Okay, so the probability of rolling a six-sided die seven times and getting a 1 at least one time is $7/6$? But probabilities can't be larger than $1$. You cannot add probabilities unless events are disjoint (i.e. can't both happen at once). Here, it can happen that you get a 1 on more than one roll. Your calculation is double counting sequences with more than one $1$ in them. In the simple case where we roll the six-sided die twice, write out each possible pair of rolls explicitly. There are $36$ options. $10$ include exactly one 1. There is $1$ option where both rolls land on 1. So the probability of getting a $1$ at least once is $11/36\neq 2/6$. This generalises to the inclusion-exclusion principle. Well, the question is more complex than it seems at first glance, but you'll soon see that the answer isn't that scary! It's all about maths and statistics. First of all, we have to determine what kind of dice roll probability we want to find. We can distinguish a few which you can find in this dice probability calculator. Before we make any calculations, let's define some variables which are used in the formulas. n - the number of dice, s - the number of an individual die faces, p - the probability of rolling any value from a die, and P - the overall probability for the problem. There is a simple relationship - p = 1/s, so the probability of getting 7 on a 10 sided die is twice that of on a 20 sided die.
In our example we have n = 7, p = 1/12, r = 2, nCr = 21, so the final result is: P(X=2) = 21 * (1/12)² * (11/12)⁵ = 0.09439, or P(X=2) = 9.439% as a percentage.
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