Comparing Spock's predictions to a coin flip, yielding a Brier score of 0.57

spock.py

import numpy as np

# impossible 0
# v unlik 10
# unlik 25
# lik 75
# vv likely 99.5
preds = [
	[0, 1],
	[0.75, 1],
	[0.75, 0],
	[0, 1],
	[0, 0],
	[0, 1],
	[0, 1],
	[0, 1],
	[0.995, 1],
	[0.995, 0],
	[0.995, 0],
	[0.25, 1],
	[0.1, 1],
	[0, 1],
	[0.1, 0],
	[0.1, 0],
	[0.1, 0],
	[0.995, 0],
	[0.995, 0],
	[0.81, 1],
	[0.75, 0],
	[0.825, 1],
	[0.75, 1],
	[0, 1],
]



def brier(ps):
	ses = [(p[0] - p[1])**2 for p in ps] 
	return np.mean(ses)

print(brier(preds))  # 0.5699

def flip() :
	return np.random.randint(0, 2)

cs = []
for i in range(10000):
	coin = [[flip(), p[1]] for p in preds]
	cs.append(brier(coin))

print(np.mean(cs))