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Jack Zander
2008-11-05, 10:29 AM
If I roll 3d20s, what are the odds of me rolling at least two 20s and how do you calculate that?

Keld Denar
2008-11-05, 10:56 AM
If I roll 3d20s, what are the odds of me rolling at least two 20s and how do you calculate that?

At least 2 20s? 1/20 x 1/20 x (20/20=1) or 00.25%
Exactly 2 20s? 1/20 x 1/20 x 19/20 or 00.2375%

Basically, what ever the odds of rolling each of the desired results multiplied together.

In other words, not good.

Jack Zander
2008-11-05, 11:06 AM
At least 2 20s? 1/20 x 1/20 x (20/20=1) or 00.25%

Yes, that one, thank you.

For my homebrew system I am thinking about rolling 3d20 and taking the median result, but I want to see how much of a bell curve that will give me.

So now my statistics get more complicated. To roll a 10 you would have to roll one 10, one would have to be a result of 1-10 and the other would have to be a result of 10-20. How do I figure that out? As soon as I get a formula I can calculate each one to make a graph and see for myself if the curve is good enough for me.

Let me try: (exactly 10) x (1-10) x (10-20)
1/20 x 1/2 x 11/20 = 1.375%

Is that correct?

jcsw
2008-11-05, 11:08 AM
Doesn't 3d6 work better for this?

You never get 1,2,19 and 20, but in comparison with the median of 3d20 it seems a bit less confusing.

Jack Zander
2008-11-05, 11:09 AM
Doesn't 3d6 work better for this?

I want to still be able to roll 1s and 20s and I find it much much faster if you take the median than if you have to add a bunch of numbers together.


You never get 1,2,19 and 20, but in comparison with the median of 3d20 it seems a bit less confusing.

No way. Median of 3d20 may sound complicated but it really isn't. Look at your dice. Which one is in the middle of the three numbers? Super easy, 0 math involved.

Jack Zander
2008-11-05, 11:13 AM
At least 2 20s? 1/20 x 1/20 x (20/20=1) or 00.25%

Hang on, that can't be right. That is the same chance of me rolling 2 dice and getting 20s on both. The chances should be better when rolling three dice and only needing two of the three.

Fax Celestis
2008-11-05, 11:15 AM
Hang on, that can't be right. That is the same chance of me rolling 2 dice and getting 20s on both. The chances should be better when rolling three dice and only needing two of the three.

Not really. His math assumes you don't care about the third result.

Kurald Galain
2008-11-05, 11:18 AM
Hang on, that can't be right. That is the same chance of me rolling 2 dice and getting 20s on both. The chances should be better when rolling three dice and only needing two of the three.

It's not right. He didn't account for the fact that one of the other dice could also be the not-20.

Odds for 2 or more 20s
= (1/20) * (1/20) * (19/20)
+ (1/20) * (19/20) * (1/20)
+ (19/20) * (1/20) * (1/20)
+ (1/20) * (1/20) * (1/20)

= about 0.7%

Jack Zander
2008-11-05, 11:20 AM
Not really. His math assumes you don't care about the third result.

True yet false. The first, second, or third die can come up a 20. One of them doesn't matter, but it won't always be the third die.

If I roll 2d6 there are 36 different outcomes. 11 of these outcomes involve rolling at least 1 6 so the odds for that are 11/36 or 30.556%, not simply 1/6 x 6/6 or 16.667%

Jack Zander
2008-11-05, 11:22 AM
It's not right. He didn't account for the fact that one of the other dice could also be the not-20.

Odds for 2 or more 20s
= (1/20) * (1/20) * (19/20)
+ (1/20) * (19/20) * (1/20)
+ (19/20) * (1/20) * (1/20)
+ (1/20) * (1/20) * (1/20)

= about 0.7%

Awesome, thanks. I got to go to class now but I'll finish the math for all 20 numbers later.

phoenixcire
2008-11-05, 11:32 AM
The first, second, or third die can come up a 20. One of them doesn't matter, but it won't always be the third die.

That's bad logic. If you were to roll 3d20 and get two 20's and a 13, the 13 would be the third die no matter which order you rolled it.

Gaiwecoor
2008-11-05, 11:41 AM
That's bad logic. If you were to roll 3d20 and get two 20's and a 13, the 13 would be the third die no matter which order you rolled it.

Actually, you do care about the order. Yes, 13 would be the third die once you rank them by value, but when determining the probability of rolling it, you don't rank them that way. It could very well have come out 20-13-20, in which case it was the second die. Kurald has it right.

