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WOIN probabilities of success


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TheHirumaChico

Villager
I just downloaded it, and I know it will be tremendously helpful for me, many thanks CitH! I've been using an online dice calculator up until now, but having this as a handy reference is great!
 

Morrus

Well, that was fun
Staff member
It’s pretty cool. But the chances of rolling 22 on 10d6, while high, aren’t 100%!
 






Cat in the Hat

Villager
It’s pretty cool. But the chances of rolling 22 on 10d6, while high, aren’t 100%!
Yeah I should have mentioned that it it only accurate to +/- 1%. But rolling 22 on 10d6 is closer to 100% than 99% once you take into account an auto success from three sixes
 

Cat in the Hat

Villager
It does not include auto-successes. I'm not even sure how to calculate that but it's definitely something I'll look into.
I calculated it by treating the chance of a normal success and the chance of three sixes as independent events. Which they’re not. Which is why it’s only accurate to +/- 1%
 

The principle of how to combine the two ways of succeeding is easy basic stats. The actual implementation is going to be complicated and fiddly

(P Success by rolling more than the target) + ((P Chance of Failure by Rolling less than the Target)*(P Chance of success by rolling 3 6's on some of the dice combinations that failed))

For 22 on 10d6 though the chance of success by rolling 3 6's on some of the dice combinations that failed is 0. If 3 of the dice add up to 18 the minimum total is 25 if all the other dice roll 1's

Which means for any number of dice N and target T Where T <= 18 + N-3 then you can ignore the chance of success from 3 6'S
Likewise whne T>6N then you can ignore the chance of success from rolling more than T (That one is obvious I admit)
The inbetween cases are tricky as not all results containing 3 6's matter due some of them being normal successes, It may be easier to calculate as

(P Chance of success by rolling 3 6's on some of the dice combinations that failed)
+ ((P Chance of Failure by Rolling less than the Target)*(P Success by rolling more than the target))
 

TheHirumaChico

Villager
Maybe this is the right formula?
The probability of the union of two events can be obtained by adding the individual probabilities and subtracting the probability of their intersection: P(A∪B)=P(A)+P(B)−P(A∩B). [Additive and Multiplicative Rules for Probability]
Seems the tricky part is calculating P(A∩B), while P(A) and P(B) seem pretty straightforward. I'm no math whiz, so I haven't figured out how to determine P(A∩B) yet.
 

I don't think thats quite the right set of maths. However this is nearly 40 years to late for me to do the maths for this, back when I left school this would have been about half an hours work.Know I would have to relearn everything I have forgotten
 

Mythological Figures & Maleficent Monsters

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