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Message: <blockquote data-quote="ichabod" data-source="post: 759912" data-attributes="member: 1257">If you are using a fair number of dice at once, and are just curious (don't need exact numbers), the normal approximation is probably the way to go. A die with s sides has a mean of (s+1)/2, and a variance of (s-1)^2/12 (assuming it's numbered 1 to s). Figure out the mean and variance for all of the dice you are rolling at once, and add them up to get the mean and variance for the total of all the dice. Now take the square root of the variance to get the standard deviation.Now, approximately 68% of the rolls will fall within 1 sig (standard deviation) of the mean. Approximately 95% of the rolls will fall within 2 sigs of the mean, and approximately 99.5% of the rolls will fall within 3 sigs of the mean. If you are more interested in specific numbers, you can do one of two things. You can use a programming language that gives probabilities for normal random variables (a lot of them do), or you can standardize your values and use a table of the standard normal distribution, like the one  <a href="http://www.stat.psu.edu/~herbison/stat200/stat200_model_demo/supplements/NormalTable.html" target="_blank">here</a> . To figure the probability of rolling n or less, look up ((n+1/2)-mean)/standard deviation on the table. To figure the probability of rolling n or more, look up ((n-1/2)-mean)/standard deviation on the table, then subtract it from one. To figure the probability of rolling n,  subtract the second value you got of the table (before subtracting from one) and subtract it from the first value.Again, these are going to be approximate, and more dice you are rolling at once, the better the approximation will be. If you want exact values, it gets either slow or messy. Slow is having the computer figure out the sum of every possible roll, and then determining the distribution from that. Fine for small number of dice with a low number of sides, but it can bog down when you get into high values of either. Messy is to figure the combinatorics. I'm having too frustrating a day to even get into that.</blockquote>

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