Probability Question, game related

>>7948034
shit, clarification

A player rolls a number of 10-sided dice, any die that comes up *EQUAL TO OR* higher than the "target number" is a success

Also, can anyone explain to me the (11-TN)/10^Number of Dice? I understand that (11-TN)/10 gives the probabilities of the 10 sided dice, but the exponent is all thats required to calculate extra dice?

File: binom dist discrete random variable.gif (723B, 244x49px) Image search: [Google]

723B, 244x49px

>>7948042

What I believe you are looking for with

1-(Successes_required * (((11-Target Number)/10)^Dicepool))

Is described by pic related formula, the 'binomial distribution' for a discrete random variable. This describes, in your dice roll example, the probability of x successes out of n rolls, where p is the probability of a success and q is the probability of failure, which is 1-p.

For this question, "how likely is it that B loses", we first need to look at the probability of A getting successes and B getting failures.


"If player A rolls eight dice against TN 7, and player B rolls eight dice against a TN 9, how likely is it that B loses? Similarly, if B rolls against TN 5, how likely is it that he succeeds?"

[math]p_{A}(1)={1\choose1}(\frac{4}{10})^1(\frac{6}{10})^{1-1}=\frac{4}{10}[/math]


[math]p_{B}(0)={1\choose0}(\frac{2}{10})^0(\frac{8}{10})^{1-0}=\frac{8}{10}[/math]

To spare you the long answer for 'why', because the two rolls are independent events, the probability of both occurring is given by [math]P_{A}(x)\cdot P_{B}(x)=\frac{4}{10}\cdot\frac{8}{10}[/math].

This is your player A succeeding TN 7 and player B failing TN9. The B succeeding TN 5 is given in the same manner.

You probably noticed how unnecessary some of the terms of this formula are, and that's because we're dealing with single roll against single roll. In the event of

"Player A gets 2 successes over the target number, then B needs to get at least 2 or more successes on their end."

you need those seemingly unnecessary terms.

File: binomial-distribution.jpg (6KB, 273x98px) Image search: [Google]

6KB, 273x98px

>>7948329
Here's a non-shit picture.

>>7948329
>>7948331
Shit. I actually might be misunderstanding because I don't play table-top roleplaying games, but what I posted will work regardless. When you say

"So if Player A gets 2 successes over the target number, then B needs to get at least 2 or more successes on their end."

If you mean that Player A must get 2 successes in two rolls, consecutively, you can probably see that [math]{n\choose x}[/math] and [math]q^{n-x}[/math] will both reduce to 1, meaning the probability will simply be [math]p^x[/math]. As said in my previous post, this is due to the fact that the two rolls are independent events, so the probability is defined as [math]p\cdot p[/math] for the two-consecutive roll example.

>>7948329
>>7948331

Hey wow. Thanks for the reply.

I have to say... I really don't understand this math.

Let me spell it out in game terms to maybe help answer your questions?

Player A has a Dodge "Dice Pool" of say 7 dice. Player B has an Attack "Dice Pool" of 7 dice. If they both have target number 7 (any die roll of 7 or more is considered 1 success), then they're evenly matched.

BUT, some weapons have higher or lower Target Numbers. A hard to use weapon, might have a Target Number 9, an easy to use might have a Target Number 5. The Target Numbers for weapons range between 5 and 9. The reason for this is that high Target Number weapons deal more "levels of damage". I realize the cost and benefit of weighing the situation of another outcome complicates things. Suffice to say, you want to deal more damage, but you don't want to miss attacks. So lets just talk about attacking and dodging.

Now, if the Dodger gets more successes than the attacker, the attack misses.

SO, given two evently matched opponents (same dice pool), how much of a bad idea is it to use the Target Number 9 weapon. How advantagous is it to use the Target Number 5 weapon?

More Questions:
Does the math change at exceptionally large numbers of dice pools? Say I'm rolling 20 attack dice and the dodger rolls 20 dodge dice. How do the different Target Numbers play then?

