Tsavo says:
twotonkatrucks says:
twotonkatrucks says:
or I think it’s slightly higher than the average
I just quickly checked (because I'm a pedant). It is indeed exactly the average (assuming fair die), (n+1) where n is the number of sides of the HD.
Or, another way...
It is intuitive to think of 4 as half of 8. So when you get 5 as the avg, it seems high. But there are 8 facets on the die, the first 4 and the last four. So technically, half, or the "middle value", is 4.5 (between 4 and 5). So, we round.
Not as technical, but that's how I've viewed the rationale behind it anyway.
PS: You use "pedant" as though it were a bad thing ;)
You're correct, it's slightly greater than the average, as I originally said. I made a really stupid arithmetic error. So not exactly the average.
The average or expected value for n-sided die is computed as
(1/n)*(sum from i=1 to n of i) and sum from i=1 to n of i is given by n(n+1)/2 (not hard to prove by induction).
so in the end the average is (n+1)/2.