Maths!

[codeGlaze]

Just a quick question, more or less to educate myself.

Each side of a d20 has a 5% chance of landing up.
So if you land a 1 or 20, would a SECOND (confirmation) roll aiming for a 17-20 (or 1-4) effectively bring the chance of a "crit" to 1%?

Or does that second roll skew the probability completely?

edit: grammar

[ishy]

Each side of a d20 has a 5% chance of landing up.
So if you land a 1 or 20, would a SECOND (confirmation) roll aiming for a 17-20 (or 1-4) effectively bring the chance of a "crit" to 1%?

Or does that second roll skew the probability completely?

edit: grammar

17-20 is 20%, [5% * 20% = 1%]

[Ancient History]

Here's how you look at it: if the rolls are independent of each other, the probabilities are added. If the rolls are dependent on each other, the probabilities are multiplied.

The probability to roll a 20 on a d20 is 1/20 = 0.05

If you roll two d20s, the chance that one of them will roll a 20 is 1/20 + 1/20 = 0.1

If you roll two d20s, the chance that both of them will roll 20 is 1/20 * 1/20 = 0.0025

The probability to roll a 17, 18, 19, or 20 on a d20 is 4/20 = 0.2

So if you roll two d20s, the chance that one will roll 20 (0.1) and one will roll 17-20 (0.2) is 0.05 * 0.2 = 0.01

So you have a 1% chance to roll a crit.

Of course, not all dice are made very well, so the odds of rolling a 20 or a 1 are actually usually less than 5% on average.

[codeGlaze]

Thank you very much, Ancient.

Probability was one of the areas of math that has always blurred together in my mind's eye. (Always seeming more complex than it is.

So thank you, again. Especially for running it out like that.

[Koumei]

If you roll two d20s, the chance that one of them will roll a 20 is 1/20 + 1/20 = 0.1

Not quite. The chance is actually 100%, minus "the chance of you rolling 1-19 on the first, multiplied by the chance of you rolling 1-19 on the second". Otherwise you'd be guaranteed, rather than likely, to roll a natural 20 on 20d20.

So it's 100%-(95%^2)
=100%-90.25%
=9.75%

Which is almost 0.1 but not quite.

[Whatever]

An even more nitpicky nitpick. Koumei has the odds of rolling at least one 20 when you roll two d20s. The odds of rolling exactly one 20 on two dice is 9.5%.

[Ancient History]

You are right, in my haste I neglected my maths. Serves me right for being a boob.

Naturally of course, you can see how some of the crit-modifiers affect this - opening up the threat range, allowing re-rolls, etc. can significantly improve the odds of a crit. But "significantly" is still often less than, say, 10%.

[codeGlaze]

If you roll two d20s, the chance that one of them will roll a 20 is 1/20 + 1/20 = 0.1

Not quite. The chance is actually 100%, minus "the chance of you rolling 1-19 on the first, multiplied by the chance of you rolling 1-19 on the second". Otherwise you'd be guaranteed, rather than likely, to roll a natural 20 on 20d20.

So it's 100%-(95%^2)
=100%-90.25%
=9.75%

Which is almost 0.1 but not quite.

Wait... does this cover only rolling the second die if the first die meets a criteria? (Such as rolling a 20 or a 1)

The initial thought process was to parse down a 5% chance to a 1% chance. While still keeping d20s (so basically a spin on crit confirm). Stemming from a thought experiment to reward a person for actually scoring in that 1% range.

And are you using 0.1 (in your final sentence) on a 100.0, 10.0: 100, or 1.0 : 100 scale?

[Koumei]

The chance of you rolling a 20 on 1d20 is 5%.
Therefore the chance of you not doing so is 95%.

So, having rolled a 20, your chance of rolling another twenty is 5% - for a total chance of 5%*5% = 0.25%

On the other hand, if the first die comes up as not twenty, you still have a 5% chance of rolling a twenty on the second, and a 95% chance of rolling "anything else". So your total chance of rolling "not twenty" (for both dice) is 95%*95% = 90.25%, and your total chance of rolling "One or more twenties" (for both dice, not caring which comes up 20) is 9.75%

Now if you only care about their second roll if the first roll is a twenty, you can ignore the above paragraph because it assumes you rolled a 1-19. In that case, just look at the one above, where a 20 is 5%, and if you wanted 1% you'd basically go "You need a 20, then a 17+" (20%*5% = 1%).

By 0.1 I meant "where 1.0 is 100%" so "almost 0.1" meant "almost 10%". I should have been clearer, given I was switching between percentages and decimals.

