AbdulAlhazred
Legend
What.
Now.
Alright. Work with me here.
1d6+1 has the following outcomes, each with the same possibilty of occuring:
2, 3, 4, 5, 6, 7. The mean of all outcomes is 4.5.
1d8 has eight outcomes, each with the same possibility of occuring:
1, 2, 3, 4, 5, 6, 7, 8. The mean of all outcomes is 4.5.
Now if you look very carefully, you'll notice that any outcome 1d6+1 could come up with, 1d8 also comes up with. 2 thru 7 are there. The difference then, is the 1 and the 8 on the end.
The mean damage over any number of rolls is therefore going to be 4.5. No matter what.
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The reason the calcuation (min + max) / 2 works is because of some interesting mathematical properties.
Basically, the mean of any number of die rolls is equal to the sum of all the die rolls divided by the number of outcomes.
This looks like this:
m = [1 + 2 + .... + n-1 + n]/n where n is the size of the die.
This can also be expressed as:
m = [1 + n + 2 + n-1 + .... 1 + n/2 + n - n/2] / n
or
m = [1 + n + 1+1 + n-1 + ..... 1 + n/2 + n - n/2] / n
m = [1 + n + [1+n] + [1-1] + ...... [1+n] + [n/2 - n/2]] /n
m = [1 + n + 1 + n + ..... 1 + n] /n
and because half of these are 1s and half of them are ns
m = [1(n/2) + n(n/2)]/n
m = [1/2 + n/2]
m = (1+n)/2
In the case of a dieroll, where it's 1d6+1, the math becomes:
m = (1+6)/2 + 1
This is because over the entire series of outcome to be meaned, you add 1 to each outcome from 1d6. This means you're adding 1 six times, then dividing by six. That means you raise the average by 1X6/6 = 1.
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Dag, you do have a point with crits tho. That said, the difference from a crit between 1d6+1 and 2d8 is .05 Dpr. This will make very little practical difference in the number of rounds it takes to down an enemy.
Ohhhh, DS, even with a math degree it all makes my head hurt nowadays, lol. I could probably drag out my number theory text and add a whole bunch to your fun list of ways to analyze that, but oh the brain cells don't really like to work that hard anymore. lol.