CRGreathouse
Community Supporter
Graf said:
Using that logic, a commoner with standard stats Str 11, Dex 10, Con 11, Int 10, Wis 11, Cha 10 is in the top .5^3*.625^3 = 3.0517578125% of the population.
Probability Distribution of Dice
- Thread starter Riveneye
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Using that logic, a commoner with standard stats Str 11, Dex 10, Con 11, Int 10, Wis 11, Cha 10 is in the top .5^3*.625^3 = 3.0517578125% of the population.
CRGreathouse
Community Supporter
Re: Math vs Brute Force
That is, as long as you're keeping more than one die at the end. It takes only basic math, really, to figure out the distribution for 5d6 drop lowest 4.
Lord Zardoz said:
That is, as long as you're keeping more than one die at the end. It takes only basic math, really, to figure out the distribution for 5d6 drop lowest 4.
CRGreathouse, didn't we debate about such a question (4d6, drop lowest, etc., drop crap characters) on these boards some months ago?
I cannot search the boards but maybe someone else could dig deep into this board and recover the former thread. We ended up with some nice tables and formulas.
I cannot search the boards but maybe someone else could dig deep into this board and recover the former thread. We ended up with some nice tables and formulas.
morpheous1777
First Post
So has anyone created a program to figure it out (instead of frying my brain trying to comprehend lazurus's formula and doing it by hand). Im pretty decent at C++. if someone could break down that formula into laymans terms, then i could, give it a try.
ichabod
Legned
Graf said:
My calculations show that with a 3d6, the probability of having at least one 15 or better is 44.18%. Double checked it too. The probability of someone having a 14 or better for their second stat is 25.21%. Someone with the default array is actually more like top 25%, given that the third quartile is close to that set of numbers.
RogerShrubber
First Post
A fast and easy solution
I know this thread has been dead for several months, but Frisbeet pointed it out to me, and I thought I'd provide an easy way to accomplish this, without going into the discrete behind it.
This works great if you have access to Matlab, a good TI, or some other 'calculator' with an expand function.
Where:
n: number of dice
s: number of sides per die
expand((x+x^2+x^3...x^s)^n)
For example, for s = 6 and n = 2, this results in:
x^12+2x^11+3x^10+4x^9...6x^7+5x^6....
The coefficient before each x is the number of ways to make the exponent. In this case, there is 1 way to make a 12, 2 ways to make an 11, 3 ways to make a 10, 6 ways to make a 7, 5 ways to make a 6, and so on.
This can be very tedious when you have to multiply and recombine the terms by hand, but is quite fast when done by a calculator.
When you combine these results with the total number of possible solutions (s^n), you can easily find the percent chance of each result.
Further, this formula can be used to determine the results if several different dice are involved.
Example: d4+d6
Expand((x+x^2+x^3+x^4)^1*(x+x^2+x^3+x^4+x^5+x^6)^1)
x^10+2x^9+3x^8+4x^7+4x^6+4x^5+3x^4+2x^3+x^2
I have found this very useful in my modification of Frisbeet's Weapon Analysis, and think you may appreciate them as well.
Roger Shrubber
I know this thread has been dead for several months, but Frisbeet pointed it out to me, and I thought I'd provide an easy way to accomplish this, without going into the discrete behind it.
This works great if you have access to Matlab, a good TI, or some other 'calculator' with an expand function.
Where:
n: number of dice
s: number of sides per die
expand((x+x^2+x^3...x^s)^n)
For example, for s = 6 and n = 2, this results in:
x^12+2x^11+3x^10+4x^9...6x^7+5x^6....
The coefficient before each x is the number of ways to make the exponent. In this case, there is 1 way to make a 12, 2 ways to make an 11, 3 ways to make a 10, 6 ways to make a 7, 5 ways to make a 6, and so on.
This can be very tedious when you have to multiply and recombine the terms by hand, but is quite fast when done by a calculator.
When you combine these results with the total number of possible solutions (s^n), you can easily find the percent chance of each result.
Further, this formula can be used to determine the results if several different dice are involved.
Example: d4+d6
Expand((x+x^2+x^3+x^4)^1*(x+x^2+x^3+x^4+x^5+x^6)^1)
x^10+2x^9+3x^8+4x^7+4x^6+4x^5+3x^4+2x^3+x^2
I have found this very useful in my modification of Frisbeet's Weapon Analysis, and think you may appreciate them as well.
Roger Shrubber
Unther the unready
First Post
Nice, but...
How well can it handle 4d6, drop the lowest, or 20d10, or 25d8?
Doing the calculations for a small number of dice wasn't really the issue, but doing them for a lot of dice, or for a complex roll.
How well can it handle 4d6, drop the lowest, or 20d10, or 25d8?
Doing the calculations for a small number of dice wasn't really the issue, but doing them for a lot of dice, or for a complex roll.
RogerShrubber
First Post
It can handle 20d10 or 25d8 just fine, provided you don't try this by hand or with a wussy calculator. Matlab will probably run either calculation in less than a second.
This could be used for 4d6: drop, but it would be a great deal more tedious than without. Essentially, you would have to incorporate multiple variables (x,y,z,t).
I haven't yet found a simple way to do calculations with 4d6: drop, but it is not a type of calculation I've had to do more than once. From what I can tell, nobody else on this thread has come close to making something as simple as what I've put forth for this calculation.
On the other hand, I have had to find the probability on every possible result for every possible damage or critical damage roll possible in 3e for tables used in the Weapon Analysis. Stupid DR and possible negative str modifiers mean that we can't just use straight averages for damage rolls but have to consider each possible result and it's probability of occuring.
Roger Shrubber
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