SR4 Net Hits Probability Distribution Options

[p00ya]

I'm new to SR4's net hits system and I was finding it hard to estimate the probability of meeting particular thresholds on opposed tests. I decided to make these nethits statistical tables as an aid.

I couldn't derive an elegant closed form expression for either the PDF or CDF here. The expected value is the intuitive (n-m)/3. The variance seems to be cubic in n, asymptotically. I generated the exact entries in Mathematica and then rounded them off for presentation. I've only listed positive entries.

You can get a PDF at a point k by taking the difference of vertically adjacent entries.

[Ol' Scratch]

I think it's a case of over-thinking things, personally.

On average, you can usually expect one hit for every three dice you have to throw and a guarantee (under many circumstances) of one hit for every four dice you have via a semi-optional rule. If you have a threshold of 3, you'll want at least 9 dice to throw and if you have 12 or more you can ask to just buy the hits. Simple. And for every three dice below that "requirement," your chance of succeeding lowers by one hit. Do you really need any sort of complicated math beyond that in most circumstances?

I mean, if you have the time to stop the game in order to crunch some numbers, and there's a chance you won't even bother with an action, do you really need to waste the time to crunch those numbers to begin with? Because if it's something important where life and death are on the line, you're probably going to try it regardless of the math. And if it's not, what's the harm in trying it to see how Lady Luck treats you?

[Saint Sithney]

I think it's a case of over-thinking things, personally.

Basically it's as the Doctor says. You should be able to grok your estimated chances quickly enough, and, if it's important, weight them with Edge.

I did however one time calculate the chances of creating a lvl 6 Fake SIN using Forgery from Unwired and capped, depleating dice pools from SR4a. The probability was so astronomically small that it would take Aeons to bang out a single lvl 6 SIN. Though, I didn't factor in the fact that the time between tests is so long that you could probably spend Karma on all 20 rolls and have it refreshed each time...

[p00ya]

On average, you can usually expect one hit for every three dice you have to throw and a guarantee (under many circumstances) of one hit for every four dice you have via a semi-optional rule. If you have a threshold of 3, you'll want at least 9 dice to throw and if you have 12 or more you can ask to just buy the hits. Simple.

Of course it's simple, that's just a binomial distribution. It gets harder for net hits because that's not a threshold test on your DP. The rule of thumb above just isn't useful.

Say you have a DP of 18 and you figure your opponent has a DP of 9. Sure, you can guess that you'll get one hit for every 3 dice E(X)=6 and your opponent will have the same E(Y)=3. Using that logic you probably figure your chances are high that you can get 3 net hits or more. How high? Well, the answer is less than 1%. Because of the high variance, you can't do linear subtraction with the expected values of X and Y and figure that the difference has a good probability.

Edit: the above 'answer' is just wrong. The answer is about 58%, as Dr Funkenstein would expect.

This post has been edited by p00ya: Apr 12 2010, 03:38 PM

[Heath Robinson]

The table seems to be incomplete to me. The probability of a 1 dice pool getting 1 net hit against a 2 dice pool should be listed as ~0.148 by my calculations.

I also have completely different numbers for practically every k=1 entry (i.e. success on an opposed test) that has a value of m greater than 1. Either your calculations are wrong, or mine are.

I think it's a case of over-thinking things, personally.

I don't. Looking at the statistical properties of the RNG is always useful, especially when creating characters.

Did you know that a 1 dice pool wins against a 5 dice pool about 4% of the time? Did you know that a 5 dice pool wins against a 9 dice pool (i.e. the same difference in absolute dice terms as 1 against 5) about 14% of the time? A 9 dice pool beats a 13 dice pool (same dice imbalance as 1 vs 5 again) about 20% of the time.

Knowing your odds is never so simple as comparing your E(X) values. To think otherwise is foolish.

[Ol' Scratch]

I don't see how that can possibly be less than 1%. Just using a random online dice roller, I get the following results from ten sets of rolls.

