GTSchemer wrote: ↑08 November 2023, 16:29 For any game where this occurs, you're certainly welcome to post the distribution of numbers from the whole game, and put it in a curve fitting website to check whether the bell curve probably follows random distribution. Or, if you just post the distribution, we can do the curve fitting.
However, if you do not post the full distribution from the game, we can't adequately perform curve fitting to demonstrate (or disprove) the distribution for the game in question.
As you may know, random numbers in the Catan source simply use the BGA random number generator and have no capacity for manipulation.
I appreciate you trying to fight the good fight, but I've yet to see someone respond by saying, "oh, I understanding now and withdraw my conspiracy theory about how the site is rigged." They are only ever looking for someone to agree with them. I suppose these responses still help guide people who are willing to learn.
Code: Select all
from math import comb
def run_analysis_(replaytranscript):
if type(replaytranscript) != str:
raise TypeError
L,histogram = len(replaytranscript),[0 for _ in range(13)]
for i in range(L-13):
if replaytranscript[i:i+13] != " rolls dice: ":
continue
if not ((i+16 < L) and (replaytranscript[i+13] in ('1','2','3','4','5','6')) and (replaytranscript[i+14] == ',') and (replaytranscript[i+15] in ('1','2','3','4','5','6')) and (replaytranscript[i+16] == '\n')):
raise ValueError
histogram[int(replaytranscript[i+13])+int(replaytranscript[i+15])] += 1
numberofrolls = sum(histogram)
if not all(sum(int(replaytranscript[i:i+3] == threecharstring) for i in range(L-3)) == numberofrolls for threecharstring in ("rol","lls","dic","ce:")):
raise ValueError
print()
print(str(numberofrolls)+" total rolls")
print("frequency of (4,5,6,8,9,10): "+(','.join(str(histogram[i]) for i in (4,5,6,8,9,10))))
print()
inner,middle,outer = histogram[6]+histogram[8],histogram[5]+histogram[9],histogram[4]+histogram[10]
frequencyofwhatwasobserved = (2-int(histogram[6] == histogram[8]))*comb(inner,histogram[8])*(2-int(histogram[5] == histogram[9]))*comb(middle,histogram[9])*(2-int(histogram[4] == histogram[10]))*comb(outer,histogram[10])
partitionintothree,numberofties = [0,0,0],0
for x in range((inner//2)+1):
for y in range((middle//2)+1):
for z in range((outer//2)+1):
frequencyofxyz = (2-int(2*x == inner))*comb(inner,x)*(2-int(2*y == middle))*comb(middle,y)*(2-int(2*z == outer))*comb(outer,z)
if frequencyofwhatwasobserved < frequencyofxyz:
partitionintothree[2] += frequencyofxyz
elif frequencyofxyz < frequencyofwhatwasobserved:
partitionintothree[0] += frequencyofxyz
elif (x,y,z) == (min(histogram[6],histogram[8]),min(histogram[5],histogram[9]),min(histogram[4],histogram[10])):
partitionintothree[0] += frequencyofxyz
else:
partitionintothree[1] += frequencyofxyz
numberofties += 1
if sum(partitionintothree) != 2**sum(histogram[i] for i in (4,5,6,8,9,10)):
raise RuntimeError
denominator = sum(partitionintothree)
print()
if numberofties == 0:
print("The p-value is "+str(partitionintothree[0]/denominator)+" .")
elif numberofties == 1:
print("There is exactly one other way to get the same frequency.")
print("If that tie is broken in the way that makes the observed event more surprising, then the p-value is "+str(partitionintothree[0]/denominator)+" .")
print("Otherwise, the p-value is "+str((partitionintothree[0]+partitionintothree[1])/denominator)+" .")
else:
print("There are "+str(numberofties)+" other ways to get the same frequency.")
print("If there is no tiebreaker, then the p-value is "+str((partitionintothree[0]+partitionintothree[1])/denominator)+" .")
print("Otherwise, the p-value is at most that and at least "+str(partitionintothree[0]/denominator)+" .")
print()