EDIT:

I did some brute-force calculation (Yay spreadsheets!) and here are the stats for each roll:

{table]Median Roll|Probability
1|0.725%
2|2.075%
3|3.275%
4|4.325%
5|5.225%
6|5.975%
7|6.575%
8|7.025%
9|7.325%
10|7.475%
11|7.475%
12|7.325%
13|7.025%
14|6.575%
15|5.975%
16|5.225%
17|4.325%
18|3.275%
19|2.075%
20|0.725%[/table]

Keld Denar
2008-11-05, 11:55 AM
Awww crap. I knew it looked too simple when I typed it out. I blame the Peas-In-The-Straw Theory. I got a 4.0 in calculus and have since forgotten "less pure" forms of math.

:P

Gaiwecoor
2008-11-05, 12:02 PM
Awww crap. I knew it looked too simple when I typed it out. I blame the Peas-In-The-Straw Theory. I got a 4.0 in calculus and have since forgotten "less pure" forms of math.

:P

I know what you mean. I did the same in Calc I-III and Thermal Physics (essentially a statistical variation course) in undergrad. Yet what did I have to do? Brute-force look at all 8000 possible rolls, and count 'em up. [See edit to my post above for results]

Jack Zander
2008-11-05, 01:10 PM
{table]Median Roll|Probability
1|0.725%
2|2.075%
3|3.275%
4|4.325%
5|5.225%
6|5.975%
7|6.575%
8|7.025%
9|7.325%
10|7.475%
11|7.475%
12|7.325%
13|7.025%
14|6.575%
15|5.975%
16|5.225%
17|4.325%
18|3.275%
19|2.075%
20|0.725%[/table]

Oh wow, that's another step forward. Thanks a ton!

http://i226.photobucket.com/albums/dd109/sexaypattimon/3d20Median.jpg

Is that a good enough curve if I am looking for a more realistic system?

Gaiwecoor
2008-11-05, 01:56 PM
It looks good, I think. However, keep in mind that the threat range is drastically reduced, unless you modify it.

I would suggest having items that threaten a critical on 20 in d20 threaten on 18-20 in the 3d20M system. This slightly increases their threat to 6% from 5%. A weapon with a range of 19-20 (d20) could be 17-20 (3d20M), again slightly increasing the range from 10% to 10.325%. This can be expanded for other ranges, as well.

If you're not worried about that, want to keep ranges as they are, be aware that weapons with larger crit ranges (typically piercing/slashing and keen) will be much more valuable than others (typically blunt instruments).

Edit:
This new curve is actually much more narrow (for critical threat ranges) than the standard 3d6 system. See here (http://www.d20srd.org/srd/variant/adventuring/bellCurveRolls.htm#threatRange) for more details about that.

Duke of URL
2008-11-05, 02:10 PM
It looks good, I think. However, keep in mind that the threat range is drastically reduced, unless you modify it.

I would suggest having items that threaten a critical on 20 in d20 threaten on 18-20 in the 3d20M system. This slightly increases their threat to 6% from 5%. A weapon with a range of 19-20 (d20) could be 17-20 (3d20M), again slightly increasing the range from 10% to 10.325%. This can be expanded for other ranges, as well.

If you're not worried about that, want to keep ranges as they are, be aware that weapons with larger crit ranges (typically piercing/slashing and keen) will be much more valuable than others (typically blunt instruments).

Edit:
This new curve is actually much more narrow (for critical threat ranges) than the standard 3d6 system. See here (http://www.d20srd.org/srd/variant/adventuring/bellCurveRolls.htm#threatRange) for more details about that.

Alternatively, you can leave the ranges alone and just automatically confirm crits, since the threats will be less often. Even a 15-20 threat range (18-20 doubled) would only crit 21.6% of the time, which is less than the old 30% threaten chance.

{table] Threat Range | Original Threat Chance | New Crit Chance
15-20 | 30% | 21.600%
16-20 | 25% | 15.625%
17-20 | 20% | 10.400%
18-20 | 15% | 6.075%
19-20 | 10% | 2.800%
20 | 5% | 0.725%[/table]

Of course, what this lacks is any mapping of how the confirmation chance affects the actual number of crits.

Person_Man
2008-11-05, 02:11 PM
Yes, that one, thank you.

For my homebrew system I am thinking about rolling 3d20 and taking the median result, but I want to see how much of a bell curve that will give me.

Why?

You're just making it harder for PCs who use attack rolls to fight enemies with high AC (bosses) and easier for them to fight mooks (mid-low AC). That makes balancing encounters harder, and gives spell using builds more powerful (unless your homebrew system is like 4E, where everything needs an attack roll).

