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document updated 16 years ago, on Dec 19, 2007

Range

To determine the min and max possible values for dice tosses such as 2d3+3, simply take the lowest (or highest) value on one die toss, and run it through the equation. So:

Expected value

As far as I can tell, the expected value is always the midpoint of the range. So that's easy, it's just (min+max)/2. For the above, that'd be (5+9)/2 = 14/2 = 7.

Resources

If you're doing complicated things, Games::Dice::Probability is useful. This is a friendly wrapper around it. I've used it to double-check the above though, so perhaps just stick with the easier method...