09-01-2020, 05:12 PM
Some stuff.
Math for Monsters
So I wrote a python script to calculate the probability of getting N Successes. It was an attempt to "learn" python and intentionally used a brute force algorithm. But while it's nice to know what the probability is to get three Successes with a TN of 5 and a pool of {d6,2d8,d12}, it's actually more useful to know how many Successes one can expect to roll (1.5). So a battle between a character with the given pool can expect to defeat a monster with a Defense of 5 with 3 Health in two Turns.
To get that 1.5 Successes is not difficult. If all the dice are the name size we're looking at a binomial distribution, Probability 101 chapter 1 stuff, and E(X) is np where n is the number of dice and p is the probability of a die rolling a Success: (s-TN)/s, or 1-TN/s, where s is the size of the die. But the dice won't necessarily be the same size, so we need to generalize. All np means is p + p + ... + p, where p is added n times. So far so good. But p for each die isn't necessarily the same so we need to add the different probabilities of each die in the pool: p1 + p2 + ... + pn, where pi is the probability that die i rolls a Success, and there we are. In the example above it was 1/6 + 3/8 + 3/8 + 7/12 = 1.5.
Hooray! Oh, but the mechanics say dice "smaller" than the TN can be added together, so what would be the expected value of {4d4} against a Defense of 5? This is trickier and I haven't looked at adding more than two dice, which changes the probabilities. The advantage for our purposes, however, is that by only working with pairs of small dice we undercount the Successes meaning in actual practice, the game will be slightly easier. Without getting into the messy stuff, p(s1+s2>TN)={(T[s1]-1)/s1s2 if TN=s2; and T[s1+s2-TN]/s1s2 if TN>s2, where s1 is the size of the smaller die and s2 is the size of the larger die (s1 and s2 can be the same size), and T[n] is the nth triangular number: (1+2+3+...+n)=n(n+1)/2. If you aren't frightened by arithmetic and find my calculations wrong, keep in mind that dice that roll a 1 cannot be used. Also keep in mind that this is only for two given dice, not for two out of some number of dice, which also undercounts the number of Successes. So the (undercounted) expected number of Successes from {4d4} against TN 5 is, after all calculations, 0.75.
Why does this matter? Because I was curious. Because monsters use full pools binomial distribution is all that's needed, and is a good enough approximation for PCs in monster creation.
Math for Monsters
So I wrote a python script to calculate the probability of getting N Successes. It was an attempt to "learn" python and intentionally used a brute force algorithm. But while it's nice to know what the probability is to get three Successes with a TN of 5 and a pool of {d6,2d8,d12}, it's actually more useful to know how many Successes one can expect to roll (1.5). So a battle between a character with the given pool can expect to defeat a monster with a Defense of 5 with 3 Health in two Turns.
To get that 1.5 Successes is not difficult. If all the dice are the name size we're looking at a binomial distribution, Probability 101 chapter 1 stuff, and E(X) is np where n is the number of dice and p is the probability of a die rolling a Success: (s-TN)/s, or 1-TN/s, where s is the size of the die. But the dice won't necessarily be the same size, so we need to generalize. All np means is p + p + ... + p, where p is added n times. So far so good. But p for each die isn't necessarily the same so we need to add the different probabilities of each die in the pool: p1 + p2 + ... + pn, where pi is the probability that die i rolls a Success, and there we are. In the example above it was 1/6 + 3/8 + 3/8 + 7/12 = 1.5.
Hooray! Oh, but the mechanics say dice "smaller" than the TN can be added together, so what would be the expected value of {4d4} against a Defense of 5? This is trickier and I haven't looked at adding more than two dice, which changes the probabilities. The advantage for our purposes, however, is that by only working with pairs of small dice we undercount the Successes meaning in actual practice, the game will be slightly easier. Without getting into the messy stuff, p(s1+s2>TN)={(T[s1]-1)/s1s2 if TN=s2; and T[s1+s2-TN]/s1s2 if TN>s2, where s1 is the size of the smaller die and s2 is the size of the larger die (s1 and s2 can be the same size), and T[n] is the nth triangular number: (1+2+3+...+n)=n(n+1)/2. If you aren't frightened by arithmetic and find my calculations wrong, keep in mind that dice that roll a 1 cannot be used. Also keep in mind that this is only for two given dice, not for two out of some number of dice, which also undercounts the number of Successes. So the (undercounted) expected number of Successes from {4d4} against TN 5 is, after all calculations, 0.75.
Why does this matter? Because I was curious. Because monsters use full pools binomial distribution is all that's needed, and is a good enough approximation for PCs in monster creation.
Getting me free admission into gaming conventions for a decade