Is there any instance where you'd rather have 1D8+1 over a D10, or vice versa?
Obviously when rolling a one sided die. But what about with a D4? Would you maybe want the D8+1 instead of the D10?
For D4, you'd want D8+1. For D4+5, you'd want D10. That's because the numbers 1-10 are symmetric in such a way that D10 has the advantage for the higher numbers and D8+1 for the lower numbers (and it all adds up so that they're both just as good against anything that rolls numbers between 1 and 10 symmetrically).
To generalize:
d10 is more likely to roll less than 5, and more likely to roll greater than 6.
So considering static numbers: (define "win" as roll better and "lose" as not "win")
d8+1 "wins" against 1, 2, 3, and 4 more frequently.
d10 "wins" against 6, 7, 8, and 9 more frequently.
They win equally often against 5.
And they both always win for 0 or less, and always lose for 10 or more.
Then for any given probability distribution including numbers from 1-9, you can see how it matches up against those values. They aren't all equivalent probability wins, but they're symmetric about 5. Basically, if comparing against other symmetric distributions, you'll just have to consider if one side is more likely than the other. Against a uniform distribution that's just counting:
Here are some examples.
1d4, 1-4 are lots more likely (only possible) 1d8+1 wins more.
1d4+5 6-9 lots more likely (only possible) 1d10 wins more.
1d6 more 1-4 counts than 6-9 counts, 1d8+1 wins more.
1d6+2 has 3, 4, 6, 7, 8...more 6-9 counts so 1d10 wins more.
How about, 2d4? It has values 2-8 symmetric about the mean (5)...and so, it compares equally to d8+1 (winning more frequently on 2, 3, and 4), and d10 (winning more frequently on 6, 7, and 8).
Aside: What if we include ties? well, if we adjust our definition of "win" to be "at least tie" then:
d8+1 "wins" against 2, 3, 4, and 5 more frequently.
d10 "wins" against 7, 8, 9, and 10 more frequently.
So, d8+1 will at least tie against 2d4 much more frequently than d10. Since they're equal for "win", we see that if getting a tie has any advantage then d8+1 is better in this case.
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What about 2d6? 2d6 has a symmetric probability about 7. If you know much about rolling 2d6, you know 6, 7, 8 are more likely than 2,3,4...and it doesn't include 1, so clearly d10 is better here.
What about 2d6-1 though? this includes all the numbers in both ranges, but it's still more likely to roll in the 6-9 range than the 1-4 range. (4 and 8 have the same likelihood and it goes down as you get farther from 6).
Finally, looking at 2d6-2. Mean is 5, probability goes down as you get away from it. So probability of 4 is same as 6, etc...makes them equal for win again.
(and again if we consider ties, 1d8+1 wins again)