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C11H17N2NaO2S
2008-01-30, 04:36 PM
what's the breakdown?

is it 50%

d4 = 2
d6 = 3
d8 = 4
d10 = 5
d12 = 6

or is it higher?

thanks

Shhalahr Windrider
2008-01-30, 04:40 PM
In general, the average of a set of results equal the (sum of (all results times their probability)).

For a single standard die roll that always results in half the maximum result plus 0.5.

So...
1d4 averages to 2.5
1d6 averages to 3.5
1d8 averages to 4.5
and so on

When applying it to a campaign where the DM asks for "Max at 1st level and Average HP afterwards", it usually comes down to
Level 1: Max
Level 2: Max / 2
Level 3: Max / 2 + 1
Level 4: Max / 2
Level 5: Max / 2 +1
And so on
So a 5th level barbarian would have 12 + 6 + 7 + 6+7 + 5*Con mod = 38 + 5 * Con mod hp.

Mando Knight
2008-01-30, 04:43 PM
It's higher. For average HPs, you add the maximum and minimum values of the die (i.e., highest die number +1, usually), then multiply that by how many dice you're rolling, then divide by two.

Or, where X = the highest number on the die, and N is the number of dice:
N*(X+1)/2

This only works for HPs after the first level if the character is a PC (the first level die is always maxed).

Basically, it's

d4: 2.5
d6: 3.5
d8: 4.5
d10: 5.5
d12: 6.5

Round down for all decimals.

Severus
2008-01-30, 04:47 PM
The average is all the results divided by the number of results.

d4 can 1 or 2 or 3 or 4. so 1+2+3+4 = 10 divided by 4 = 2.5

The other, perhaps easier way to think of it, is to split the number at the point where half is higher or lower. so for the d4, 3-4 is higher, 1-2 is lower. so the average is 2.5.

The only way to get a whole number as an average is if your dice is odd. like a d7. 1-2-3 are low, 5-6-7 is high, 4 is average.

so d4 is 2.5
d6 is 3.5
d10 is 5.5
d12 is 6.5
d20 is 10.5

and so on.

mroozee
2008-01-30, 06:55 PM
In general, the average of a set of results equal the (sum of (all results times their probability)) divided by the total number of results.

Actually, it is just the sum of (all results times their probability).

For 2d6, this would be
2x1/36 + 3x2/36 + 4x3/36 + ... + 12x1/36 = 7

For 1dN it is either:
"the sum of (all results times their probability)" or, since each result is equiprobable (=1/N), equivalently "the (sum of all results) divided by the total number of results." This further simplifies to (N+1)/2.

Shhalahr Windrider
2008-01-30, 07:12 PM
Actually, it is just the sum of (all results times their probability).
You're right. I'm confusing shortcut methods with the full method. :smallredface:

Rad
2008-01-31, 03:37 AM
If that helps, consider having half Hit Point like having half a rank in a cross-class skill. The half rank has no effect at all but when you ad up another half HP at next level they add up to give you one full HP.