1chance of rolling a 1
3 chances of rolling a 2
5 chances of rolling a 3
7 chances of rolling a 4
9 chances of rolling a 5
11 chances of rolling a 6
So the numbers on the left there are all odd numbers, starting at 1 and increasing by 2 every time. That's interesting, and it'll help us work out how many times a number is likely to be rolled on larger dice , but besides that it doesn't help us all that much.
The next pattern I want you to notice though, is Squares.
The chance of rolling a 1 or lower is 1, or 1 squared
The chance of rolling a 2 or lower is 4, or 2 squared
The chance of rolling a 3 or lower is 9, or 3 squared
The chance of rolling a 4 or lower is 16, or 4 squared
The chance of rolling a 5 or lower is 25, or 5 squared
The chance of rolling a 6 or lower is 36, or 6 squared
Now we know these are all square numbers because if you look at THE PICTURE they're all squares. They all take up a square amount of space.
This is important and much more useful information. First of all, every number you roll is a fraction of a square - specifically a large square minus a small square. If you want to know the odds of rolling a 3 or higher on the dice you just take the big square and minus the numbers below a 3.
So 6^2 - (3-1)^2
= 6^2 - (2)^2
= 36 - 4
= 32
Now there are 2 useful things we can do with this information.
1. We can work out the odds of rolling with DISadvantage, using the same numbers but reversed:
The chance of rolling a 1 6 or lower higher is 1, or 1 squared
The chance of rolling a 2 5 or lower higher is 4, or 2 squared
The chance of rolling a 3 4 or lower higher is 9, or 3 squared
The chance of rolling a 4 3 or lower higher is 16, or 4 squared
The chance of rolling a 5 2 or lower higher is 25, or 5 squared
The chance of rolling a 6 1 or lower higher is 36, or 6 squared
That's not necessarily super-relevant right now, but it's good to know anything we do with these numbers can be done both ways.
2. You can work out hit-chances, crit-chances, fumble-chances, etc.
If I know I need an 18+ to get a critical threat, then that's a 3/20 chance on a d20 . If I roll with advantage that's 111/400 . Here's how I worked that out:
To work out how often I roll 18 or higher, I just take the big square and minus the little square , and then divide the number by the total amount .
( 20^2 - [18-1]^2 ) ÷ 20^2
( 20^2 - [17]^2 ) ÷ 20^2
( 400 - 289 ) ÷ 400
( 111 ) ÷ 400
= 0.2775
So again, that might look complicated, but literally all I did was:
[BIG SQUARE - little square] ÷ BIG SQUARE.
If you want a "ROOT" of the equation, it's SQUARES.
With a single die the pattern is 1-dimensional, and we get a line.
With 2 dice pick the highest the pattern is 2-dimensional, and we get a square.
With 3 dice pick the highest the pattern is 3-dimensional, and we get - you guessed it - a cube.
If we look again at our d6's, but instead roll 3 and pick the highest ...
The chance of rolling a 1 or lower is 1, or 1 cubed
The chance of rolling a 2 or lower is 8, or 2 cubed
The chance of rolling a 3 or lower is 27, or 3 cubed
The chance of rolling a 4 or lower is 64, or 4 cubed
The chance of rolling a 5 or lower is 125, or 5 cubed
The chance of rolling a 6 or lower is 216, or 6 cubed
.
the difficulty I experienced earlier with the equation you supplied (and later corrected), derklord, was that even though it was in bold font, it was incomplete.
No, the bolded equation is complete*, you just misread it. For mathematical operations on the same order of operation, as multiplication and division are, you read them from left to right, so for 2/3*n, the division is done before the multiplication. In other words, 2/3*n is (2/3)*n rather than 2/(3*n), as you did it. You can see that in the second part, where the the n is multiplied with the divisor, I wrote it in brackets.
If you replace the n with a cell designation, the formula 2/3*n-1/(6*n)+1/2 is ready to be put into a spreadsheet as it is. For example, for a die value in A1, the formula would be =2/3*A1-1/(6*A1)+1/2.
*) Apart from my mistake with switching a + for a -, of course. Note that despite written slightly different, the two equations in my first and second post are functionally identical, as is MrCharisma's version with the square brackets.
I don't see how the "exploded view" process correlates to the equation that Derklord has spelled out.
I have no idea what you mean with "exploded view".
This is the third daggum time the boards have eaten my reply . . . .
ctrl+a ctrl+c all your posts before sending them, that's how I protect myself from such issues. Edits, too.
Edit:
If I'm completely honest I didn't follow exactly how Derklord's equation got related to the numbers either
Which step(s)?
The "exploded view" is my term for the d6 graphic that MrCharisma has so often and so helpfully linked. :)
Er, did you read my first post on this page?
