How do I get a diminishing return curve besides using a dice pool?

torbenm

Registered User
Validated User
#12
It is not really easy to define precisely what "diminishing returns" means in the context of dice mechanics. The normal meaning is that the return from added investment is not as large as the return from the original investment. If return is measured as money or another open-ended measure, this is easy enough to define: R(2x) < 2R(x), where R(x) is the return from investing x. If you count successes, this carries over: You have diminishing returns if the number of successes on 2x dice is less than twice the number of successes on x dice. This is not usually the case with simple dice pools: If every die yield a success if it is above a threshold, the returns is N*p, where p is the probability that a single die gives a success.

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 any of the dice succeeds is 1-(1-p)^N. For example, if p is 0.5, the probability of getting at least one success with N dice is 1-0.5^N, so for N = 1,2,3,4,..., the probability is 1/2, 3/4, 7/8, 15/16, ..., i.e. (2^N-1)/2^N.

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.
 

Marius B

Euro-Trash
Validated User
#13
Roll two dice (let's say d12 but it could be any type). If the lowest die is equal to or lower than your ability, it's a 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%
As you can see, each increase is less than the previous one.
 

torbenm

Registered User
Validated User
#14
Marius wrote:

Roll two dice (let's say d12 but it could be any type). If the lowest die is equal to or lower than your ability, it's a 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%
As you can see, each increase is less than the previous one.
Let us try to apply my suggestion for converting success probability into an open-ended value: -log(1-p).

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
So by this measure, it is not diminishing returns. Whether or not it is the right measure is debatable, though.
 
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