Got any Dice Question? II

[Thrawn4]

Well I do.

(There used to be a thread like this. )

So I've been diving into the debts of probabilty, and I am in over my head. Maybe somebody would be kind enough to enlighten me?

Long story short, at the moment I am tinkering with the WOD mechanics, specifically reducing the target numbers in order to achieve a more realistic outcome. But my math is... worse than I thought.


Premises:
4D10 are rolled (average guy with some experience)
target numer is 4 or higher (meaning a result from 4-10 on a D10 is necessary to get 1 success)
3 successes or more are necessary (decent result as per the default rules)
simplicity: willpower and 1s decreasing successes are ignored for now



So by my estimation, an average person with some experience should have a chance of

0,7 x 0,7 x 0,7 x 1 = 0,343 = 34,3%

of getting the job done?

I am confused, because that is the same probabilty that you have using 3 dice, but if you split the probablities up into all possible outcomes:
0,7 x 0,7 x 0,7 x 0,7 = 0,2401
0,7 x 0,7 x 0,7 x 0,3 = 0,1029
--------
0,343

you have the same result.

I feel like you have to add 34,3% and 24,01% so that the actual probabilty is 58,31%.
But I am very confused.
Any help would be appreciated.

[jayem]

Well I do.

Close,
For the case where you have 3 successes and one failure, you have the case where die 1 is the failure, die 2, is the failure, die 3 is the failure and die 4 is the failure.
So
0,7 x 0,7 x 0,7 x 0,7 = 0,2401
0,3 x 0,7 x 0,7 x 0,7 = 0,1029
0,7 x 0,3 x 0,7 x 0,7 = 0,1029
0,7 x 0,7 x 0,3 x 0,7 = 0,1029
0,7 x 0,7 x 0,7 x 0,3 = 0,1029
--------
0,652

______
Nice to have a thread, (the pretty die thread was also nice, but always fooled my expectations)

Spoiler: More generally (for similar dice)

Show

Firstly we can save the repitition in the rows
0,7 x 0,7 x 0,7 x 0,7=0,7⁴
And more generally the odds of x passes followed by y failures
p^x*(1-p)^y
In the case of p=0.7, x=3, y=1 we get 0,1029

As you'll notice the probability of each of the last 4 is the same. So where we don't care about orders we can easily just repeat that for each combination of x passes and y failures.
Going back to the dice, A,B,C,D. The first dice in our list could be any of them...(if we reverse the problem and say the failed dice (D) could be any of them we get 4 and this is the answer we want), however if we carry on the second could be any of the remaining...etc. So in this case we have 4*3*2*1=24 ways of arranging the die. This as you'll have noticed is not 4. We have overcounted ABCD (PPPF) and CABD (PPPF) are the same so now we have to divide by 3*2*1.

This seems like a waste of time but there exists a calculator button for doing this kind of multiplication, and if we have two fails we need to do something like this anyway.

Going back to the question. We now have
the odds of x passes and y failures
p^x*(1-p)^y * factorial(x+y) / factorial(x) / factorial(y)

The way I have the repeated division is ambiguous,
p^x*(1-p)^y * factorial(x+y) / (factorial(x) * factorial(y))

For the odds of at least x passes, the quickest way is to do each sum individually, exactly like you were doing but with the combinational factor.
If you were doing silly numbers and had no computer you could probably find an approximation with clever tricks.

[EricAlvin]

0.3 * 0.7 * 0.7 * 0.7 | = 0.1029 = 10.29% [First die fails]
+ 0.7 * 0.3 * 0.7 * 0.7 | = 0.1029 = 10.29% (Current Total = 0.2058 = 20.58%) [Second die fails]
+ 0.7 * 0.7 * 0.3 * 0.7 | = 0.1029 = 10.29% (Current Total = 0.3087 = 30.87%) [Third die fails]
+ 0.7 * 0.7 * 0.7 * 0.3 | = 0.1029 = 10.29% (Current Total = 0.4116 = 41.16% Chance of any 1 die failing and 3 die succeeding) [Fourth die fails]
+ 0.7 * 0.7 * 0.7 * 0.7 | = 0.2401 = 24.01% (Final Total - See Below) [No die fails]
======================
0.6517 = 65.17% total chance of success

I was going to type up a longer/better explanation, but it seems jayem posted a good explanation while I was getting this typed/figured out.

[Imbalance]

Uh-oh. What happened to Kevin Cook?

[jayem]

Uh-oh. What happened to Kevin Cook?

I don't know. Monthly posts, so only a little late. I hope it's nothing serious.

(but yes we should probably ask about getting this title changed)

[Thrawn4]

I don't know. Monthly posts, so only a little late. I hope it's nothing serious.

(but yes we should probably ask about getting this title changed)

I thought it was appropriate. But I guess we could change it if it violates nettiquette?

[jayem]

I thought it was appropriate. But I guess we could change it if it violates nettiquette?

Did things make sense on your question? And where do you want to take things.*

I'm not sure on the netiquette (tbh it's what I always want to expect from the other thread, and I'm left thinking where's the question. Before I start appreciating the cool dice)

___When there aren't any other questions
*I'd quite like to know how to halve/double the odds of succeeding on each given system
and the mean/median and mode are always useful.

[Thrawn4]

0.6517 = 65.17% total chance of success

I was going to type up a longer/better explanation, but it seems jayem posted a good explanation while I was getting this typed/figured out.

Nevertheless very helpful, thank you.

Did things make sense on your question? And where do you want to take things.*

Yes they did, thank you very much.

I'm not sure on the netiquette (tbh it's what I always want to expect from the other thread, and I'm left thinking where's the question. Before I start appreciating the cool dice)

Okay, so you are basically concerned because the former thread was about the aesthetics of dice, whereas this one is about probability, and people might be annoyed by that. In other words, keeping the old title is.... probably dicy

Yes, I have no shame.

___When there aren't any other questions
*I'd quite like to know how to halve/double the odds of succeeding on each given system
and the mean/median and mode are always useful.

Not sure how I can help with that one, I am still trying to understand the basics....

But as to my questions:

What if we allow for the expenditure of willpower? That means we have one additional success, and we only need to generate two successes with the dice.

Seeing as we already have 0,652 probability to generate three or more successes, I only have to add the probability of generating two successes (I think):
0,7²x0,3²= 0,0441

but because we have 6 possible outcomes/combinations, we multiply it by six (0,0441x6=0,2646)

which results in (0,2646+0,652=) 91,66 probability.

Right?

Again, thanks in advance for putting up with sub-par maths.

[Jay R]

Let n= the number of dice = 4,
p= the probability of one success = 4,
q = the probability of a failure on one die = p - 1,
C(n,x) = the number of combinations of n things taken c at a time.

0,0081 Zero successes
0,0756 One success.
0,2646 Two successes.
0,4116 Three successes.
0,2401 Four successes.
———-
1,0000 Total.

The probability of at least 3 successes is 0,7517.

The mean = np = 2,8 successes. The median is 3.
The variance = npq = 0,8400.
The standard deviation = sqrt(npq) = approximately 0,916515.

[jayem]

I think the final total read the table backwards?
That looks like the odds of 0,1,2,3 successes or three or less.

Thanks for the stat things, I thought they'd be vaguely usefull (to e.g. make things easier/harder in a controlled way). Though not sure how yet.

[Ooh, Npq for variance. That looks like there's something to play with]