[ Spoiler ]

Dice Roll: 9d6s5
d6 Results: 2, 6, 4, 2, 5, 4, 3, 5, 1 (Total Successes = 3)
Dice Roll: 18d6s5
d6 Results: 4, 4, 1, 5, 6, 2, 1, 1, 6, 4, 3, 6, 3, 4, 1, 2, 3, 5 (Total Successes = 5)

Dice Roll: 9d6s5
d6 Results: 3, 2, 5, 6, 5, 4, 5, 4, 3 (Total Successes = 4)
Dice Roll: 18d6s5
d6 Results: 4, 4, 4, 4, 2, 5, 5, 2, 5, 5, 6, 3, 1, 5, 5, 3, 5, 1 (Total Successes = 8)

Dice Roll: 9d6s5
d6 Results: 3, 2, 6, 6, 2, 6, 3, 3, 3 (Total Successes = 3)
Dice Roll: 18d6s5
d6 Results: 1, 6, 1, 6, 2, 6, 4, 6, 4, 2, 3, 2, 4, 4, 3, 5, 5, 1 (Total Successes = 6)

Dice Roll: 9d6s5
d6 Results: 3, 5, 1, 4, 5, 5, 4, 5, 6 (Total Successes = 5)
Dice Roll: 18d6s5
d6 Results: 6, 5, 2, 4, 2, 4, 5, 3, 3, 4, 2, 6, 4, 5, 6, 2, 1, 4 (Total Successes = 6)

Dice Roll: 9d6s5
d6 Results: 3, 1, 2, 6, 4, 6, 1, 1, 2 (Total Successes = 2)
Dice Roll: 18d6s5
d6 Results: 3, 3, 4, 4, 5, 5, 5, 4, 3, 4, 5, 6, 1, 4, 2, 5, 3, 1 (Total Successes = 6)

Dice Roll: 9d6s5
d6 Results: 4, 4, 4, 5, 4, 1, 1, 1, 5 (Total Successes = 2)
Dice Roll: 18d6s5
d6 Results: 6, 2, 6, 6, 4, 6, 4, 3, 2, 3, 3, 5, 6, 2, 2, 3, 2, 4 (Total Successes = 6)

Dice Roll: 9d6s5
d6 Results: 3, 6, 2, 3, 2, 6, 2, 1, 2 (Total Successes = 2)
Dice Roll: 18d6s5
d6 Results: 5, 2, 3, 1, 5, 2, 1, 4, 3, 2, 2, 4, 4, 6, 3, 1, 4, 6 (Total Successes = 4)

Dice Roll: 9d6s5
d6 Results: 3, 5, 1, 5, 6, 6, 2, 6, 3 (Total Successes = 5)
Dice Roll: 18d6s5
d6 Results: 2, 4, 4, 1, 1, 5, 6, 2, 5, 5, 5, 2, 6, 4, 4, 6, 6, 3 (Total Successes = 8)

Dice Roll: 9d6s5
d6 Results: 3, 4, 4, 6, 3, 6, 5, 3, 2 (Total Successes = 3)
Dice Roll: 18d6s5
d6 Results: 5, 3, 6, 2, 5, 5, 1, 5, 1, 2, 5, 5, 3, 6, 5, 2, 5, 5 (Total Successes = 11)

Dice Roll: 9d6s5
d6 Results: 6, 1, 1, 6, 1, 5, 5, 4, 6 (Total Successes = 5)
Dice Roll: 18d6s5
d6 Results: 1, 2, 2, 6, 2, 6, 2, 6, 1, 4, 2, 5, 6, 5, 2, 6, 1, 1 (Total Successes = 7)

Not a single time did the 18d6 fail to out roll the 9d6, and only once was it even a close call. I'm no math wiz, not by a long shot, but I'm pretty sure 100% (of the sample) is more than "less than 1%" by quite a huge margin. Are your sure your math is right? Or am I not understanding what you're trying to say?

[Heath Robinson]

I don't see how that can possibly be less than 1%. Just using a random online dice roller, I get the following results from ten sets of rolls.

Not a single time did the 18d6 fail to out roll the 9d6, and only once was it even a close call. I'm no math wiz, not by a long shot, but I'm pretty sure 100% (of the sample) is more than "less than 1%" by quite a huge margin. Are your sure your math is right? Or am I not understanding what you're trying to say?

He said that the chance of getting 3 or more Net Hits was less than 1%.

So, getting 3 or more hits if the opposing DP rolled no hits, getting 4 or more if the opposing DP rolled 1 hit, etc.

[Ol' Scratch]

Knowing your odds is never so simple as comparing your E(X) values. To think otherwise is foolish.

