

Hi all,
I am interested in creating an excel spreadsheet to calculate the probability, given x number of dice, y number of successes required, z number of rerolls as well as blessed/cursed effects, Mandy etc.
I am aware of this link: http://www.arkhamhorrorwiki.com/Probability but it doesn't explain how the numbers are calculated. I would much prefer to have some sort of a formula so that I can test it myself and understand how it works.
So far I understand how the 1success row works although I haven't managed to turn this into a single excel formula. 1. 1/3 = 0.33 2. 0.33 + (1  0.33)*(1/3) = 0.55 3. 0.55 + (1  0.55)*(1/3) = 0.70 etc.
I also haven't figured out how the formulas involving two or more successes in a single roll would work.
On a side note, I saw another thread on BGG that said this could be done using the binomial distribution formula in excel, but I tried this and the numbers didn't match up so not sure what I'm doing wrong there... Eg. BINOMDIST(1,3,1/3,TRUE) = 0.74 (It should be 0.70 as above)
Can anyone help me out with any of this?
Thanks


Dominic Lauke
Germany Görlitz Sachsen
Sponsored by AR147

What is the probability that n = 4 dice with 6 sides (k = 6) to achieve at least one "six"? The probability is calculated as 1 minus the probability to achieve no "six" with 4 dice:
P = 1 – (1 – 1/k)^n
For a success on 5 and 6 you have to take k = 3.


MC Crispy
United Kingdom Basingstoke Hampshire

Jordan Kane wrote: What is the probability that n = 4 dice with 6 sides (k = 6) to achieve at least one "six"? The probability is calculated as 1 minus the probability to achieve no "six" with 4 dice:
P = 1 – (1 – 1/k)^n
For a success on 5 and 6 you have to take k = 3. And for Blessed you would put k=2
Blessed k=2 Normal k=3 Cursed k=6
I'm afraid my probability maths isn't up to the task of 2 or more successes.


Bobby Ramsey
United States Grove City OH
Hush

jaredmason wrote: On a side note, I saw another thread on BGG that said this could be done using the binomial distribution formula in excel, but I tried this and the numbers didn't match up so not sure what I'm doing wrong there... Eg. BINOMDIST(1,3,1/3,TRUE) = 0.74 (It should be 0.70 as above)
For multiple dice, the probabilities are based on the binomial distribution. There is an issue with the example you give, though. BINOMDIST( 1, 3, 1/3, TRUE ) The TRUE parameter here returns the cumulative probability. It is the probability of 0 or 1 successes. To get the probability of exactly 1 success you need BINOMDIST(1,3,1/3,FALSE).
For the probability of multiple successes, that TRUE parameter will be needed. Suppose you need N successes with M dice. The probability of passing the check is:
1  BINOMDIST(N1, M, 1/k, TRUE)
where k is as in the above posts. The BINOMDIST(N1,M,1/k,TRUE) calculates the probability of at most N1 successes in those M dice. 1BINOMDIST(N1,M,1/k,TRUE) calculates the probability of everything else. That is, of at least N successes.


Brian Mc Cabe
United States Arizona
There are those who look at things the way they are and ask why . . . I dream of things that never were and ask why not

The way it was explained to me is that you calculate the odds of failing. With one die, there is a 2/3 chance that the roll will be a failure, or a 33% to succeed. That's pretty simple, but I have to start somewhere.
With two dice 2/3 x 2/3 would be a 4/9 chance to fail, giving a 5/9 chance to succeed, or approximately 56% chance of success.
If you're blessed or blessed with a +1 Skill or Cursed, your odds vary and so will your equation.
Since they're dice, these are just odds. I'm playing a game right now and have gone 0/5 three times. Trying to kill a Formless Spawn, I had three rounds of five dice at 1/5.
That's three hits in thiry dice. :)
Brian


Craig H
United States Missouri

Ah you have pissed off lady luck in someway, L=0 where you divide that whole mess of equations above by L.




Hey, thanks so much guys. All responses so far have been very helpful.
I have got my spreadsheet working with the BINOMDIST function for Base rolls. @Brian, your explanation of it being the calculation of failure (inversed) explains a lot to me, given Bobby's formulas with the 1x... and all.
A few things I am still searching for are: 1. A formula so I can calculate multisuccess BINOMDIST's manually (this is just for my own understanding of what is going on) 2. Some explanation of how to manage additional rolls from clues and card effects. Are these different to if those dice were just part of the original roll? 3. Mandy. How does her reroll effect fit in?
Thanks again!


Bobby Ramsey
United States Grove City OH
Hush

jaredmason wrote: Hey, thanks so much guys. All responses so far have been very helpful.
I have got my spreadsheet working with the BINOMDIST function for Base rolls. @Brian, your explanation of it being the calculation of failure (inversed) explains a lot to me, given Bobby's formulas with the 1x... and all.
A few things I am still searching for are: 1. A formula so I can calculate multisuccess BINOMDIST's manually (this is just for my own understanding of what is going on)
The basic formula for working with a binomial distribution is:
( (n!)/( (k!)((nk)!)) ) * p^k * (1p)^(nk)
(Edit: I think I went a little overboard with parentheses to ensure order of operations... )
This gives you the probability of rolling k successes on n dice, where each die has a probability for success p. ( For completeness, I'll say that the '!' denotes a factorial. 2! = 2*1, 3! = 3*2*1, 4!=4*3*2*1, ... For everything to work out, you need to remember that 0! = 1 )
So, that is if you want exactly k successes. What if you want the probability of no more than k successes? Add up what you get for all smaller numbers of successes.
i.e.: sum( ( (n!)/( (j!)((nj)!)) ) * p^j * (1p)^(nj) ) where j goes from 0 to k.
Then if you want the probability of "k or more successes" (as with skill checks), you first find the probability of (k1) successes as a sum, then take 1  that sum.
Quote: 2. Some explanation of how to manage additional rolls from clues and card effects. Are these different to if those dice were just part of the original roll?
Don't treat those dice as part of the original roll. You have already rolled those, so they do not affect the probability of what you still have to roll. See how many successes you still need after the original roll, and setup a new binomial trial with just these new dice and the new number of successes needed.
Quote: 3. Mandy. How does her reroll effect fit in?
Thanks again!
Mandy's ability allows you to reroll an entire check. By this point you know what the probability was for the original check. It will be the same for the reroll. There is probably some a priori probability you can run using a binomial trial with two experiments (the first check and the reroll check), but after the first is rolled it's back to the way it was before.




I use a little great app for Android I found a few days ago:
https://play.google.com/store/apps/details?id=com.kolita.ark... I tried it at a game last time and it worked great.



