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diceprobs.py
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import pandas as pd
import numpy as np
from sympy import Function, Rational, Sum, exp, oo, series, solve, Indexed, Poly
from sympy.abc import n, x, m, i
# Old function that only works for d6 and specified expoding count
#def sym_probs(num_dice, success=4, open_ended=False):
# initial = Rational(5/2)
# p = initial*6**(-n)
# g = Sum(p * x**n, (n, 0, 10)).doit()
# print(g)
# pol = Poly(g - 2, x)**num_dice
# coeffs = np.array(pol.all_coeffs())
# print(pol)
# print(coeffs)
# print(float(coeffs.sum()))
# return coeffs.cumsum()[::-1]/coeffs.sum()
def pad_cut_probs(probs, length):
probs = probs[:length]
probs = np.pad(probs, (0, length - len(probs)), mode='constant', constant_values=(0,0)).astype(float)
return probs
# Return a table of probabilities
def get_probs_table(die_range=(1,10,), die_faces=6, success_count=3, explode_count=0, obstacle_limit=10, log=False):
# make die_range inclusive
die_range = (die_range[0], die_range[1] + 1)
columns = {'obstacle': [i for i in range(obstacle_limit + 1)]}
col_length = obstacle_limit + 1
for i in range(*die_range):
column = get_probs(num_dice=i, die_faces=die_faces, success_count=success_count, explode_count=explode_count, obstacle_limit=obstacle_limit, log=log)
column = pad_cut_probs(column, col_length)
columns[f"{i}D"] = column
return pd.DataFrame(data=columns).set_index('obstacle')
# Return an array of probabilties
def get_probs(num_dice=1, die_faces=6, success_count=3, explode_count=0, obstacle_limit=10, log=False):
# Are we using exploding dice?
if explode_count == 0 or explode_count == None:
return sym_probs(num_dice=num_dice, die_faces=die_faces, success_count=success_count, obstacle_limit=obstacle_limit, log=log)
else:
return sym_explode_probs(num_dice=num_dice, die_faces=die_faces, success_count=success_count, explode_count=explode_count, obstacle_limit=obstacle_limit, log=log)
def sym_probs(num_dice=1, die_faces=6, success_count=None, obstacle_limit=10, log=False):
# This is so much easier when not exploding, just discreet normal
if success_count == None:
success_count = die_faces // 2
# Probability of a die roll being a success
success_chance = Rational(str(success_count) + "/" + str(die_faces))
# Probability of a die roll failing
failure_chance = 1 - success_chance
if log: print(f"Die faces: {die_faces}, success count: {success_count}, success chance: {success_chance}")
pol = Poly(failure_chance + success_chance*x, x)**num_dice
coeffs = np.array(pol.all_coeffs())
if log:
print("Generating function: ", pol)
print("Cummulative prob (low means raise obstacle limit): ", float(coeffs.sum()))
cum_coeffs = coeffs.cumsum()[::-1] + (1 - coeffs.sum())
return cum_coeffs
def sym_explode_probs(num_dice=1, die_faces=6, success_count=None, explode_count=1, obstacle_limit=10, log=False):
# This is so much easier when not exploding, just discreet normal
if success_count == None:
success_count = die_faces // 2
# Probability of a die roll being a success
success_chance = Rational(str(success_count) + "/" + str(die_faces))
# Probability of a *success* exploding
explode_chance = Rational(str(explode_count) + "/" + str(success_count))
# Probability of a die roll failing
failure_chance = 1 - success_chance
# Probability of a *success* not exploding
dud_chance = 1 - explode_chance
if log: print(f"Die faces: {die_faces}, success count: {success_count}, success chance: {success_chance}, explode count: {explode_count}, explode_chance: {explode_chance}")
# There is probably a way to do this that doesn't involve calculating the
# first two terms and finding the ratio this is how I did it by hand though
# and I don't remember enough combinatorics to know a better way
p_0 = failure_chance
p_1 = success_chance*(dud_chance + explode_chance*failure_chance)
p_2 = success_chance*explode_chance*success_chance*(dud_chance + explode_chance*failure_chance)
if log:
print("p_0 = ", p_0)
print("p_1 = ", p_1)
print("p_2 = ", p_2)
term_relation = p_2 / p_1
# We will need p_0 to be the failure chance, that doesn't always work out, so we force it
# I don't really know why this works, but it does in the BW case
initial = p_1 / term_relation
correction = p_0 - initial
if log:
print("term relation = ", term_relation)
print("initial = ", initial)
print("correction = ", correction)
p_n = initial*(success_chance * explode_chance)**(n)
g = Sum(p_n * x**n, (n, 0, obstacle_limit)).doit()
pol = Poly(g + correction, x)**num_dice
coeffs = np.array(pol.all_coeffs())
if log:
print("Generating function: ", pol)
print("Cummulative prob (low means raise obstacle limit): ", float(coeffs.sum()))
print("Coefficients: ", coeffs[::-1])
# We do this here so our cummulative sum includes higher order terms
cum_coeffs = coeffs.cumsum()[::-1] + (1 - coeffs.sum())
return cum_coeffs
#print(list(map(float,sym_probs(2))))
#print(list(map(float,get_probs(num_dice=3))))
#print(list(map(float,get_probs(num_dice=3, explode_count=1))))
#print(list(map(float,sym_explode_probs(num_dice=3))))
#print(list(map(float,sym_probs_flex(1, die_faces=4))))
#print(list(map(float,sym_probs_flex(1, success_count=4))))
# Given N dice to roll, return number of successes
def explode(N, explode_min=6, success=4):
if N == 0:
return 0
rolls = np.random.randint(1,7, size=N)
exploders = (rolls >= explode_min).sum()
successes = (rolls >= success).sum()
return successes + explode(exploders, explode_min=explode_min, success=success)
# This function produces the probabilities using simulation, only works for burning wheel dice system right now
# I do not recommend using this as it took 10m iterations to get an accuracy of +/- 0.001, but it's good to verify the symbolic version
def simulate_probs(num_dice, success=4, open_ended=False, iterations=10000):
rolls = np.random.randint(1, 7, size=(iterations, num_dice))
print(rolls)
successes = (rolls >= success).sum(axis=1)
# Reroll 6
if open_ended:
sixes = (rolls >= 6).sum(axis=1)
additional_successes = np.vectorize(explode)(sixes)
successes = successes + additional_successes
freq_count = np.bincount(successes)
print(freq_count/freq_count.sum())
return freq_count[::-1].cumsum()[::-1]/freq_count.sum()
#np.random.seed(0)
#print(get_probs(1))
#print(get_probs(1, open_ended=True, success=4, iterations=100000))
#
#np.random.seed(0)
#print(explode(12))
#print(get_probs_table(explode_count=1))
#print(get_probs_table(explode_count=0))