Idle Scalability Notion

[prestidigitator]

Re: Idle Scalability Notion

 

Oh, dash it all! Let's just roll 2d(8.5)+1 and be done with it!

[PhilFleischmann]

Re: Idle Scalability Notion

 

Yes. I realize that my last post marks me as not merely a gaming nerd, but as a math nerd. It's my problem and I'm dealing with it the best I can.

 

And before any bigger math nerd than myself points it out, I was going from memory. The actual standard deviation on 3d6 is just a little less than 3. The Greek letter sigma is usually used as a symbol for standard deviation.

 

----- sigma

3d6 ~2.95

3d8 ~3.9

3d10 ~4.9

3d12 ~5.9

3d20 ~9.9

 

1d6 ~1.73

2d6 ~2.42

3d6 ~2.95

4d6 ~3.46

5d6 ~3.84

6d6 ~4.22

7d6 ~4.46

8d6 ~4.84

 

More information than you needed, I know.

[prestidigitator]

Re: Idle Scalability Notion

 

Yes. I realize that my last post marks me as not merely a gaming nerd' date=' but as a math nerd. It's my problem and I'm dealing with it the best I can. [/quote']

We're getting math-nerdy? Cool! Well, standard deviation (sigma) is actually defined as the square root of the variance (written sigma^2). The variance is the mean of the square of the difference from the mean. For a distribution which is the sum of two other distributions, the variance is also the sum of their variances (thus the standard deviation is the square root of the sum of the squares of the standard deviations).

 

For a dn, the mean is (n+1)/2. The variance is (n^2-1)/12. Thus, for the sum of N dice dn(1) + dn(2) + ..., the mean is:

 

avg = [n(1) + n(2) + ... + N]/2

and the standard deviation is:

 

sigma = sqrt( [n(1)^2 + n(2)^2 + ... - N]/12 )

Waaay more than you wanted to know?

[DangerousDan]

Re: Alternatives to 3d6

 

No' date=' that's not the idea. You don't change the cost, you just change the roll required. A 14- Activation for instance, is worth the same thing as always, it's just a 27- now instead of 14- (or whatever it should be).[/quote']

 

This actually changes the shape of the bell curve dramatically! I'll leave it as an exercise for the student to determine the odds of rolling an 18 with 6d6/2 as opposed to 3d6. Advanced students can work out the standard deviations of 6d6/2 and 3d6 to see how different they are.

You woke up a few brain cells that have been asleep for a few years. I thought that I'd check this out. I took the standard 3d6 distribution and a 6d6/2, with adjacent 'bins' added together (p(3) + p(3.5)) and graphed that. You are right, and I'm wrong. 6d6/2 does narrow the graph. See attached graph (if I actually manage to attach the graph)

[Trebuchet]

Re: Alternatives to 3d6

 

You woke up a few brain cells that have been asleep for a few years. I thought that I'd check this out. I took the standard 3d6 distribution and a 6d6/2' date=' with adjacent 'bins' added together (p(3) + p(3.5)) and graphed that. You are right, and I'm wrong. 6d6/2 does narrow the graph.[/i']

I'm no math whiz (Now there's a generous 1000% understatement), but it looks to me like 6d6/2 markedly increases the probability of getting a midrange result and a corresponding decrease in results at the upper and lower ends of the curve. What would be the percent chance of rolling an 11- on 6d6/2 as compared to on 3d6? (My rough estimate based on your graph is about 73%) And how much lower would be the probabilities of rolling a 3 or 18 to produce an automatic success or failure?

 

This does not look like a good idea. Is the graph for 3d8 as distorted?

[prestidigitator]

Re: Alternatives to 3d6

 

I'm no math whiz (Now there's a generous 1000% understatement), but it looks to me like 6d6/2 markedly increases the probability of getting a midrange result and a corresponding decrease in results at the upper and lower ends of the curve. What would be the percent chance of rolling an 11- on 6d6/2 as compared to on 3d6? (My rough estimate based on your graph is about 73%) And how much lower would be the probabilities of rolling a 3 or 18 to produce an automatic success or failure?

 

This does not look like a good idea. Is the graph for 3d8 as distorted?

Well, given that the dice you are rolling at any given time are of the same kind (all d6s, all d8s, etc.), the mean is proportional to the product of the number and type of dice. So if the number of dice is N and each die has n sides, the mean is (approximately) proportional to:

avg ~ N*n

The standard deviation is proportional to the square root of the number of dice and roughly the number of sides on each die:

sigma ~ sqrt(N)*n

So the standard deviation relative to the mean (and also relative to the maximum possible roll) is approximately proportional to the reciprocal of the square root of the number of dice:

sigma/avg ~ sqrt(N)*n/(N*n) = 1/sqrt(N)

This is independent of the type of die, so the general shape of the distribution will not change in character if you go from d6s to d8s but leave the number of dice constant. You must realize that there will be slight differences because the distributions are integral (you can roll a 1, 2, etc., but not a 1.5 or a 1.324987342).

