Solution: Bayesian Analysis

1.) Suppose you have two coins in a box. One is a normal coin with heads on one side and tails on the other, and one is a trick coin with heads on both sides. You choose a coin at random and see that one of the sides is heads. What is the probability that you chose the trick coin?

(From Think Bayes By Allen Downey )

1.

from fractions import Fraction
import pandas as pd

coinTable = pd.DataFrame(index=['fair', 'trick'])
coinTable["prior"] =  Fraction(1, 2) #prior probability is that you have 1/2 chance of choosing the trick coin
coinTable

	prior
fair	1/2
trick	1/2

For our likelihoods we have to consider probability of our data (seeing heads) given each hypothesis:

• If we chose our fair coin, then the probability of getting heads would be 1/2
• if we chose our trick coin, then the probability of getting heads would be 1

coinTable["likelihood"] = Fraction(1, 2), 1
coinTable

	prior	likelihood
fair	1/2	1/2
trick	1/2	1

def update(table):
    """Compute the posterior probabilities."""
    table['unnorm'] = table['prior'] * table['likelihood']
    prob_data = table['unnorm'].sum()
    table['posterior'] = table['unnorm'] / prob_data
    return prob_data

update(coinTable)
coinTable

	prior	likelihood	unnorm	posterior
fair	1/2	1/2	1/4	1/3
trick	1/2	1	1/2	2/3

The probability that we chose the trick coin is 2/3

2

There are many variations of the Monty Hall problem. For example, suppose Monty always chooses Door 2 if he can, and only chooses Door 3 if he has to (because the car is behind Door 2). If you choose Door 1 and Monty opens Door 2, what is the probability the car is behind Door 3? If you choose Door 1 and Monty opens Door 3, what is the probability the car is behind Door 2?

(From Think Bayes By Allen Downey)

Solution

Our prior is the same

MHtable = pd.DataFrame(index=['Door 1', 'Door 2', 'Door 3'])
MHtable['prior'] = Fraction(1, 3)
MHtable

	prior
Door 1	1/3
Door 2	1/3
Door 3	1/3

But now our likelihood is different. We go through the possible hypothesis:

First scenario: You choose door 1, monty opens door 2

• If the car is behind door 1, then monty chooses door 2 (hecause he always does if he can), then probability of monty choosing door 2 is 1
• if the car is behind door 2, then monty would not open door 2, thus the probability of him choosing door 2 is 0
• if the car is behind door 3, then monty would open door 2, so again the probability is 1

MHtable["likelihood"] = 1,0,1

update(MHtable)

Fraction(2, 3)

MHtable

	prior	likelihood	unnorm	posterior
Door 1	1/3	1	1/3	1/2
Door 2	1/3	0	0	0
Door 3	1/3	1	1/3	1/2

So in this case, we really do have a 50/50 chance

Second scenario: You choose door 1, monty opens door 3

• If the car is behind door 1, then monty would only choose door 3 if he had to (the car is behind door 2), so this has 0 probability

• If the car is behind door 2, then monty would choose door 3 with probability 1

• If the car is behind door 3, then monty would not open door 3, so 0 probability

MHtable = pd.DataFrame(index=['Door 1', 'Door 2', 'Door 3'])
MHtable['prior'] = Fraction(1, 3)
MHtable['likelihood'] = 0,1,0
update(MHtable)

Fraction(1, 3)

MHtable

	prior	likelihood	unnorm	posterior
Door 1	1/3	0	0	0
Door 2	1/3	1	1/3	1
Door 3	1/3	0	0	0

In this case, we are 100 percent certain that car is behind door 2

3 Decision Analysis: 9.8 (Regression and Other Stories Exercise 9.8, p.129)

Simulation for decision analysis: An experiment is performed to measure the efficacy of a television advertising program. The result is an estimate that each minute spent on a national advertising program will increase sales by € 500 000 (you can assume no variable costs for this product), and this estimate has a standard error of € 200 000. Assume the uncertainty in the treatment effect can be approximated by a normal distribution. Suppose ads cost € 300 000 per minute.

• What is the expected net gain for purchasing 20 minutes of ads?

• What is the probability that the net gain is negative?

import numpy as np

nsim = 10000
salesPerMin = np.random.normal(500000,200000, nsim)

netGain = salesPerMin - 300000 

netGain.mean()

201621.25464952042

mean netGain of ~ 200 000

netGain[netGain<0].size/nsim

0.1578

There is an approximately 15% chance that you get a net loss