I didn't say it was that simple. I said was it really that important to stop in the middle of a game to calculate your odds when the situation is either 1) so important you're going to try it anyway despite the odds or 2) so indifferent that you may as well try the action anyway just to see what happens. And in most circumstances, you can guestimate if an action will be successful based upon how many dice are being rolled.

I just don't see why an exact percentage number is required or even really helpful. "I can succeed at this action 80% of the time." What does that knowledge actually do for you in practice?

[Ol' Scratch]

He said that the chance of getting 3 or more Net Hits was less than 1%.

So, getting 3 or more hits if the opposing DP rolled no hits, getting 4 or more if the opposing DP rolled 1 hit, etc.

Ah, I guess I missed that part. Still, 3/5 successes in that sample kind of suggests that "less than 1%" isn't accurate. I even spammed the random roller a few times and didn't see a fail rate anywhere close to that number. In fact, it comes a lot closer to matching my "dumb guy" math; he had 9 more dice than his opponent, so he could expect getting those 3 net hits a little more than half the time. No?

And yeah, I just did a whole mess of those rolls (using a different roller, even) and it still came out that way.

I'm not trying to harsh on anyone. I just don't understand where these numbers are coming from or the desired result.

[Faraday]

I just don't see why an exact percentage number is required or even really helpful. "I can succeed at this action 80% of the time." What does that knowledge actually do for you in practice?

Theoretically speaking, it would allow you to optimize your actions if you knew the probability of success for every action you took before taking said action. Not really worth the hassle to get those probabilities though, IMO.

[p00ya]

The table seems to be incomplete to me. The probability of a 1 dice pool getting 1 net hit against a 2 dice pool should be listed as ~0.148 by my calculations.

I get the same, but yes my table is incomplete (it doesn't show negative k values, or where m > n). Your example there would be equilavent to n = 2, m = 1, k = -1, and I get 4/27 for the PDF.

[Heath Robinson]

I didn't say it was that simple. I said was it really that important to stop in the middle of a game to calculate your odds when the situation is either 1) so important you're going to try it anyway despite the odds or 2) so indifferent that you may as well try the action anyway just to see what happens. And in most circumstances, you can guestimate if an action will be successful based upon how many dice are being rolled.

Who cares about doing it in the middle of a game?

[Ascalaphus]

We had a related discussion yesterday;

At what point is it better to spend edge on a roll before rolling (to get exploding 6's and make the pool bigger), or after (to re-roll failed dice)?

[DWC]

We had a related discussion yesterday;

At what point is it better to spend edge on a roll before rolling (to get exploding 6's and make the pool bigger), or after (to re-roll failed dice)?

Obviously, the inflection point moves around based on your Edge score and your dice pool. I did an approximation (only open ended the 6s like 6 times) a little while ago and the results were pretty close to what I expected. If I can dig them up, I'll share them, but it worked out that unless your Edge was big and your dice pool was fairly small you were way better off rerolling failures.

[Mongoose]

In you chart, you say to treat blank entries as 0.000. The n=1,k=1,m=2+ entries are blank, but its obviously possible to score a success with one die when the opposition has 2+ dice. Granted, its only really an issue when both pools are fairly small, but it seems an odd oversight.
Also, you say you only listed positive entries. How could you have calculated any negative probabilities?

[Banaticus]

Say you have a DP of 18 and you figure your opponent has a DP of 9. Sure, you can guess that you'll get one hit for every 3 dice E(X)=6 and your opponent will have the same E(Y)=3. Using that logic you probably figure your chances are high that you can get 3 net hits or more. How high? Well, the answer is less than 1%.

So far, I've come to 2.25%: http://i9.photobucket.com/albums/a66/bubbajoe12345/18to9.png which I don't think is right, it seems like it should be more.
I used the formula (n!/r!(n-r)!)*p^r*(1-p)^(n-r) with n as the number of hits, r the number of rolls and p as 1/3 (1 or 2 on a 6-sided die).