If I'm completely honest I didn't follow exactly how Derklord's equation got related to the numbers eitherWhich step(s)?
Well I just hadn't fully read through everything you wrote as it was a bit dense, but in reading through it I got through the following fine ...
Well, It's a bit complicated, but I'll try to explain.
There're three different parts to the base term.
A) The number of different die combinations that result in a specific end value is always twice the value minus one. The picture MrCharisma linked shows that fairy well, but I can try to explain it further if it's unclear.
B) The number of different possible numbers (N) are always equal to the sides of the dice. This should be self-explanatory.
C) The total number of combinations is always the dice maximum value squared (N^2). I hope this is self-explanatory, too.
To get the average, first, we write out the above part A as a mathematic term, for a given number x: 2*x-1.
Next, since we care about the value rather than the chance for a given result, we multiply this by the value of the given number: x*(2*x-1).
Third, as we care about all possible results, we do the above for all the different possible values for x, i.e. six times for a six-sided die, and add them all together.
Fourth, to get the average, divide the sum produced in the last step by the number of possible results, i.e. N^2.The above lets you do it by hand, but an eight sided dice would take nine different calculations. Thankfully, it's possible to put the different terms together into a single term by using summation, by using an index number instead of a regular variable: sum(i*(2*i-1)), where i ranges from 1 to N. Now all we need to add in the division by N^2 to finish the formula: The average value of 2dN take higher is sum(i*(2*i-1))/N^2.
But then had trouble with the sudden introduction of this equation ...
The sum term can be spelled out, although I honestly forgot how to do that by hand, as 2/3*N^3+N^2/2-N/6 ...
I'm not 100% sure where a lot of this comes from.
Did you click the link? I used the site I linked to transform the summation formula into a closed-form expression, because spreadsheets can't handle summations of variable length. It is possible to do it by hand, and if you use the standard formuae for ∑i and ∑i², not actually hard to do. I can do a step-by-step process later, but I'll wait for Syrus to catch up to that point, else it will only cause confusion, I think. Alternatively, click this link and learn it yourself.
Did you click the link?
Honestly I didn't even notice the links until I copied that post to reply last night .
I also think using more complex mathematical equations like that one isn't necessarily helpful in explaining the maths to someone who's less mathematically-minded. If you have to explain what a symbol means it's probably too complex for this kind of explanation. If you want to put it in as a formula that can be used that's fine, but using it as an explanation is going go require a lot more explanation.
Suffice to say - that equation does work, but the explanation of how we get from odd numbers and square patterns to 2/3*N^3+N^2/2-N/6 isn't really explanatory.
(NOTE: This is why in my posts I always post links in CAPITAL LETTERS, it helps the link be a little more obvious for the reader)
I know, but I think capitalizing links looks ugly (from a typesetting viewpoint) and I'd have thought it redundant. Maybe it has something to do with monitor settings and background light level or whatever, but I never had an issue seeing a link in text - they are in a different color, after all.
Or maybe I just read posts more carefully? *shrugs*
I also think using more complex mathematical equations like that one isn't necessarily helpful in explaining the maths to someone who's less mathematically-minded.
But unless I'm missing something it's the only way to do it. Your stuff for basically doing it all by hand only works for a specific die value you want to calculate for, but as far as I understand it, this is for a spreadsheet, where I presume different dice are fillable in. And for that, the partial sum that forms the central part of the formula must have a variable length - which means we need the capital-sigma expression.
the explanation of how we get from odd numbers and square patterns to 2/3*N^3+N^2/2-N/6 isn't really explanatory.
There is no explanation yet. I didn't do the transformation the first time (instead using the linked site), and I don't want to explain it until Syrus Terrigan understands the previous steps and also arrives at the "sum(i*(2*i-1))" part.
" . . . and on the eighth day, Satan said 'Now we're going to put the alphabet into math.' And there was wailing and gnashing of teeth."
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Waste no more of your time, fellas. I think I've gotten all I'm going to get.
Thanks for your patience.
The last thing to go with, If all else fails, coming from someone who's dealt with it for a long long time, is just not to worry about it. Big number is usually better when it comes to things involving damage and that's only as good as its efficiently applied. Which is the simulation versus boots on the ground conclusion I ended up with in the end.
So yeah, if you dont understand the math, but do understand basic strategy and tactics and swing a big enough axe to do the job, you're probably still better off thab many that lack one or both.
Understanding the math is useful from an analysis standpoint, but limited from a practical one. So don't sweat too hard.
Analysis is a centerpiece of game development. I simply rezzed this thread because math was the focus; the goal has changed, though. Tactics are easy; abandoning the bland comfortableness of d20 mechanics isn't (for me, anyway).