In this project, we are going to contribute to the development of a mobile app by writing a couple of functions that are mostly focused on calculating probabilities. The app is aimed to both prevent and treat lottery addiction by helping people better estimate their chances of winning.
The app idea comes from a medical institute which is specialized in treating gambling addictions. The institute already has a team of engineers that will build the app, but they need us to create the logical core of the app and calculate probabilities. For the first version of the app, they want us to focus on the 6/49 lottery and build functions that can answer users the following questions:
The scenario we're following throughout this project is fictional — the main purpose is to practice applying probability and combinatorics (permutations and combinations) concepts in a setting that simulates a real-world scenario.
Below, we're going to write two functions that we'll be using frequently:
factorial() — a function that calculates factorials
combinations() — a function that calculates combinations
def factorial(n):
result = n
while n > 1:
n -= 1
result *= n
return result
def combinations(n, k):
result = factorial(n)/(factorial(n-k)*factorial(k))
return int(result)
We need to build a function that calculates the probability of winning the big prize for any given ticket. For each drawing, six numbers are drawn from a set of 49, and a player wins the big prize if the six numbers on their tickets match all six numbers.
The engineer team told us that we need to be aware of the following details when we write the function:
Below, we write the one_ticket_probability() function, which takes in a list of six unique numbers and prints the probability of winning in a way that's easy to understand.
def one_ticket_probability(user_numbers: list) -> float:
n_combinations = combinations(49, 6)
probability_one_ticket = 1 / n_combinations
percentage_form = probability_one_ticket * 100
print(f'''Your chances to win the big prize with the numbers {user_numbers}
are {percentage_form:.7f}%. In other words, you have a 1 in {n_combinations:,}
chances to win.''')
We now test a bit the function on two different outputs.
test_input_1 = [2, 43, 22, 23, 11, 5]
one_ticket_probability(test_input_1)
Your chances to win the big prize with the numbers [2, 43, 22, 23, 11, 5] are 0.0000072%. In other words, you have a 1 in 13,983,816 chances to win.
test_input_2 = [9, 26, 41, 7, 15, 6]
one_ticket_probability(test_input_2)
Your chances to win the big prize with the numbers [9, 26, 41, 7, 15, 6]
are 0.0000072%. In other words, you have a 1 in 13,983,816
chances to win.
import pandas as pd
lottery_canada = pd.read_csv('649.csv')
lottery_canada.shape
(3665, 11)
lottery_canada.head(3)
| PRODUCT | DRAW NUMBER | SEQUENCE NUMBER | DRAW DATE | NUMBER DRAWN 1 | NUMBER DRAWN 2 | NUMBER DRAWN 3 | NUMBER DRAWN 4 | NUMBER DRAWN 5 | NUMBER DRAWN 6 | BONUS NUMBER | |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 649 | 1 | 0 | 6/12/1982 | 3 | 11 | 12 | 14 | 41 | 43 | 13 |
| 1 | 649 | 2 | 0 | 6/19/1982 | 8 | 33 | 36 | 37 | 39 | 41 | 9 |
| 2 | 649 | 3 | 0 | 6/26/1982 | 1 | 6 | 23 | 24 | 27 | 39 | 34 |
lottery_canada.tail(3)
| PRODUCT | DRAW NUMBER | SEQUENCE NUMBER | DRAW DATE | NUMBER DRAWN 1 | NUMBER DRAWN 2 | NUMBER DRAWN 3 | NUMBER DRAWN 4 | NUMBER DRAWN 5 | NUMBER DRAWN 6 | BONUS NUMBER | |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 3662 | 649 | 3589 | 0 | 6/13/2018 | 6 | 22 | 24 | 31 | 32 | 34 | 16 |
| 3663 | 649 | 3590 | 0 | 6/16/2018 | 2 | 15 | 21 | 31 | 38 | 49 | 8 |
| 3664 | 649 | 3591 | 0 | 6/20/2018 | 14 | 24 | 31 | 35 | 37 | 48 | 17 |
The engineering team tells us that we need to write a function that can help users determine whether they would have ever won by now using a certain combination of six numbers. These are the details we'll need to be aware of:
We're going to begin by extracting all the winning numbers from the lottery data set. The extract_numbers() function will go over each row of the dataframe and extract the six winning numbers as a Python set.
