Copyright 2018 Allen Downey
MIT License: https://opensource.org/licenses/MIT
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# Configure Jupyter so figures appear in the notebook
%matplotlib inline
# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'
import numpy as np
import pandas as pd
import thinkplot
from thinkbayes2 import Pmf, Suite
from fractions import Fraction
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class BayesTable(pd.DataFrame):
def __init__(self, hypo, prior=1, **options):
columns = ['hypo', 'prior', 'likelihood', 'unnorm', 'posterior']
super().__init__(columns=columns, **options)
self.hypo = hypo
self.prior = prior
def mult(self):
self.unnorm = self.prior * self.likelihood
def norm(self):
nc = np.sum(self.unnorm)
self.posterior = self.unnorm / nc
return nc
def update(self):
self.mult()
return self.norm()
def reset(self):
return BayesTable(self.hypo, self.posterior)
Suppose I have a box that contains one each of 4-sided, 6-sided, 8-sided, and 12-sided dice. I choose two dice at random and roll them without letting you see the die or the outcome. I report that the sum of the dice is 3.
1) What is the posterior probability that I rolled each possible pair of the dice?
2) If I roll the same dice again, what is the probability that the sum of the dice is 11?
Solution
I'll start by making a list of possible pairs of dice.
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sides = [4, 6, 8, 12]
hypo = []
for die1 in sides:
for die2 in sides:
if die2 > die1:
hypo.append((die1, die2))
hypo
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Here's a BayesTable that represents the hypotheses.
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table = BayesTable(hypo)
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Since we didn't specify prior probabilities, the default value is equal priors for all hypotheses. They don't have to be normalized, because we have to normalize the posteriors anyway.
Now we can specify the likelihoods: if the first die has n1 sides and the second die has n2 sides, the probability of getting a sum of 3 is
2 / n1 / n2
The factor of 2 is there because there are two ways the sum can be 3, either the first die is 1 and the second is 2, or the other way around.
So the likelihoods are:
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for i, row in table.iterrows():
n1, n2 = row.hypo
table.loc[i, 'likelihood'] = 2 / n1 / n2
table
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Now we can use update to compute the posterior probabilities:
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table.update()
table
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The second part of the problem asks for the (posterior predictive) probability of getting a total of 11 if we roll the same dice again.
For this, it will be useful to write a more general function that computes the probability of getting a total, k, given n1 and n2.
Here's an example with the 4 and 6 sided dice:
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n1, n2 = 4, 6
d1 = Pmf(range(1, n1+1))
d2 = Pmf(range(1, n2+1))
total = d1 + d2
thinkplot.Hist(total)
And here's the general function:
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def prob_total(k, n1, n2):
d1 = Pmf(range(1, n1+1))
d2 = Pmf(range(1, n2+1))
total = d1 + d2
return total[k]
To check the results, I'll compare them to the likelihoods in the previous table:
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for i, row in table.iterrows():
n1, n2 = row.hypo
p = prob_total(3, n1, n2)
print(n1, n2, p, p == row.likelihood)
Now we can answer the second part of the question using the law of total probability. The chance of getting 11 on the second roll is the
$\sum_{n1, n2} P(n1, n2 ~|~ D) \cdot P(11 ~|~ n1, n2)$
The first term is the posterior probability, which we can read from the table; the second term is prob_total(11, n1, n2).
Here's how we compute the total probability:
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total = 0
for i, row in table.iterrows():
n1, n2 = row.hypo
p = prob_total(11, n1, n2)
total += row.posterior * p
total
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This calculation is similar to the first step of the update, so we can also compute it by
1) Creating a new table with the posteriors from table.
2) Adding the likelihood of getting a total of 11 on the next roll.
3) Computing the normalizing constant.
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table2 = table.reset()
for i, row in table2.iterrows():
n1, n2 = row.hypo
table2.loc[i, 'likelihood'] = prob_total(11, n1, n2)
table2
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table2.update()
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table2
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We can solve this problem more concisely, and more efficiently, using a Suite.
First, I'll create Pmf object for each die.
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dice = {}
for n in sides:
dice[n] = Pmf(range(1, n+1))
And a Pmf object for the sum of each pair of dice.
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pairs = {}
for n1 in sides:
for n2 in sides:
if n2 > n1:
pairs[n1, n2] = dice[n1] + dice[n2]
Here's a Dice class that implements Likelihood by looking up the data, k, in the Pmf that corresponds to hypo:
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class Dice(Suite):
def Likelihood(self, data, hypo):
"""Likelihood of the data given the hypothesis.
data: total of two dice
hypo: pair of sides
return: probability
"""
return pairs[hypo][data]
Here's the prior:
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suite = Dice(pairs.keys())
suite.Print()
And the posterior:
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suite.Update(3)
suite.Print()
And the posterior probability of getting 11 on the next roll.
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suite.Update(11)
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