Say the attacker is more skilful, has 15 dice, and the dodger only has 7 dice to dodge. how do the Target Numbers stack up in these kinds of situations.

>>7948354
Also, sorry for my lack of probability knowledge. I realize terms like "how advantagous is it" isn't very specific.

Essentially, it boils down to this. I'm worried that a Target Number of 9 will make attack successes so unlikely, that those weapons should never be used and that Target Number 5 weapons are so good they should always be used.

And I'm also curious to how larger numbers of dice play into this. I realize that two 10-sided dice being rolled have a high variability, while twenty of those same dice will lead to very predictable results. I think this is called Standard of Deviation? I dropped out of AP Statistics, forgive me, I'm a Painting/Art History graduate.

>>7948354
So, this kind of brings into play 'expected values' which boils down to the average. It also turns the formula into something else when you bring in stuff like damage for success. So in your example of 7 dice, that is, I assume, 7 rolls. So the function

[math]P(x)={7\choose x}p^xq^{7-x}[/math]

Of course, this the probability of success, p, and failure, q, will depend on your target value. Let's judge in your example a Target Number 9 weapon.

[math]P(x)={7\choose x}(\frac{2}{10})^x(\frac{8}{10})^{7-x}[/math]. I won't go through the 'proof' of why it is, but this type of probability distribution (binomial) has an expected value of [math]np[/math]. Meaning, the expected value for a TN9 in 7 is [math]\frac{2}{10}\cdot7[/math]. This means that the average amount of successes in 7 rolls for TN9 will be [math]\frac{14}{10}=1\frac{2}{5}[/math].

Now, when we're judging something like damage, I would need to know exactly how it works, but if your total damage in 7 rolls for TN9 is like (20)x where x is number of successes, and TN5 is about (10)x, the expected value (mean) for the two compared total damages would be

[math]E_{TN9}(Y)=20\cdot\frac{14}{10}=28[/math]
[math]E_{TN5}(T)=10\cdot7\cdot\frac{6}{10}=42[/math]

If you don't have enough knowledge in probability, I can't exactly explain why this is the case, but it IS provably the case.

So what this says is, the average total damage for using the lower TN weapon is going to be higher, meaning it's more advantageous to use it. Of course, these are my bullshitted numbers for damage, so it depends on what the actual numbers are. Essentially, in my numbers example, many people would consider it smarter to use TN5 because on average it will do more total damage in 7 rolls.

So, Attack Dice Pool vs Dodge Dice Pool is the first step of an attack. If the attacker has more "successes" than the dodger, then two more rolls are made. The Attacker rolls damage and the defender rolls his "soak" or protection. These values are balanced with each other. Both are always against Target Number 7. And the Attacker has much more opportunity to gain extra damage dice on this roll, than the defender has possibilities of getting extra "soak" protection dice.

So it really boils down to the Attack vs Dodge roll, not the Damage vs Soak roll. If you're hitting, you're winning.

What concerns me is the variable target numbers on different weapons. Given the game effects of taking damage (which includes lowering all future dice pool rolls), you want to hit first, even if lightly. That much is clear.

So, can you give me a rough kind of "estimate" kind of a dumbing down that a statistician would give a journalist? Are Target Number 5 weapons X times better than Target Number 9 weapons?

Can you tell me something like "If you've only got 2 dice to roll, don't use a TN 9" or "if you've got 20 dice to roll, TN 9 is fine"?

>>7948360
So, when you're talking about standard deviation, I'm more familliar using variance, which is just the SD squared. They both represent the 'spread' around the average, meaning how far distributed various event's values can be from the mean. For this binomial distribution the variance is defined by [math]npq[/math]. Calculating the variance of total damage is a little different, but I can just type it out for my example.

[math]Var_{TN9}(Y)=(20)^{2}(7)(\frac{2}{10})(\frac{8}{10})=448[/math]
[math]Var_{TN5}(Y)=(10)^{2}(7)(\frac{6}{10})(\frac{4}{10})=168[/math]

This means that the TN5 total damage output will typically be closer to your mean.