I figured that getting at least one hit would be the same chance as getting at least two hits, plus the chance or rolling only one; rolling at least two was the same as rolling at least three plus two.
n=1,r=1 chance of rolling a 1, this + 2 = at least one
...
n=1,r=9
n=2,r=2 chance of rolling a 2, this + 3 = at least one two
...
n=2,r=9
n=3,r=3 chance of rolling a 3, this + 4 = at least one three
...
n=8,r=8 chance of rolling an 8, this + 9 = at least one eight
n=8,r=9
n=9,r=9 = chance of rolling at least one nine

On the top row of the table are the percentages that "at least" one hit will be rolled with 18 dice, at least two hits, etc.
On the left hand columwn are the percentages that "at least" one hit will be rolled in 9 dice, at least two hits, etc.
In the middle, I multiplied the two percentages together. For instance, there's a 66.20% chance of rolling at least 1 hit in 18 dice and a 66.18% chance of rolling at least 1 hit in 9 dice, so there's a 43.81% chance of rolling at least 1 hit in 9 dice at the same time that at least 1 hit is rolled in 18 dice.
Then I added up all those percentages that the 18 pool has 3 or more net hits over the 9 pool.

I ended up with 2.25%.

[Banaticus]

Say you have a DP of 18 and you figure your opponent has a DP of 9. Sure, you can guess that you'll get one hit for every 3 dice E(X)=6 and your opponent will have the same E(Y)=3. Using that logic you probably figure your chances are high that you can get 3 net hits or more. How high? Well, the answer is less than 1%.

So far, I've come to 2.25%: http://i9.photobucket.com/albums/a66/bubbajoe12345/18to9.png which I don't think is right, it seems like it should be more.
I used the formula (n!/r!(n-r)!)*p^r*(1-p)^(n-r) with n as the number of hits, r the number of rolls and p as 1/3 (1 or 2 on a 6-sided die).

I figured that getting at least one hit would be the same chance as getting at least two hits, plus the chance or rolling only one; rolling at least two was the same as rolling at least three plus two.
n=1,r=1 chance of rolling a 1, this + 2 = at least one
...
n=1,r=9
n=2,r=2 chance of rolling a 2, this + 3 = at least one two
...
n=2,r=9
n=3,r=3 chance of rolling a 3, this + 4 = at least one three
...
n=8,r=8 chance of rolling an 8, this + 9 = at least one eight
n=8,r=9
n=9,r=9 = chance of rolling at least one nine

On the top row of the table are the percentages that "at least" one hit will be rolled with 18 dice, at least two hits, etc.
On the left hand columwn are the percentages that "at least" one hit will be rolled in 9 dice, at least two hits, etc.
In the middle, I multiplied the two percentages together. For instance, there's a 66.20% chance of rolling at least 1 hit in 18 dice and a 66.18% chance of rolling at least 1 hit in 9 dice, so there's a 43.81% chance of rolling at least 1 hit in 9 dice at the same time that at least 1 hit is rolled in 18 dice.
Then I added up all those percentages that the 18 pool has 3 or more net hits over the 9 pool.

I ended up with 2.25%.

[Ol' Scratch]

Both of those numbers are still clearly wrong. Yeah, the samples I included early in the thread were just ten rolls of each, but I kept doing it using two different dice rollers, too. Not once was the result anywhere close to such a tiny percentage. I don't care what some spreadsheet says, either your math has to be wrong or I'm really not understanding what your numbers are supposed to represent. Because, in the real world, your chance of getting 3 or more net hits in a contest between 18d6 vs. 9d6 is not less than 1% or even 2.25%. It's well over 50%. I seriously doubt the numbers will change that much after millions or even billions of rolls. Maybe you're doing it backwards or something, and calculating the chances of the 9d6 winning? I just dunno.

Heck, feel free to try it yourself. The dice syntax is 18d6.hits(5) and 9d6.hits(5). You can roll up to 50 times each with that roller. The sample I used earlier came from a private forum I'm on, so I can't really link to that roller (you have to be a member and an active member of a game in order to use it). The spoiler below contains yet another sample using the linked roller, just to drive the fact home some more.

[ Spoiler ]

