Random Fiver Option?

[Zeke]

Ah good, I did enable tables in posts. That'll make this easier.

Now then: consider the question of 1d8+1 versus 2*(1d4)+1. The former has eight equally likely outcomes (2 to 9), each with 1/8 probability. The latter has four possible outcomes (3, 5, 7, 9), each with 1/4 probability. Let's make a chart....

1d8+1 2*(1d4)+1

2   3 3

4   5 5

6   7 7

8   9 9

Each "box" here has a 1/4 probability. Now, if one die is in a higher box than the other, it wins, guaranteed. There's a 3/4 chance of that. But if both dice land in the same box, the d4 will win half the time and tie the other half. Thus it has a clear advantage. (Specifically, its probability of winning is 1/2, compared to 3/8 that the d8 will win and 1/8 that they'll tie.)

Now, Sa'ar's method -- using expected values -- is very straightforward and it'll always tell you which die has the better chance of winning. For instance, by comparing expected values, we can instantly find that 5d6+3 (expected value 20.5) will beat 2d20-1 (expected value 20) in the long run. The drawback is that expected values don't tell you exactly what those odds of winning are. That's generally a tougher problem.

More on random fivers later.

Wow, it really is that serious. :P

Assuming you mean serious pain, yes, yes it is.