i cut this baby out from dragon magazine as a kid, and could never quite believe why they didn't catch on more... http://mageofthestripedtower.blogspot.com/2012/02/ad-1ed-combat-computer.html

- Thread starter Extrakun
- Start date

i cut this baby out from dragon magazine as a kid, and could never quite believe why they didn't catch on more... http://mageofthestripedtower.blogspot.com/2012/02/ad-1ed-combat-computer.html

The situation is more complex if you have a single success/fail result. The return from a die-roll is the probability of success, so a naïve answer could be that you have diminishing returns if twice the investment gives less than twice the probability. But that does not make sense, as probability is bounded by 1 (or 100%, if you prefer), so you can never get twice the probability if it is already over 0.5 (or 50%). Also, some would say that 99% chance of success is much more than twice as good as 60% chance of success, because the chance of failure is much, much lower.

So we need a measure of "returns" of a success probability that goes from 0 to infinity. If p is the success (measured from 0 to 1), and supports our intuition of goodness of a result. To do this, we look at dice pools where we do not count successes, but just need at least one successful die to succeed. If we have a dice pool with N dice and a probability that each succeeds at probability p, the probability that

If we set R(p) = -log(1-p) and use base 2 for the logarithm, we get R(1/2) = -log(1/2) = 1, R(3/4) = 2, R(7/8) = 3, R(15/16) = 4, and so on, so for N dice, you get a return of N. If we instead have 1/4 success for each die, we get returns of 0.415, 0.830, 1.245, 1.660, ... for N=1, 2, 3, 4, ..., which is equal to 0.415×N, so still linear returns. This ties reasonably well to the case where we count successes, where the returns is the number of successes, which also grows linearly with N. So even if it does not give diminishing returns for dice pools, I think it is a reasonable baseline for the "goodness" of a success/fail result. Using natural logarithm or base-10 logarithm just scales the result by a constant factor, so it does not make much of a difference.

You can also argue that, when counting successes, that four successes is not twice as good as two successes, as the effect you get for four successes is not twice as good as the effect of two successes. So you need to apply some vague measure of goodness on top of the numerical result.

In summary, I suggest that when counting successes, the returns is measured in number of successes, and when having just a succeed/fail result, the returns is measures as -log(1-p) (using base 2 logarithms), where p is the chance of success.

Code:

```
t# Chance Increase
1 15,97%
2 30,56% 14,58%
3 43,75% 13,19%
4 55,56% 11,81%
5 65,97% 10,42%
6 75,00% 9,03%
7 82,64% 7,64%
8 88,89% 6,25%
9 93,75% 4,86%
10 97,22% 3,47%
11 99,31% 2,08%
```

Code:

```
t# Chance Increase
1 15,97%
2 30,56% 14,58%
3 43,75% 13,19%
4 55,56% 11,81%
5 65,97% 10,42%
6 75,00% 9,03%
7 82,64% 7,64%
8 88,89% 6,25%
9 93,75% 4,86%
10 97,22% 3,47%
11 99,31% 2,08%
```

This gives

Code:

```
t# Value Increase
1 0.251
2 0.526 0.275
3 0.830 0.304
4 1.170 0.350
5 1.555 0.385
6 2.000 0.445
7 2.526 0.526
8 3.170 0.644
9 4.000 0.830
10 5.169 1.169
11 7.179 2.010
```