The 18d6 Set
18d6.hits(5) → [4,3,5,1,5,3,2,1,4,2,4,2,1,6,5,3,4,6] = (5)
18d6.hits(5) → [1,3,6,2,3,4,1,6,6,6,3,6,2,6,3,5,5,3] = (8)
18d6.hits(5) → [6,1,3,5,5,5,1,3,3,3,1,2,2,1,4,3,4,4] = (4)
18d6.hits(5) → [6,4,5,3,6,6,1,3,4,3,3,5,4,3,5,3,3,6] = (7)
18d6.hits(5) → [3,5,2,5,5,6,1,1,4,5,3,2,1,1,6,6,2,1] = (7)
18d6.hits(5) → [1,3,6,6,1,1,5,1,4,6,5,1,4,4,6,1,1,4] = (6)
18d6.hits(5) → [4,4,6,3,5,6,2,5,2,2,4,2,2,3,6,3,2,3] = (5)
18d6.hits(5) → [6,4,6,5,5,4,6,6,3,1,3,4,2,1,5,3,1,4] = (7)
18d6.hits(5) → [2,4,5,2,6,1,4,1,4,4,5,2,6,3,1,1,4,4] = (4)
18d6.hits(5) → [2,1,1,1,3,6,4,3,2,3,1,4,4,1,4,3,3,4] = (1)
18d6.hits(5) → [1,4,2,4,3,3,6,5,3,6,5,4,5,5,3,1,4,1] = (6)
18d6.hits(5) → [6,2,6,4,3,5,6,5,1,2,3,2,3,5,6,5,1,4] = (8)
18d6.hits(5) → [4,6,1,2,1,4,5,3,5,3,5,3,5,2,2,3,4,6] = (6)
18d6.hits(5) → [3,6,5,1,2,6,1,3,5,3,2,5,1,4,2,5,4,1] = (6)
18d6.hits(5) → [3,3,6,6,6,6,4,2,1,3,1,3,5,4,3,6,6,3] = (7)
18d6.hits(5) → [2,4,3,2,3,2,5,4,3,3,5,1,5,1,6,4,1,1] = (4)
18d6.hits(5) → [1,4,3,6,6,4,2,6,3,1,1,4,6,2,4,1,4,2] = (4)
18d6.hits(5) → [2,1,2,2,6,3,6,5,1,3,1,1,2,2,6,1,2,5] = (5)
18d6.hits(5) → [3,2,4,3,4,4,1,4,6,3,5,4,6,6,2,1,2,4] = (4)
18d6.hits(5) → [3,1,6,1,2,5,6,1,2,6,3,2,6,2,2,4,6,2] = (6)
18d6.hits(5) → [3,6,1,6,5,6,6,1,1,5,6,4,2,6,3,1,1,4] = (8)
18d6.hits(5) → [2,5,1,4,4,4,1,5,6,2,2,4,2,1,2,4,4,6] = (4)
18d6.hits(5) → [3,4,2,3,5,5,5,6,2,5,5,5,6,1,2,3,6,6] = (10)
18d6.hits(5) → [3,3,5,6,6,5,5,3,3,5,1,5,5,2,1,6,3,3] = (9)
18d6.hits(5) → [6,1,1,1,5,6,6,3,4,4,2,1,4,6,1,4,5,1] = (6)
18d6.hits(5) → [5,1,1,4,1,1,3,3,2,6,2,5,5,6,6,1,2,4] = (6)
18d6.hits(5) → [2,3,4,2,3,5,5,5,3,3,5,2,6,4,2,6,2,1] = (6)
18d6.hits(5) → [5,6,4,4,4,1,4,6,3,6,2,1,2,6,2,6,4,6] = (7)
18d6.hits(5) → [2,6,5,1,2,2,2,5,6,4,3,1,1,6,2,2,6,2] = (6)
18d6.hits(5) → [6,2,4,6,2,5,4,4,4,3,3,5,4,6,1,2,1,1] = (5)
18d6.hits(5) → [6,4,5,3,2,2,6,3,3,4,5,5,6,6,4,6,1,2] = (8)
18d6.hits(5) → [5,1,3,2,6,1,1,6,1,1,4,2,5,2,4,3,3,2] = (4)
18d6.hits(5) → [5,4,2,5,6,5,1,1,5,2,1,6,5,6,6,1,2,1] = (9)
18d6.hits(5) → [3,1,5,4,1,5,3,3,4,1,6,3,4,6,5,2,5,4] = (6)
18d6.hits(5) → [6,1,3,6,2,6,5,1,2,6,5,6,1,1,2,5,4,4] = (8)
18d6.hits(5) → [2,3,1,3,2,3,6,4,5,4,2,5,3,4,3,1,4,3] = (3)
18d6.hits(5) → [1,3,4,2,5,6,1,2,1,6,5,4,6,5,1,6,6,5] = (9)
18d6.hits(5) → [6,5,6,3,3,5,6,5,3,3,3,2,3,3,2,3,3,5] = (7)
18d6.hits(5) → [3,3,4,5,4,5,4,3,6,5,1,2,6,6,2,1,6,2] = (7)
18d6.hits(5) → [3,2,5,3,2,3,3,5,1,3,1,2,1,3,3,5,5,3] = (4)
18d6.hits(5) → [6,1,5,2,2,4,4,6,5,6,6,3,3,1,4,5,4,1] = (7)
18d6.hits(5) → [6,2,2,2,4,4,3,1,1,2,2,2,1,4,6,1,4,3] = (2)
18d6.hits(5) → [5,6,3,6,3,1,6,5,5,3,3,3,4,3,3,6,4,2] = (7)
18d6.hits(5) → [1,5,6,3,3,2,3,6,4,3,1,4,5,4,4,5,6,2] = (6)
18d6.hits(5) → [4,3,2,1,1,1,6,4,6,1,3,3,6,1,1,5,1,5] = (5)
18d6.hits(5) → [1,3,6,3,2,3,1,2,3,1,3,3,6,2,1,1,1,3] = (2)
18d6.hits(5) → [4,3,5,5,6,2,1,4,2,6,2,5,2,4,2,6,4,6] = (7)
18d6.hits(5) → [2,2,2,3,1,3,4,4,2,5,4,2,2,5,5,5,3,5] = (5)
18d6.hits(5) → [3,4,2,4,1,2,6,3,1,2,3,1,2,6,2,6,6,3] = (4)
18d6.hits(5) → [1,1,4,1,6,4,1,5,2,5,6,2,4,6,5,3,1,4] = (6)

