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Message: <blockquote data-quote="Staffan" data-source="post: 9566620" data-attributes="member: 907">This gets into some gnarly math. It all flows from fairly simple premises, but there's a lot going on.Let's first start with the probability on a single die. If the probability of a success is P, then the probability of failure Q = 1-P. So on a target number of 4, P is 0.7 and Q is 0.3.Expanding to two dice, we have three possible results: 0, 1, or 2 successes. These probabilities are as follows:0: Q^2 = 0.3^2 = 0.091: This is a success on either one of the two dice, so it's P*Q + Q*P = 0.7*0.3 + 0.3*0.7 = 0.21 + 0.21 = 0.42.2: P^2 = 0.7^2 = 0.49.Going to three dice, things are getting complicated, but we'll soldier on:0: Q^3 = 0.027. We'll hold off on 1 and 2 because that starts to get complicated, and go straight to3: P^3 = 0.3431: We want one failure and two successes. So this can be either PQQ, QPQ, or QQP. These are all the same result, so we get 3*(0.3*0.3*0.7) = 3*(0.063) = 0.189.2: Same reasoning as above, but with two P and one Q. 3*(0.3*0.7*0.7) = 3*(0.147) = 0.441With four dice, the options for 1-3 successes are:1: PQQQ, QPQQ, QQPQ, QQQP.2: PPQQ, PQPQ, PQQP, QPPQ, QPQP, QQPP3: PPPQ, PPQP, PQPP, QPPPThis is becoming ridiculous. We need a more generic formula. What we need to know is: if you have n dice with k successes and n-k failures, how many ways are there to arrange those?Well, if you have n of anything, and want to put those in any order, you have n options for what to put in the first slot, n-1 for the second, and so on. The mathematical term for this is a factorial, and is usually written n!But in this case, there are a number of options that are functionally the same. So what you need to do is to divide the result by the number of ways to order each of the sub-divisions. In other words, if you want to know how many ways you can pick 3 out of 5, you take 5! (120) and divide it by 3! (6) and 2! (2), for 120/(6*2) = 120/12 = 10. This is usually written as C(n, k) or:[ATTACH=full]394348[/ATTACH]So for 7 dice, you get:0: 1 combination of Q^71: 7 combinations of P*Q^62: 21 combinations of P^2*Q^53: 35 combinations of P^3*Q^44: 35 combinations of P^4*Q^35: 21 combinations of P^5*Q^26: 7 combinations of P^6*Q7: 1 combination of P^7At this point, it all becomes busywork which you probably want to put in a spreadsheet, like <a href="https://docs.google.com/spreadsheets/d/1Jtsxr_zvE7KxwOjvmjR_7I9YixThunPP9Fg9rsWLpXo/edit?usp=sharing" target="_blank">this one</a> (feel free to copy and mod it to adapt it to your own needs). What you're looking for is in the "N or more" column, which tells you that there's a 0.97 probability of 4 or more successes, or 97%. If you modify P, you will see how that affects the outcome.</blockquote>

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