>>7948414
Beginning to lose me in terms of game mechanics, sorry familia. What I can say is that the TN5 will on average hit more times during any amount of rolls based on [math]np[/math], but comparing it to TN9 depends on the individual damage. If TN9 damage is higher to a certain extent than TN5, they may be considered equally optimal from a probability point of view.

>>7948422
Hey thanks for your help.

These equations you're giving me, with the [math] [/math] tags around them. Is this some special text formating code? Should I enter these in somewhere to see something else?

File: 0504f4082a8f53ee4e44e1a988504599.png (16KB, 860x162px) Image search: [Google]

16KB, 860x162px

>>7948440
wait, is it appearing to you as

[ math ]Blah blah blah[ /math ] instead of pic related?

I thought it's supposed to autoformat on /sci/. Sorry about that if it contributed to confusion.

>>7948445
I'm new to /sci/ just started reading the sticky about JSMath and Latex? Do I need to download a plugin or a program or something?

You're image really helps. I now see the different variances of TN 9 =448 and TN 5=168.

So lower target numbers yield more predictable results. That makes sense to me.

>>7948414
>>7948440
Sorry familia, gotta study for test now. Hopefully someone with more knowledge of the game and how formatting on /sci/ works will be able to help out more. I don't post enough to know if you need a plugin, I figured it was built in or something.

https://sites.google.com/site/scienceandmathguide/other/-sci-infographics/joseflatex.png?attredirects=0

^/sci/ LaTeX guide from the sticky.

>>7948445
yeah. I just see the code, not the pretty equations.

>>7948458
Thanks again. You've been very helpful.

>>7948440
Learn LaTeX, it works wonders.

\frac{a}{b} = [math]\frac{a}{b}[/math]
\int_{a}^{b} = [math]\int_{a}^{b}[/math]
\oint = [math]\oint[/math]
\sum_{n=a}^{b} = [math]\sum_{n=a}^{b}[/math]
\lim_{x \rightarrow c} = [math]\lim_{x \rightarrow c}[/math]
\mathbf{R} = [math]\mathbf{R}[/math]
Trig functions: \sin, \cos, \tan, \csc, \sec, \cot, etc., you get it.
\\ = new line
&X = align X (can be anything) to other stuff with & before them

>>7948455
If you're on an iphone, enable javascript, either that or try a different browser because JSMath works on all of my PCs and mobile devices without needing any plugins.

>>7948504
do the following posts display correctly for you?
>>7948422
>>7948388

your post at >>7948497
looks great to me

>>7948497
\frac{10}{20}

\frac{10}{20} = frac{5}{10}
testing

>>7948516
FAIL

testing again
[math]\frac{10}{20}[/math]

>>7948509
No.
Sci LaTeX is retarded and will not render if you don't add whitespaces between the end of curly brackets.

Here's my attempt at fixing:
[math]Var_{TN9} (Y) = (20)^{2} (7)( \frac{2}{10} )( \frac{8}{10} ) = 448[/math]
[math]Var_{TN5} (Y) = (10)^{2} (7)( \frac{6}{10} )( \frac{4}{10} ) = 168[/math]

Protip: use \left( ... \right) for parenthesis to autosize. it looks prettier, and yes it also works with | .. | and [ .. ].

>>7948360
chance of getting S successes with R rolls and T target number:

p(S) = (R choose S) (1-T/10)^S (T/10)^(R-S)

chance Sa < Sd:

P = sum from Sa = 1 to 9 of p(Sa) (sum from Sd = Sa+1 to 10 of p(Sd))

To answer your example it's best to implement this is an excel sheet. The answer is that the dodger will doge the attack (have more successes than the attacker) 93% of the time.

Here is the spreadsheet which can be modified for any variables you want (but if rolls are more than 10 you need to extend the rows):

https://docs.google.com/spreadsheets/d/1Q3jCHOZdmjkW4QmzvPdgWO2QdZncslSmYlctYA3ibZ4/edit?usp=sharing

>>7948414
Guy who made the spreadsheet here. I'm confused, is the attack successful if the attacker has more successes than the dodger or is the attack successful when it has more than OR the same amount of successes? In other words, what happens when the number of successes is tied? The spreadsheet is set so that that the attacker wins in a tie. If the dodger wins ties then you just have to switch "attacker" with "dodger" in the spreadsheet.