The 9d6 Set
9d6.hits(5) → [2,3,2,6,3,6,3,2,2] = (2)
9d6.hits(5) → [6,1,5,6,1,4,3,2,5] = (4)
9d6.hits(5) → [6,1,6,4,6,6,6,4,2] = (5)
9d6.hits(5) → [2,4,1,1,4,4,4,3,5] = (1)
9d6.hits(5) → [6,1,5,4,3,1,1,4,3] = (2)
9d6.hits(5) → [4,3,6,2,2,3,4,4,3] = (1)
9d6.hits(5) → [5,1,4,2,3,4,3,5,2] = (2)
9d6.hits(5) → [3,6,1,6,4,1,2,3,1] = (2)
9d6.hits(5) → [1,4,3,3,2,2,5,2,2] = (1)
9d6.hits(5) → [4,2,5,5,5,6,3,1,1] = (4)
9d6.hits(5) → [6,2,5,6,2,3,4,5,4] = (4)
9d6.hits(5) → [5,1,1,5,3,4,6,2,2] = (3)
9d6.hits(5) → [4,3,6,5,5,4,1,1,5] = (4)
9d6.hits(5) → [2,1,4,4,4,1,5,1,1] = (1)
9d6.hits(5) → [2,2,3,1,4,2,2,6,2] = (1)
9d6.hits(5) → [2,5,5,1,4,4,3,4,1] = (2)
9d6.hits(5) → [5,6,2,2,5,2,4,5,6] = (5)
9d6.hits(5) → [3,3,6,3,3,3,1,4,1] = (1)
9d6.hits(5) → [1,6,3,3,1,6,2,3,4] = (2)
9d6.hits(5) → [1,5,1,6,5,5,4,3,3] = (4)
9d6.hits(5) → [3,3,1,1,6,4,4,3,1] = (1)
9d6.hits(5) → [2,6,6,5,4,1,2,3,4] = (3)
9d6.hits(5) → [1,2,5,6,3,5,4,1,5] = (4)
9d6.hits(5) → [2,5,3,3,6,5,2,3,3] = (3)
9d6.hits(5) → [5,1,4,1,2,2,3,3,1] = (1)
9d6.hits(5) → [3,4,1,1,1,3,5,2,5] = (2)
9d6.hits(5) → [2,2,5,2,2,4,2,5,5] = (3)
9d6.hits(5) → [3,6,4,6,6,4,6,2,2] = (4)
9d6.hits(5) → [4,3,5,2,1,3,5,2,4] = (2)
9d6.hits(5) → [3,2,4,2,3,2,1,4,3] = (0)
9d6.hits(5) → [1,1,2,5,6,5,6,1,5] = (5)
9d6.hits(5) → [4,3,4,2,5,6,3,1,5] = (3)
9d6.hits(5) → [1,4,2,4,1,3,6,2,3] = (1)
9d6.hits(5) → [1,5,3,2,1,6,1,2,5] = (3)
9d6.hits(5) → [6,3,6,3,4,4,3,2,5] = (3)
9d6.hits(5) → [4,6,3,5,5,2,4,6,5] = (5)
9d6.hits(5) → [6,1,1,3,5,6,5,1,1] = (4)
9d6.hits(5) → [2,5,5,5,1,6,5,6,4] = (6)
9d6.hits(5) → [6,5,3,6,5,4,2,6,3] = (5)
9d6.hits(5) → [2,1,2,5,6,2,6,3,3] = (3)
9d6.hits(5) → [2,2,1,2,6,5,4,2,6] = (3)
9d6.hits(5) → [3,2,2,5,5,3,5,2,5] = (4)
9d6.hits(5) → [2,6,4,2,2,5,2,5,5] = (4)
9d6.hits(5) → [4,4,4,3,2,2,3,3,5] = (1)
9d6.hits(5) → [6,6,6,6,3,4,1,5,4] = (5)
9d6.hits(5) → [3,5,2,4,4,3,6,2,4] = (2)
9d6.hits(5) → [1,4,3,6,3,5,3,2,3] = (2)
9d6.hits(5) → [3,5,3,3,2,3,6,3,5] = (3)
9d6.hits(5) → [2,2,4,3,2,2,1,3,3] = (0)
9d6.hits(5) → [2,5,2,4,4,6,5,2,5] = (4)

