Results 1 to 13 of 13

20190729, 07:10 PM (ISO 8601)
 Join Date
 Apr 2012
 Gender
Help Me Correct My Probabilistic Misunderstandings
Some months ago, I posted in a roleplaying thread and made a comment about someone else's post on probability. Long story short, I found out that I had some messed up assumptions about probability that I didn't realize I had—specifically about additivity—and I want to set them right.
My confusion is this. Take two twentysided die. Rolling them simultaneously, what's the probability of rolling at least one twenty? Or, to put it in more applied terms: in D&D 5e, what're the odds of rolling a twenty with advantage? To figure this out, we'd calculate
1 − (1 − .05) × (1 − .05),
which would equal 9.75%.
But let us then take a different example. Imagine that for a given set of data D, there exist twenty scientific theories—which we'll label T1, T2, ... T20—that fully describe D. We will assume that only one theory is true. Furthermore, we currently have no reason to prefer any one scientific theory over any other, and so we say that the odds of any one theory being true is equal to any other theory. That is, we assign them equal probabilities. Picking two of those theories, what are the odds that one of them will be true? In this case, our calculations are different. We would calculate
.05 + .05,
which would be 10%.
Obviously there's a difference between these two examples. There's a reason we calculate them differently. But for whatever reason, I can't seem to figure out what it is. Can anybody help?
Thanks in advance."I kinda think I'm not supposed to go along with something that's wrong just to get food."
Spoiler: Play by Post Gamesx Through the Faerie Ring // PL + Progressing // 05/17  Present
x Chrono Isekai // PL + Progressing // 03/20  Present
x Lightning Rails and Whispers of the Vampire's Blade // DM + Completed // 05/17  08/18

20190729, 07:52 PM (ISO 8601)
 Join Date
 Jul 2014
 Location
 Avatar By Linklele!
Re: Help Me Correct My Probabilistic Misunderstandings
The first is the odds of rolling at least one 20which means that, out of 400 possible results (20 X 20) there are 39 results that are correct. That's because there's one instance of overlap.
In 19 cases, you get 20 on one die.
In 19 cases, you get 20 on the other die.
In 1 case, you get 20 on both.
For the latter, you have 20 results. You pick 2, meaning you have 2/20 results, or 1/10 of the results, which is 10%.I have a LOT of Homebrew!
Spoiler: Former AvatarsSpoiler: Avatar (Not In Use) By Professor Gnoll!
Spoiler: Avatar (Not In Use) By Cdr. Fallout!

20190729, 07:53 PM (ISO 8601)
 Join Date
 Aug 2008
Re: Help Me Correct My Probabilistic Misunderstandings
In the first example, the events are independent  rolling a 20 on one die doesn't make rolling a 20 on another die impossible, rolling a not20 on one die doesn't make rolling a 20 on another die more likely.
In the second example the events aren't independent. One of the 20 must be correct, 19 of the 20 must be wrong, and so if you get one and it's correct the next must be wrong, and if you get one and it's wrong the next is more likely to be correct. The standard example here is actually playing cards (or pulling colored marbles out of a bag), where if you draw/pull one and put it aside it's no longer there for the next pull, and so the odds are different.I would really like to see a game made by Obryn, Kurald Galain, and Knaight from these forums.
I'm not joking one bit. I would buy the hell out of that.  ChubbyRain
Current Design Project: Legacy, a game of masters and apprentices for two players and a GM.

20190730, 12:40 AM (ISO 8601)
 Join Date
 Jul 2010
Re: Help Me Correct My Probabilistic Misunderstandings
Yes. The others are right.
Your second example is the same as if you cut the 20 faces of your d20 apart, put them in a bag, and drew two of them without replacement. In that situation, you can't get doubles of the same number.

20190730, 01:26 AM (ISO 8601)
 Join Date
 Mar 2012
 Location
 UK
 Gender
Re: Help Me Correct My Probabilistic Misunderstandings
As everyone else has been saying this is all about Independence  your two scenarios are different because the two dice are independent events and the scientific theories are not.
So, to look at the maths in these cases to help illustrate what is really going on:
In the first case you are correctly looking at the chance of neither dice being a 20 and then subtracting that from 1 to get the chance that either dice will be a 20.
In the second case you simply (for good reason) just sum the seperate probabilities; however you might understand this better if you solve the second case the same way as the first  what's the chance that neither is correct?
So case 1: 1  (19/20 * 19/20) = 1  (361/400) = 0.0975 = 9.75%
Case 2 is: 1  (19/20 * 18/19) = 1  (18/20) = 0.1 = 10%
The big difference is the second term inside the brackets  the second theory you select isn't one from twenty anymore, it is one from 19.
I hope this helps.

20190730, 07:04 AM (ISO 8601)
 Join Date
 Dec 2010
 Location
 The Great White North
 Gender
Re: Help Me Correct My Probabilistic Misunderstandings
It depends if the events are independent of each other or not. If they are independent, add the probabilities together. If interdependent, use this formula:
1  ( 1  p1 ) * ( 1  p2 )How do you keep a fool busy? Turn upside down for answer.
˙ɹǝʍsuɐ ɹoɟ uʍop ǝpısdn uɹnʇ ¿ʎsnq ןooɟ ɐ dǝǝʞ noʎ op ʍoɥ

20190730, 11:34 AM (ISO 8601)
 Join Date
 Aug 2005
 Location
 Mountain View, CA
 Gender
Re: Help Me Correct My Probabilistic Misunderstandings
Like 4X (aka Civilizationlike) gaming? Know programming? Interested in game development? Take a look.
Avatar by Ceika.
Archives:
SpoilerSaberhagen's Twelve Swords, some homebrew artifacts for 3.5 (please comment)
Isstinen Tonche for ECL 74 playtesting.
Team Solars: Powergaming beyond your wildest imagining, without infinite loops or epic. Yes, the DM asked for it.
Arcane Swordsage: Making it actually work (homebrew)

20190730, 07:05 PM (ISO 8601)
 Join Date
 Nov 2010
 Location
 California
 Gender
Re: Help Me Correct My Probabilistic Misunderstandings
In simple terms, I say: If you roll 20 d20s, there's a still a chance (about 35%) that you'll get no 20s.
But if you chose 20 of the 20 theories, you would be guaranteed to get the correct one.
So the probability must rise faster in the second case.Last edited by Sermil; 20190730 at 07:06 PM.