>>7948669
OP here. If there is a tie, the Dodger Wins.

The way it actualy works is that each success on a dodge "cancels out" a success on the attack. So if the attacker gets 3 successful rolls, and the dodger gets 3, the attack misses.

>>7949216
OK, I changed the spreadsheet to reflect this

>>7949216
If you were using the spreadsheet to record values when it was set to have the attacker win in ties, just do 1-chance of dodging to get the real chance.

OK, I also found a few bugs that affected results with high amounts of dice. Everything should be fixed now and I tidied up the formatting so it's easier to use. I might use this to make charts so that you can just lookup the chance of hitting given dodger's dice and attacker's dice and target. It's kind of a pain in the ass to do that though.

>>7949382
Hey, just got off work from my dayjob. Thanks again for all this help.

Don't worry about making the charts. I can take care of that bit.

This will be very helpful for me and my friends as we begin to adjust the rules of this game.

--
I hesitate to ask you for more, but if you're enjoying it, there is a second part you could work into the sheet, now that we have an attack/dodge calculator.

If an attack is a "hit" (the dodger fails to dodge, has fewer successes than the attacker), a second set of rolls occurs.

The attacker rolls Damage and the defender rolls Soak. These rolls are also "dice pools". An attacker might roll anywhere from 1 to 30 dice. The defender might roll anywhere from 1 to 30 dice of "Soak" The "Target Numbers" for both of these dice pools is typically 7, though that could change.

Each success the defender gets on his soak roll, counts against the number of successes the attacker gets. So if the Attacker rolls 7 dice, gets 4 successes, and the defender rolls 7 dice and gets 4 successes, then no damage is done. If the attacker rolls 7 dice, and gets 4 successes and the defender rolls 7 dice and gets 2 successes, then those 2 are subtracted from the total successes of the damage roll, in this case 4-2=0.

As I stated in another post, the attacker might pick a weapon with a high Target Number, say 9, but with a higher damage bonus. For instance a Chainsaw might have target number 9, but deal +6 damage, where as a small knife might be target number 5 but deal +1 damage.

To be clear, a weapons "target number" never affects the damage dice pool, only the attack dice pool. so that chainsaw with a target number 9, only has a target of 9 on the attack. After it hits, the damage dice pool is measured against a universal target number of 7.

>>7949549
OK, so this is pretty much the same procedure as calculating the hit chance, the only difference is that it gives us an expected damage value. So what I need to know is, what is the affect of the attacker having a certain number of successes above the defender? Is that just multiplied by the damage of his weapon to get the total damage of the attack?

>>7949611
Well this gets in the the kind of complex mechanics of the game.

The Dice Pools are made of other combined "attributes".

So a character will have an Strength Rating, Dexterity Rating, Stamina Rating. But Also a Melee Skill and Dodge Skill, as well as specifc weapons and armor.

Here's an example of a combat round.

player A attacks player B. Player B chooses to dodge the attack. pA rolls Dex+Melee against pB Dex+Dodge. (this is the part we already did.

If its a hit, pA then rolls Strength+Weapon Damage against pB Stamina+Armor. These are almost always against a Target Number 7.

However many more successes the pA gets over pB's is the damage dealt.

Strength, Dexterity and Stamina all range from 1 to 5. Dodge and Melee all range from 0 to 5.

Armor Dice and Weapon Dice might range from 0 to 5 or 6.

>>7949645
OK, so you're saying the raw number of successes is the damage?

Big Shout Out to /sci/ for helping me in my inquiry. For any interested, here is the calculator that a /sci/ poster has built.

https://docs.google.com/spreadsheets/d/1Y5u4xkZRynH-Kf8VWyQEaGdw0sBZLdWaEDflTs0D1-Y/edit?usp=sharing