Post date: Oct 24, 2017 2:33:09 PM
One mechanic I typically like to use is to allow creatures to reroll damage dice. This raises the average, without the chance of having them roll obscenely high and one-shot someone unexpectedly.
But how much does this help?
First off, here is the conclusion. If you're not interested in mathematics, you can read that and stop there.
Average roll of a d6 = 3.5
Average when rerolling 1-2s = 4.1666...
Average when rerolling 1-3s = 4.25
Average when rerolling 1-4s = 4.1666...
Obviously, the optimal action is to reroll 1-3s when using a d6.
I can also assume that this can be applied to larger dice; always reroll dice that score below the average. So, below 2.5 on a d4, 4.5 on a d8, 5.5 on a d20, 6.5 on a d12, and 10.5 on a d20.
Math time.
To start off, let's begin with a d6. Each number has a 1/6 chance of being rolled, obviously. To keep things on track, it helps to write it down.
Pr (1) = 1/6
Pr (2) = 1/6
Pr (3) = 1/6
Pr (4) = 1/6
Pr (5) = 1/6
Pr (6) = 1/6
As obvious as this looks, it really helps when we get to the more complex stuff. Make sure all of the numerators added together equal the denominator (6, in this case). If you get lower, then you apparently have a chance of getting a result not listed. If you get higher, you've just messed up.
Next, we want to express these fractions as decimals. This will allow us to determine the average. To convert them to fractions, we simply divide the numerator by the denominator.
Pr (1) = 1/6 = 0.166...
Pr (2) = 1/6 = 0.166...
Pr (3) = 1/6 = 0.166...
Pr (4) = 1/6 = 0.166...
Pr (5) = 1/6 = 0.166...
Pr (6) = 1/6 = 0.166...
(If you multiply one of these results by 100, you get your percentage. It's pretty helpful for finding your percentage of correct answers on a school test.)
If you add these numbers together, you should get 1. Make sure you do NOT round your numbers.
Now, we can determine the average. Multiply each decimal with the value associated with it. Where I've written 0.1666... here, substitute (1÷6) to get the exact value. for example, 2 × 0.1666... should be input as 2 × (1÷6).
Pr (1) = 1/6 = 0.166... 》1 × 0.1666... = 0.166...
Pr (2) = 1/6 = 0.166... 》2 × 0.1666... = 0.333...
Pr (3) = 1/6 = 0.166... 》3 × 0.1666... = 0.5
Pr (4) = 1/6 = 0.166... 》4 × 0.1666... = 0.666...
Pr (5) = 1/6 = 0.166... 》5 × 0.1666... = 0.833...
Pr (6) = 1/6 = 0.166... 》6 × 0.1666... = 1
Finally, add all these values together to get the average. Because we have so many recurring numbers, our final formula looks something like this:
((1 × (1÷6)) + ((2 × (1÷6)) + ((3 × (1÷6)) + ((4 × (1÷6)) + ((5 × (1÷6)) + ((6 × (1÷6))
That's messy, but unless you can copy and paste results from previous steps, that's what you need. You can't write "0.1666" or "0.1667", because that's ever so slightly different to "1÷6". Then again, if you've read this far, you probably already know this.
Press enter, and we get 3.5, the average dice roll of a d6.
Now for the fun part.
We want to work out what the average roll of a d6 will be if we reroll all 1s and 2s.
There are two ways we can roll a 1:
-By rolling a 1, and rerolling it into a 1.
-By rolling a 2, and rerolling it into a 1.
For the first one, we have a 1/6 chance of rolling a 1, and a 1/6 chance of the reroll being a 1. Multiply those two fractions together, and we get a 1/36 chance of rerolling a 1 into a 1. This is not necessarily the same as our chance of rolling a 1.
For the second one, we have a 1/6 chance of rolling a 2, and a 1/6 chance of the reroll being a 1. Again, this has a 1/36 chance of occurring.