 

Also, while the probability of rolling within a certain, "distance," of the mean (relative to the size of a die) remains constant when you change die type, the probability of rolling any one number will go down. So for 3d8 (standard deviation of just under 4), the probability that you roll within 4 of the mean of 13.5 is about the same as the probability on 3d6 (standard deviation of just under 3) of rolling within 3 of the mean of 10.5. However, the probability of rolling 13 or 14 on 3d8 will be less than the probability of rolling a 10 or 11 on 3d6.

[Trebuchet]

Re: Alternatives to 3d6

 

Well, given that the dice you are rolling at any given time are of the same kind (all d6s, all d8s, etc.), the mean is proportional to the product of the number and type of dice. So if the number of dice is N and each die has n sides, the mean is (approximately) proportional to:

avg ~ N*n

The standard deviation is proportional to the square root of the number of dice and roughly the number of sides on each die:

sigma ~ sqrt(N)*n

So the standard deviation relative to the mean (and also relative to the maximum possible roll) is approximately proportional to the reciprocal of the square root of the number of dice:

sigma/avg ~ sqrt(N)*n/(N*n) = 1/sqrt(N)

This is independent of the type of die, so the general shape of the distribution will not change in character if you go from d6s to d8s but leave the number of dice constant. You must realize that there will be slight differences because the distributions are integral (you can roll a 1, 2, etc., but not a 1.5 or a 1.324987342).

 

Also, while the probability of rolling within a certain, "distance," of the mean (relative to the size of a die) remains constant when you change die type, the probability of rolling any one number will go down. So for 3d8 (standard deviation of just under 4), the probability that you roll within 4 of the mean of 13.5 is about the same as the probability on 3d6 (standard deviation of just under 3) of rolling within 3 of the mean of 10.5. However, the probability of rolling 13 or 14 on 3d8 will be less than the probability of rolling a 10 or 11 on 3d6.

I only understood the last two paragraphs of this dissertation, prestidigitator. The rest of it might as well have been in Klingon.

[prestidigitator]

Re: Alternatives to 3d6

 

I only understood the last two paragraphs of this dissertation' date=' prestidigitator. The rest of it might as well have been in Klingon. [/quote']

Sorry. I have a hard time presenting conclusions without their explanations.

[DangerousDan]

Re: Alternatives to 3d6

 

I'm no math whiz (Now there's a generous 1000% understatement), but it looks to me like 6d6/2 markedly increases the probability of getting a midrange result and a corresponding decrease in results at the upper and lower ends of the curve. What would be the percent chance of rolling an 11- on 6d6/2 as compared to on 3d6? (My rough estimate based on your graph is about 73%) And how much lower would be the probabilities of rolling a 3 or 18 to produce an automatic success or failure?

 

This does not look like a good idea. Is the graph for 3d8 as distorted?

 

on 3D6, you roll a three 0.463% of the time. On 6D6/2, a 3 comes up only 0.002% of the time, and even a 5 or less (5, 4.5, 4, 3.5 or 3) comes up only 0.45% of the time. Near the middle of the rolls, things are not quite so wonky.

           3D6              3d6/2
Roll  prob   cumulative    cumulative
 3   0.463    0.463        0.002
 4   1.389    1.852        0.06
 5   2.778    4.63         0.45
 6   4.63     9.259        1.968
 7   6.944   16.204        6.076
 8   9.722   25.926       14.463
 9  11.574   37.5         27.939
10  12.5     50           45.358
11  12.5     62.5         63.69
12  11.574   74.074       79.415
13   9.722   83.796       90.353
14   6.944   90.741       96.412
15   4.63    95.37        99.01
16   2.778   98.148       99.82
17   1.389   99.537       99.985
18   0.463  100          100

 

As for 3d8, 3d10 and even 3d100, the attached charts may shed further darkness on the topic. One chart simply shows the frequency of die rolls for various sets of dice. The second shows the same graphs with their scales all set to have the same width and height.

[atlascott]

Re: Idle Scalability Notion

 

I like the multiple-dice mechanic ('bell curve") because I think it is appropriate to the game. In a d20 game I run, there is neve a game session where several players fail to make several crits. Thats because they expect to do so ona regular basis. Obscene success ought to be as odd as obscene failure. It means more and is more 'fair' and sensical. See? no math!