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Message: <blockquote data-quote="AbdulAlhazred" data-source="post: 4991046" data-attributes="member: 82106">True. You roll SuccessesRequired+2 dice and if there are less than 3 fails in those dice then the residual chance of failure is the toss of the final die. In order to consider different character's with different success probabilities is actually simple enough, you just assign different probabilities to each die. It doesn't actually matter which die is that last die either since the whole challenge is essentially order independent.EDIT: Sorry. This is what happens when you think about it for 1 minute too few... On complexity 1 you never need more than 6 tosses to determine if the party succeeded or not. These dice can be thrown in parallel. Thus your code can be parallelized. Assuming you have more than one core on your CPU multiple dice can be tossed at one time, if you have a single core processor then backtracking is equally efficient. So essentially both you and Stalker0 are correct, its just that your way of looking at it can lead to a more efficient code solution if you have the proper hardware. Create a threadpool, drop SuccessesRequired+2 die rolling jobs into the pool. Sum the fails and if its <2 its a failed challenge, otherwise its a successful challenge. On a single core processor it won't be faster than backtracking, but on a multicore machine it should give you a bit quicker answer. That would be the Monte-Carlo method. You can also simply sum the probabilities and get a statistical answer, still uses the same number of dice, which is a binomial distribution. I could work up a piece of code, but then again I'm sure any of us can do that... lol.</blockquote>

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