There are no other ways we can get an end result of a 1, so we can add both together to get our probability of rolling a 1. 1/36 + 1/36 = 2/36. Easy.
Our chance of rolling a 2 is the same; we reroll a 1 into a 2, or reroll a 2 into a 2. Two 1/36 chances, for a probability of 2/36.
Our chance of rolling a 3 at first is a 1/6 chance. We don't reroll a 3, so we are certain to have a 3 if we roll it first. A certainty has a probability of 1, so we'll use 6/6 so that we have a common denominator with the others. 1/6 × 6/6 = 6/36.
However, we can also roll a 3 if we reroll a 1 into a 3, or reroll a 2 into a 3. We have a 1/36 chance of either of those. So, when we add together the probabilities of the different ways we can roll a 3, we get 6/36 + 1/36 + 1/36, which is 8/36.
Same goes for a 4, 5, or 6.
Pr (1) = 2/36
Pr (2) = 2/36
Pr (3) = 8/36
Pr (4) = 8/36
Pr (5) = 8/36
Pr (6) = 8/36
Again, make sure all the numerators added together equal the denominator. 2 + 2 + 8 + 8 + 8 + 8 = 36. Check.
Now we convert those to decimals again.
Pr (1) = 2/36 = 0.055...
Pr (2) = 2/36 = 0.055...
Pr (3) = 8/36 = 0.222...
Pr (4) = 8/36 = 0.222...
Pr (5) = 8/36 = 0.222...
Pr (6) = 8/36 = 0.222...
Multiply them by their values again.
Pr (1) = 2/36 = 0.055... 》0.055...
Pr (2) = 2/36 = 0.055... 》0.111...
Pr (3) = 8/36 = 0.222... 》0.166...
Pr (4) = 8/36 = 0.222... 》0.888...
Pr (5) = 8/36 = 0.222... 》1.111...
Pr (6) = 8/36 = 0.222... 》1.333...
Add them together and we get 4.166...
Seeing what happens when we reroll 3s as well, we do the same thing, exept:
We can get a 1 in 3 different ways now: rerolling a 1 into a 1, rerolling a 2 into a 1, and rerolling a 3 into a 1.
Similarly, we can get a 2 in 3 different ways now, and we can get a 3 in only 3 different ways now.
A 3/36 chance of getting a 1, a 3/36 for a 2, and a 3/36 for a 3.
As you can see, our chance of getting a 1 or a 2 is slowly rising, while our chance of getting a 3 drops.
Additionally, our chance of getting a 4 becomes a 9/36 (6/36 from a natural roll, and 1/36 from each of the rerolls). Same for a 5 or 6.
We do all the math, and we get an average of 4.25.
For rerolling 4s and below, our chance of getting a 1 rises again to a 4/36. 2 and 3 do the same, while our chance of getting a 4 drops to the same. Our chances of getting a 5 are now 10/36, same for a 6.
The average of this drops back to 4.166...
Rerolling higher values does increase our chance of getting a higher value, but it also brings down our average by lowering our chances of getting the value we are now rerolling.
It's late here, so I'll do the math some other time, but I expect that each time you increase the reroll threshold, you raise the chance of getting a number above the threshold by 1/(number of sides on the dice^2). So, a d4 would raise from 4/16 to a 5/16 if we rerolled 1s, a 6/16 for 2s, and a 7/16 for 3s. A d8 would got from 8/64 to 9/64 by rerolling 1s, and so on.
For values below this threshold, they drop to (the threshold)/
(the number of sides on the dice^2), and the denominator rises by 1 for each value the threshold is raised.
On a d4, the chance of rolling a 1 starts as 4/16, then 1/16 if we reroll 1s, 2/16 when we reroll 2s, 3/16 when we reroll 3s, and 4/16 when we reroll 4s. For a d8, we start with 8/64 chance of a 1, then 1/64 if we reroll 1s, a 2/64 if we reroll 2s, and so on.