Jack Zander
2008-11-05, 02:12 PM
It looks good, I think. However, keep in mind that the threat range is drastically reduced, unless you modify it.

I would suggest having items that threaten a critical on 20 in d20 threaten on 18-20 in the 3d20M system. This slightly increases their threat to 6% from 5%. A weapon with a range of 19-20 (d20) could be 17-20 (3d20M), again slightly increasing the range from 10% to 10.325%. This can be expanded for other ranges, as well.

If you're not worried about that, want to keep ranges as they are, be aware that weapons with larger crit ranges (typically piercing/slashing and keen) will be much more valuable than others (typically blunt instruments).

Edit:
This new curve is actually much more narrow (for critical threat ranges) than the standard 3d6 system. See here (http://www.d20srd.org/srd/variant/adventuring/bellCurveRolls.htm#threatRange) for more details about that.

I'm not too worried about crits. In fact, I may scrap them altogether and give bonus damage simply on how much higher you score against a target's evasion.

Also, my armor gives DR against certain types of weapons rather than an evasion bonus, so having slashing and piercing weapons be better than blunt weapons would balance things out more (most armor has less effect against blunt weapons).

Jack Zander
2008-11-05, 02:14 PM
Why?

You're just making it harder for PCs who use attack rolls to fight enemies with high AC (bosses) and easier for them to fight mooks (mid-low AC). That makes balancing encounters harder, and gives spell using builds more powerful (unless your homebrew system is like 4E, where everything needs an attack roll).

Everything does need an attack roll. Spells are done similar to how SW:SE handles the Force. And it's a skill based system. No levels, just ranks in various skills.

kbk
2008-11-05, 11:13 PM
You have to consider all the possible scenarios, so:


It's not right. He didn't account for the fact that one of the other dice could also be the not-20.

Odds for 2 or more 20s
= (1/20) * (1/20) * (19/20)
+ (1/20) * (19/20) * (1/20)
+ (19/20) * (1/20) * (1/20)
+ (1/20) * (1/20) * (1/20)

= about 0.7%

This, except the actual values come out to


=.05*.05*.95
+.05*.95*.05
+.95*.05*.05
+.05*.05*.05

= 0.002375
+ 0.002375
+ 0.002375
+ 0.000125

= .00725
= .725%

Edit: fixed my math. Now it matches the simulation posted perfectly.

The ultimate result of your system is going to be a lot more rolls in the middle range with very few extreme results. If you instead used means, you would get an even more exaggerated curve compared to the one posted (Means are affected by outliers more)

Jack Zander
2008-11-05, 11:16 PM
No means thank you. That would be terrible to calculate each time you rolled.

kbk
2008-11-05, 11:54 PM
Okay, sorry, I decided I wanted to do this anyways:

This compares median of 3d20, mean of 3d20, and a simple 1d20. The mean rolls are a result of a random simulation of 100,000 rolls.

http://i44.photobucket.com/albums/f3/kalmbob/3d20Median-1.jpg


Overall what you want to do is to simulate a normal curve if you really value an authentic system. Means would be difficult to calculate with just dice, but I could whip up a line of code in 2 minutes (in fact I did that and more to simulate it) and you could easily run this with a computer or blackberry or iphone. Hell, I could make excel do it.

Anyways, here's the skinny:
A single d20 is a uniform distribution (Every roll is equally likely)
The median of 3d20 is closer to the normal distribution.
The mean of 3d20 is a lot closer to the normal distribution.
By the way, the roll 3d6 is a lot closer to the normal distribution than median of 3d20, but it doesn't allow the flexibility of already being in the hand to use 1-20 range.

Now, if you really wanted to use a normal distribution, I could whip that up to run on a computer pretty easily.

Jack Zander
2008-11-06, 12:46 AM
Well I don't think I want as much of a curve as that anyway. 1s and 20 do need to come up sometime after all. A true bell curve will make those results come up only once in every 8000 rolls.

kbk
2008-11-06, 12:46 PM
Well I don't think I want as much of a curve as that anyway. 1s and 20 do need to come up sometime after all. A true bell curve will make those results come up only once in every 8000 rolls.

Then probably a median system is the system for you. If you look at the curves, the median system basically just truncates the tails of a normal distribution. Now, it is true you can scale a normal distribution by increasing the variance, but if you're trying to confine that in between 1 and 20, its not going to be appreciably different than your proposed median system.

There are other complex statistical transformations you could do, like mix a uniform and a normal distribution, but honestly, median rolls will approximate that sufficiently anyways.