The text formatting seems to be messed up (it did not look like that in the original HTML). I hope you can still make sense of it.
No!
The font size changes half way through becoming too small to read comfortably, which is OK, because I'd already given up as the window is wider than the screen and I don't want to keep scrolling left and right on every line.
I suspect that (part of) the problem is the size of your diagrams forcing the page width to be wider than is sensible.
Browsers other than IE might handle this better, and when I have access to a different PC I might try again...
The text formatting seems to be messed up (it did not look like that in the original HTML). I hope you can still make sense of it.
The formatting is fixed now. The font change was caused by an improper HTML end tag (my fault) and the long lines were caused by a script at RPG.net that removed all newlines from the program code (which it shouldn't, as it was inside a HTML tag that preserves line breaks).
I will be more careful with my HTML tags in the future and avoid using <pre> or <code> tags in future columns.
[T]he events "the die landed with an even number facing up'' and "the die landed with a number in the upper half facing up'' are not independent, as knowing one of these will help you predict the other more accurately. Taking any one of these events alone (on a normal d6), will give you a probability of ½ of it happening, but if you know that the result is even, there is 2/3 chance of it being in the upper half, as two of 4, 5 and 6 are even.
I think you have a confusion of the inverse problem here. Knowing that the result is even tells you that there is a 2/3 chance that the number is in the upper half because the result must be either 2, 4, or 6 -- two of which are in the upper half. What you have described is the probability that the number is even given that it is in the upper half, not that it is in the upper half given that it is even. In this case, both probabilities are the same, but that will not always be the case with conditional probabilities.
In my system, I use additional dice to alter the difficulty of a result. I was too lazy to program it but used formulas in an excel spreadsheet. I am missing something because it is at first gave me results above 100%. That was fine I suppose, it was letting me know that I could get multiple dice that matched the result. It was not a true probability, I tried reconstructing it a few times but eventually switched to something more productive. That part of my system does not count multiple successes, can you give me some pointers here!
How likely is someone to succeed if one of the dice has to be bellow a given number? (for easier than average tasks)
How likely is someone to succeed if all of the dice has to be bellow a given number? (for harder than average tasks)
I already have a good idea but was hopping to get a more specific understanding of the probabilities for various skill values and number of dice.
__________________ Paul R. DuPont
Chronic Thinker and RPG Designer Live life as if enchanted!
(for hard tasks: must have all numbers bellow given value)
I calculated the total posible combinations by dictating the number needed on the die roll and giving it a power equal to the number of dice. The same was done for the dice in total and then I simply divided the needed number of posibilities by the total number of posibilites to get a percentage. Simply repeated that formula on the spreadsheet to get all my values.
(for easier tasks: at least one number must be bellow a given value)
These were troublesome but I simply calculated as above for the likelyhood that I would not get the result then inverted it (1-p).
Thanks for the column, by the way! Jogged loose a few odd mathematical facts that had gotten burried deep in my brain.
__________________ Paul R. DuPont
Chronic Thinker and RPG Designer Live life as if enchanted!
I think you have a confusion of the inverse problem here. Knowing that the result is even tells you that there is a 2/3 chance that the number is in the upper half because the result must be either 2, 4, or 6 -- two of which are in the upper half. What you have described is the probability that the number is even given that it is in the upper half, not that it is in the upper half given that it is even. In this case, both probabilities are the same, but that will not always be the case with conditional probabilities.
-Eric
My formulation was not very clear. What I meant was "two of the three even numbers from 1-6 are in the upper half", not "two of the three numbers in the upper half are even". I agree that these are two quite different statements, and it is unfortunate that my phrasing didn't distinguish properly between them.
Re: #2: Probability Calculation
Linear Mode