winning_numbers = lottery_canada.iloc[:, 4:10].apply(set, axis =1)
# def extract_numbers(row):
# row = row[4:10]
# row = set(row.values)
# return row
# winning_numbers = lottery_canada.apply(extract_numbers, axis = 1)
winning_numbers.head()
0 {3, 41, 11, 12, 43, 14}
1 {33, 36, 37, 39, 8, 41}
2 {1, 6, 39, 23, 24, 27}
3 {3, 9, 10, 43, 13, 20}
4 {34, 5, 14, 47, 21, 31}
dtype: object
Below, we write the check_historical_occurrence() function that takes in the user numbers and the historical numbers and prints information with respect to the number of occurrences and the probability of winning in the next drawing.
def check_historical_occurrence(user_numbers, historical_numbers):
'''
user_numbers: a Python list
historical numbers: a pandas Series
'''
user_numbers_set = set(user_numbers)
check_occurrence = historical_numbers == user_numbers_set
n_occurrences = check_occurrence.sum()
if n_occurrences == 0:
print('''The combination {} has never occured.
This doesn't mean it's more likely to occur now. Your chances to win the big prize in the next drawing using the combination {} are 0.0000072%.
In other words, you have a 1 in 13,983,816 chances to win.'''.format(user_numbers, user_numbers))
else:
print('''The number of times combination {} has occured in the past is {}.
Your chances to win the big prize in the next drawing using the combination {} are 0.0000072%.
In other words, you have a 1 in 13,983,816 chances to win.'''.format(user_numbers, n_occurrences,
user_numbers))
test_input_3 = [33, 36, 37, 39, 8, 41]
check_historical_occurrence(test_input_3, winning_numbers)
The number of times combination [33, 36, 37, 39, 8, 41] has occured in the past is 1. Your chances to win the big prize in the next drawing using the combination [33, 36, 37, 39, 8, 41] are 0.0000072%. In other words, you have a 1 in 13,983,816 chances to win.
test_input_4 = [3, 2, 44, 22, 1, 44]
check_historical_occurrence(test_input_4, winning_numbers)
The combination [3, 2, 44, 22, 1, 44] has never occured. This doesn't mean it's more likely to occur now. Your chances to win the big prize in the next drawing using the combination [3, 2, 44, 22, 1, 44] are 0.0000072%. In other words, you have a 1 in 13,983,816 chances to win.
For the first version of the app, users should also be able to find the probability of winning if they play multiple different tickets. For instance, someone might intend to play 15 different tickets and they want to know the probability of winning the big prize.
The engineering team wants us to be aware of the following details when we're writing the function:
The multi_ticket_probability() function below takes in the number of tickets and prints probability information depending on the input.
def multi_ticket_probability(n_tickets):
n_combinations = combinations(49, 6)
probability = n_tickets / n_combinations
percentage_form = probability * 100
if n_tickets == 1:
print('''Your chances to win the big prize with one ticket are {:.7f}%.
In other words, you have a 1 in {:,} chances to win.'''.format(percentage_form, int(n_combinations)))
else:
combinations_simplified = round(n_combinations / n_tickets)
print('''Your chances to win the big prize with {:,} different tickets are {:.7f}%.
In other words, you have a 1 in {:,} chances to win.'''.format(n_tickets, percentage_form,
combinations_simplified))
Below, we run a couple of tests for our function.
test_inputs = [1, 10, 100, 10000, 1000000, 6991908, 13983816]
for test_input in test_inputs:
multi_ticket_probability(test_input)
print('------------------------') # output delimiter
Your chances to win the big prize with one ticket are 0.0000072%. In other words, you have a 1 in 13,983,816 chances to win. ------------------------ Your chances to win the big prize with 10 different tickets are 0.0000715%. In other words, you have a 1 in 1,398,382 chances to win. ------------------------ Your chances to win the big prize with 100 different tickets are 0.0007151%. In other words, you have a 1 in 139,838 chances to win. ------------------------ Your chances to win the big prize with 10,000 different tickets are 0.0715112%. In other words, you have a 1 in 1,398 chances to win. ------------------------ Your chances to win the big prize with 1,000,000 different tickets are 7.1511238%. In other words, you have a 1 in 14 chances to win. ------------------------ Your chances to win the big prize with 6,991,908 different tickets are 50.0000000%. In other words, you have a 1 in 2 chances to win. ------------------------ Your chances to win the big prize with 13,983,816 different tickets are 100.0000000%. In other words, you have a 1 in 1 chances to win. ------------------------
In most 6/49 lotteries, there are smaller prizes if a player's ticket match two, three, four, or five of the six numbers drawn. This means that players might be interested in finding out the probability of having two, three, four, or five winning numbers — for the first version of the app, users should be able to find those probabilities.