[sn0mm1s]

I calculate that 18 dice should get 3 or more net hits than 9 dice about 57-58% of the time

[p00ya]

Okay so I screwed up the layout of my tables (I've uploaded a new PDF but all that I've changed is the position of the k and the m at the top left). *curses mathematica's export formats*

So the actual value for n = 18, m = 9, k = 3 is 0.578. This seems a lot more correct because you expect the CDF to be 0.5 either side of the expected value.

Sorry for the confusion, the intuitive guess is a good one. I just saw the tables and went "wow that's surprising" when I first read them (the wrong way around).

[Banaticus]

I'm sorry, but I'm not in the mood to trust tables unless the method used to generate them is listed, especially when they just changed (no offense intended). (IMG:style_emoticons/default/wink.gif)

So, how are you coming up with that answer?

[Draco18s]

I'm sorry, but I'm not in the mood to trust tables unless the method used to generate them is listed, especially when they just changed (no offense intended). (IMG:style_emoticons/default/wink.gif)

So, how are you coming up with that answer?

His answer is correct, he just read 0.5 as "less than 1%" rather than the correct 50%.

See: Verizon can't tell the difference between 0.002 dollars and 0.002 cents.

[sn0mm1s]

I'm sorry, but I'm not in the mood to trust tables unless the method used to generate them is listed, especially when they just changed (no offense intended). (IMG:style_emoticons/default/wink.gif)

So, how are you coming up with that answer?

Well, I can tell from your post that your numbers are off. The odds of at least 1 hit in 9 dice is equal to 1 - odds of no hits.

The odds of no hits in 9 dice is: (9!/9!0!) * [(2/3)^9] * [(1/3)^0].
The first and last parts of that equation = 1 so, the odds of no hits can be simplified to (2/3)^9 = 2.6%
So the odds of rolling at least 1 hit in 9 dice = 97.4%

p00ya seems to have corrected his numbers and his result for 18 dice vs. 9 dice match my results so I am pretty sure his table is now correct.

[Draco18s]

Well, I can tell from your post that your numbers are off. The odds of at least 1 hit in 9 dice is equal to 1 - odds of no hits.

1 or more hits is not one or more net hits.

[sn0mm1s]

1 or more hits is not one or more net hits.

No kidding? Read his post and look at his linked image - his math is wrong. I posted the correct number several posts up - before p00ya.

This is what he said:

For instance, there's a 66.20% chance of rolling at least 1 hit in 18 dice and a 66.18% chance of rolling at least 1 hit in 9 dice, so there's a 43.81% chance of rolling at least 1 hit in 9 dice at the same time that at least 1 hit is rolled in 18 dice.

Which is not correct. I was correcting that math.
There is a 97.4% chance of rolling at least 1 hit in 9 dice... not 66.18%
There is a 99.9% chance of rolling at least 1 hit in 18 dice... not 66.20%