20190806, 08:50 PM (ISO 8601)
 Join Date
 Apr 2012
 Gender
Re: Help Me Correct My Probabilistic Misunderstandings
Thanks to everyone for their help! The independent/dependent divide was especially helpful, and seems obvious in retrospect. I was working in the probabilities of scientific theories, and so I completely neglected the logic of the most basic independent probabilities, like dice rolls and coin flips.
"I kinda think I'm not supposed to go along with something that's wrong just to get food."
Spoiler: Play by Post Gamesx Through the Faerie Ring // PL + Progressing // 05/17  Present
x Chrono Isekai // PL + Progressing // 03/20  Present
x Lightning Rails and Whispers of the Vampire's Blade // DM + Completed // 05/17  08/18

20190816, 07:58 AM (ISO 8601)
 Join Date
 Jun 2008
Re: Help Me Correct My Probabilistic Misunderstandings
Think of it this way: When you are determining probabilities and the chances of something to happen, what you really want is the chances that you will keep repeating the trial. If you are rolling a d20 and you get a 20 on the first try, then that's it. You're done with rolling. But if you roll a d20 and you get a 19, then you keep on rolling. So it's the "get anything but a 20" that keeps the rolls going, and it becomes the easiest thing to work with mathematically. Looking at it this way, your chances of not getting a 20 on 2d20 is:
0.95 * 0.95 = 0.9025
And your 90.25% chance of not getting a single 20 is identical to your 9.75% chance of getting at least one 20 in 2d20. It's the exact same situation, just presented slightly differently. All those "(1  x)" things that are getting thrown around are really just formatting; it's a way to present the formula in a way that you can enter the value you have  chances of getting a 20 in one roll  and the value you expect  chances of getting a 20 in all rolls. But the above does give you the same answer, effectively, just with a more direct way of using the numbers. You just need to understand how to look at the probabilities properly to make it easier.
I will note that this trick only works on "simple" probabilities. That is, the chances of one thing happening, only once, out of so many attempts. There is also "What are the chances of getting two 20s when rolling three d20 dice?" which uses permutations to calculate, but that's a bit more complicated to explain and I don't think I could simplify it as easy as the example above. The math isn't necessarily too hard, it's just not as easy of an explanation as just multiplying the chances of not succeeding together.Thank you to zimmerwald1915 for the Gustave avatar.
The full set is here.Spoiler
Air Raccoon avatar provided by Ceika
from the Request an OotS Style Avatar thread
A big thanks to PrinceAquilaDei for the gryphon avatar!
original image

20190816, 10:31 AM (ISO 8601)
 Join Date
 Apr 2013
Re: Help Me Correct My Probabilistic Misunderstandings
The general form is this
P(A or B) = P(A) + P(B)  P(A and B)
Thus for rolling one 20 on two d20:
P(A)=1 /20
P(B)=1 /20
P(A and B): probability of both dice coming up 20
P(A and B)=1 /400
Therefore
P(A or B) = 1/20 + 1/20  1/400
= 9.75%
For any mutually exclusive events P(A and B)=0
The two theories are mutually exclusive, they cannot both be right.
Therefore, the probability of one of the 2 theories being right:
P(A or B) = 1/20 + 1/20  0
= 10%

20190816, 04:07 PM (ISO 8601)
 Join Date
 Apr 2009
 Location
 Germany
Re: Help Me Correct My Probabilistic Misunderstandings
I always think that the most important thing to know about statistics is that the human brain has absolutely no instinctive understanding of probability.
Any time your intuition is correct, or you think something is obvious when it comes to probability, it's because someone taught you the correct answer to the problem at some point.
When you're new at statistics and get things completely wrong, it's not becausw you're bad at statistics. I think pretty much everyone got those things really wrong before they learned the correct math behind it.
And when you want to go deeper into statistics, you have to train yourself to never trust your intuition on anything. Because when its about probability, your brain always wants to lie to you.
Statistical logic is only logical when you've been trained in it.We are not standing on the shoulders of giants, but on very tall tower of other dwarves.
Beneath the Leaves of Kaendor  Writing Sword & Sorcery
Spriggan's Den Heroic Fantasy Roleplaying

20190819, 04:51 PM (ISO 8601)
 Join Date
 Aug 2008
Re: Help Me Correct My Probabilistic Misunderstandings
I wouldn't go that far. It's a learned skill, but you don't necessarily need to be taught the correct answer to every given problem to get a sense of things  numerical intuition is a general thing, statistical intuition is a general thing, and you can have an intuition that is often accurate (albeit not truly reliable, so do the math) just because you've done a bunch of the math before.
I would really like to see a game made by Obryn, Kurald Galain, and Knaight from these forums.
I'm not joking one bit. I would buy the hell out of that.  ChubbyRain
Current Design Project: Legacy, a game of masters and apprentices for two players and a GM.