[codeGlaze]

Okay, so then my initial math was right.

With the second die only being cast if the first result is a natural 20 as long as the second d20 roll is a range (17-20), which I posited earlier, would allow for a 1% "filter" so to speak.

Thanks a bunch, guys.

[Koumei]

Here's my math question: who the fuck decided that x^0 = 1, and why did everyone go along with this? How does it work? For the main part is seems entirely theoretical anyway, and the only thing it works nicely for is 10^x, where it's 1, followed by x zeroes.

[TarkisFlux]

Here's my math question: who the fuck decided that x^0 = 1, and why did everyone go along with this? How does it work? For the main part is seems entirely theoretical anyway, and the only thing it works nicely for is 10^x, where it's 1, followed by x zeroes.

I'm pretty sure it's a limit thing. X^(1/Y) is the Yth root of X (so X^(1/2) == SqRt(X) ). Do some limit magic and send Y -> infinity, and you can see that 1/Y -> 0 and the Yth root of X -> 1. Thus X^0 = 1.

There may be another reason why this is the case, but it's not coming to me off hand.

[Edit] Since negative exponents basically throw the thing into the denominator (i.e. X^(-12) == 1/(X^12) ), setting X^0 = 1 also patches the function nicely for X^Y, where you hold X constant and vary Y from positive infinity to negative infinity. Which is probably a side effect of the definition and not a reason for it, but still.

[name_here]

Probably because X^n * X^-n = X*(1/X) = 1 and X^(n + (-n))

Multiplicative reciprocals equal one, and multiplying powers of a number gets you that number to the sum of the powers.

[Username17]

Here's my math question: who the fuck decided that x^0 = 1, and why did everyone go along with this? How does it work? For the main part is seems entirely theoretical anyway, and the only thing it works nicely for is 10^x, where it's 1, followed by x zeroes.

Let's say you have X^Y. That's some number with X multiplied by itself Y times. And now let's say you have X^(Y-1). That is some number with X multiplied by itself (Y-1) times. But it's also the previous number divided by X. So it would be fair to say that for any X and Y, X^(Y-1) is simply X^Y/X, right?

Now let's consider the case where Y is 1. X^1 is of course X itself. And X^(Y-1) is now X^0. And so we can now show that X^0 = X/X, which is 1 for any X that doesn't have stupid properties.

-Username17

[Ancient History]

Technically, 10^X is only 1 with X zeroes because we use base ten. If you used base two, for example, then 2^X would be 1 with X zeroes as well. Of course, you would still write it as 10^X.

[tussock]

It works "nicely" for everything X > 0. X^n is ...

(5) ..., 1/125, 1/25, 1/5, 1, 5, 25, 125, ....
(4) ..., 1/64, 1/16, 1/4, 1, 4, 16, 64, ....
(3) ..., 1/27, 1/9, 1/3, 1, 3, 9, 27, ....
(2) ..., 1/8, 1/4, 1/2, 1, 2, 4, 8, ....
(1) ..., 1, 1, 1, 1, 1, 1, 1, ....
(1/2) ..., 8, 4, 2, 1, 1/2, 1/4, 1/8, ....

X^0 = 1 because you have to multiply something by X to get X in the first place. Note that 0^0 is undefined, and negative bases involve imaginary numbers under the hood.

[Koumei]

Thanks, that really cleared that up.

[codeGlaze]

What's the name of the equation that goes something like this...
y = 2;
X = ( 1*y )+( 2*y )+( 3*y )+( 4*y );

The equation solves for x.

[Juton]

Isn't that equation just X=10y? Do the number of terms change dependent on y?

[John Magnum]

He probably means the formula for the sum of some finite arithmetic sequence.

If your sum is i=1..k of (b + ai), then the total is equal to [b + (k + 1)a] * [k * (k + 1)]/2.

The way of getting that is that you look at 1y + 2y + 3y + 4y. You note that 1y + 4y = 2y + 3y = 5y, and there are two pairs of value 5y each, so the sum is 10y. This works for an arithmetic series of any length. Taking pairs from opposite ends of the sequence and counting inward is a nice way of summing it up.

[Username17]

You're taking triangular numbers and then multiplying them by the scalar value Y. So it's just triangular progression.

1, 3, 6, 10, 15, 21, 28...

For the Nth value, the value is: N*(N+1)/2. To scale it by Y, you'd just have it be: N*(N+1)/2*Y

-Username17

[codeGlaze]

Thanks guys!