These are the details we need to be aware of when we write a function to make the calculations of those probabilities possible:
To calculate the probabilities, we tell the engineering team that the specific combination on the ticket is irrelevant and we only need the integer between 2 and 5 representing the number of winning numbers expected. Consequently, we will write a function named probability_less_6() which takes in an integer and prints information about the chances of winning depending on the value of that integer.
The function below calculates the probability that a player's ticket matches exactly the given number of winning numbers. If the player wants to find out the probability of having five winning numbers, the function will return the probability of having five winning numbers exactly (no more and no less). The function will not return the probability of having at least five winning numbers.
def probability_less_6(n_winning_numbers):
n_combinations_ticket = combinations(6, n_winning_numbers)
n_combinations_remaining = combinations(43, 6 - n_winning_numbers)
successful_outcomes = n_combinations_ticket * n_combinations_remaining
n_combinations_total = combinations(49, 6)
probability = successful_outcomes / n_combinations_total
probability_percentage = probability * 100
combinations_simplified = round(n_combinations_total/successful_outcomes)
print('''Your chances of having {} winning numbers with this ticket are {:.6f}%.
In other words, you have a 1 in {:,} chances to win.'''.format(n_winning_numbers, probability_percentage,
int(combinations_simplified)))
Now, let's test the function on all the three possible inputs.
for test_input in [2, 3, 4, 5]:
probability_less_6(test_input)
print('--------------------------') # output delimiter
Your chances of having 2 winning numbers with this ticket are 13.237803%. In other words, you have a 1 in 8 chances to win. -------------------------- Your chances of having 3 winning numbers with this ticket are 1.765040%. In other words, you have a 1 in 57 chances to win. -------------------------- Your chances of having 4 winning numbers with this ticket are 0.096862%. In other words, you have a 1 in 1,032 chances to win. -------------------------- Your chances of having 5 winning numbers with this ticket are 0.001845%. In other words, you have a 1 in 54,201 chances to win. --------------------------
def probability_at_least_winning(winning_at_least_n):
n_combinations_total = combinations(49, 6)
successful_outcomes = 1
if winning_at_least_n == 6:
probability = 1/n_combinations_total
else:
for i in range (5, winning_at_least_n-1, -1):
n_combinations_ticket = combinations(6, i)
n_combinations_remaining = combinations(43, 6 - i)
successful_outcomes += n_combinations_ticket * n_combinations_remaining
probability = successful_outcomes / n_combinations_total
probability_percentage = probability * 100
combinations_simplified = round(n_combinations_total/successful_outcomes)
print('''Your chances of having at least {} winning numbers with this ticket are {:.7f}%.
In other words, you have a 1 in {:,} chances to win.'''.format(winning_at_least_n, probability_percentage,
int(combinations_simplified)))
for test_input in [2, 3, 4, 5, 6]:
probability_at_least_winning(test_input)
print('--------------------------') # output delimiter
Your chances of having at least 2 winning numbers with this ticket are 15.1015574%. In other words, you have a 1 in 7 chances to win. -------------------------- Your chances of having at least 3 winning numbers with this ticket are 1.8637545%. In other words, you have a 1 in 54 chances to win. -------------------------- Your chances of having at least 4 winning numbers with this ticket are 0.0987141%. In other words, you have a 1 in 1,013 chances to win. -------------------------- Your chances of having at least 5 winning numbers with this ticket are 0.0018521%. In other words, you have a 1 in 53,992 chances to win. -------------------------- Your chances of having at least 6 winning numbers with this ticket are 0.0000072%. In other words, you have a 1 in 13,983,816 chances to win. --------------------------
For the first version of the app, we coded four main functions:
one_ticket_probability() — calculates the probability of winning the big prize with a single ticketcheck_historical_occurrence() — checks whether a certain combination has occurred in the Canada lottery data setmulti_ticket_probability() — calculates the probability for any number of of tickets between 1 and 13,983,816probability_less_6() — calculates the probability of having two, three, four or five winning numbers exactlyPossible features for a second version of the app include:
one_ticket_probability() and check_historical_occurrence() to output information on probability and historical occurrence at the same time