Standard deviation of a die roll
I assume everybody knows how to find the average value of 2d4, 3d10+6, 6d6, etc., but even though 7d4 has the same average value (17.5) as 1d20+7, the actual results will tend to be more spread out in the latter case.
The method for determining the amount of spread (called the standard deviation) is a bit tricky to derive, so here it is, in 5 easy steps (this method only works for a simple ?d?+?d?+...+? roll, not a roll where all 1s count as 2s or only the 3 highest dice are counted or a negative final result counts as 0 or anything like that):
1. For each die rolled, square the number of sides to the die, and subtract one.
2. Multiply the result by the number of times the die is rolled.
3. If multiple die sizes are rolled (e.g. 1d4+2d6), add together the results from steps 1-2 for each size (constant modifiers are ignored).
4. Divide by 12
5. Take the square root (a calculator will probably be needed here).
This result is the standard deviation.
Example: 2d4+3d6+1d20+5:
-For the 2d4, square 4 for 16, subtract 1 and multiply by 2 for 30.
-For the 3d6, square 6 for 36, subtract 1 and multiply by 3 for 105.
-For the 1d20, square 20 for 400, subtract 1 and multiply by 1 for 399.
-The 5 is ignored, since it does not involve a roll
-Add 30+105+399 for 534, divide by 12 for 44.5, and take the square root for ~6.7 This is the standard deviation.
So what does this mean? A roll with a lower standard deviation will be more dependable, but also less likely to go far above the average. Furthermore, for a roll involving a lot of dice (10d6, for example), there will be about a 68% chance that the result will differ from the average by less than the standard deviation, and about a 95% chance that it will differ by less than twice the standard deviation. There is a less than 1% chance of it being more than 3 times the standard deviation away from the average. Therefore, the standard deviation gives you a better idea of the chances that that fireball will kill your enemy.
I assume everybody knows how to find the average value of 2d4, 3d10+6, 6d6, etc., but even though 7d4 has the same average value (17.5) as 1d20+7, the actual results will tend to be more spread out in the latter case.
The method for determining the amount of spread (called the standard deviation) is a bit tricky to derive, so here it is, in 5 easy steps (this method only works for a simple ?d?+?d?+...+? roll, not a roll where all 1s count as 2s or only the 3 highest dice are counted or a negative final result counts as 0 or anything like that):
1. For each die rolled, square the number of sides to the die, and subtract one.
2. Multiply the result by the number of times the die is rolled.
3. If multiple die sizes are rolled (e.g. 1d4+2d6), add together the results from steps 1-2 for each size (constant modifiers are ignored).
4. Divide by 12
5. Take the square root (a calculator will probably be needed here).
This result is the standard deviation.
Example: 2d4+3d6+1d20+5:
-For the 2d4, square 4 for 16, subtract 1 and multiply by 2 for 30.
-For the 3d6, square 6 for 36, subtract 1 and multiply by 3 for 105.
-For the 1d20, square 20 for 400, subtract 1 and multiply by 1 for 399.
-The 5 is ignored, since it does not involve a roll
-Add 30+105+399 for 534, divide by 12 for 44.5, and take the square root for ~6.7 This is the standard deviation.
So what does this mean? A roll with a lower standard deviation will be more dependable, but also less likely to go far above the average. Furthermore, for a roll involving a lot of dice (10d6, for example), there will be about a 68% chance that the result will differ from the average by less than the standard deviation, and about a 95% chance that it will differ by less than twice the standard deviation. There is a less than 1% chance of it being more than 3 times the standard deviation away from the average. Therefore, the standard deviation gives you a better idea of the chances that that fireball will kill your enemy.
I have a question about the second one. Does the formula give its results in terms of hours between encounters or some other time unit?
One additional thing: What would be your interpretation of the part about tha average time between encounters? Wouldn't this depend on the number of encounters to be had each day, over the course of the journey and then averaged out by the number of hours in each day?
Basically, I am not good with statistics (hence why I really liked your first section on deviation), and was wondering how to put all of this in perspective.
I can and do these things now without the formula - er... forget it.. but I'd like a little more randomization and less sense of fiat on my end.
Thanks,
AzarDM
And the average roll on a d6 is 3.5, not 3.
How did you determine the percentages you used in the first example? The standard deviation is a handy little formula, but you don't describe how to use it.
On a slightly different note, I am taking a Pre-Calculus course in school, and am learning about logarithms. Now as I listen to the teacher drone on, I can only think that there is an actual use for this crap. Thank you for showing me that.
Some (Relatively) Simple Mathematical Formulas .... requires a scientific calculator or equivalent
Hrm?!? Relatively simple formulas to help my game ... I will need a scientific calculator to run 50% of the formulas you propose we use in your thread here?!
This is even before I open your spoiler blocks.
Looking in the first option, standard deviation for rolling multiple dice (do we even care to know standard deviation for dice spreads?), there are 5 steps. I'm required to square the number of sides of a dice, subtract one, then multiple by the number of dice used to roll the number and then divide by twelve (((S^2)-1)*N)/12). Simple if I want to set it up in a spread sheet, or something similar, before the game - but this isn't something to just whip out in the middle of the game.
There is virtually no purpose for calculating standard deviation in D&D. Perhaps only if a DM wanted to create a 'standard deviation' from a typical monster by having it do weird damage or something. Something like ... I don't know if its claw should do 4d4 or 2d8, maybe I should calculate the standard deviation for this. Yet again, most DMs with experience will realize without calculating standard deviation that the 4d4 is more threatening to the party not because of standard deviation but because the claw attack will always do a minimum of 4+str dmg with 4d4 as opposed to a minimum of 2+str for the 2d8 roll. You have shown us that you enjoy calculating probability and standard deviation to control dice selection, but there is no relevance for a DM.
The second one requires a scientific calculator to calculate the probability of random encounters. This is even less relevant than the standard deviation/dice spread formula. It even includes a footnote saying it will waste time in game so do all calculations before hand.
The proposed purpose of the second formula is, according to you, that the chance of a random encounter from one hour to the next is independent of the amount of random encounters from previous hours. Instead of trying to set up an elaborate calculation to find out when the next random encounter is, take out your trusty old d2 and roll it for each hour (roll it before game if you want to save time or have players roll them for you during the game). Each roll is independent of other hours on the journey - thus, each hour has the same chance of a random encounter as the preceding hours whether the party has gone weeks without a random encounter or has had 5 for the day so far. By the by, a trusty d2 is a coin, heads=1, tails=2 (for some of you out there, this can be heads=0 (no), tails=1 (yes)).
Here is a suggestion to make the random encounter determination a bit easier by eliminating the calculator:
- Unchanged: Decide the journey's total length of time, T.
- Unchanged: Decide the average time between encounters, t.
- Roll d%. Let X be a number that depends on this roll as follows: