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I never understood why people play the same numbers every week

Posted: 6 February 2022
Updated: 24 March 2022
Stats: 1067 words / ~6 minutes

I had a busy week. It seems that the busier the week, the easier it is to get distracted by everything.

As is the case with nearly every person I have ever met, my head is filled with information that, albeit cool, is less than useful on a daily basis. I remember a large quantity of the dialogue from Transformers: The Movie (1986), and bits and pieces of my high school AP math classes, for example.

This week, a student asked about the chances of randomly guessing their way to a perfect score on a multiple choice test. I don’t know why they would ask the English teacher, but my brain gave them the answer:

If there are 200 multiple choice questions with 4 options each…

$$\frac{1}{4^{200}} = \frac{1}{2.58224987809e+120}$$

That is a big number, and it put them on the path to studying.

Unfortunately, it put me on the path to thinking about the lottery.

I come from the Maritimes, in Canada. We have a lottery there called Lotto 6/49. As the name suggests, you choose 6 numbers out of 49 possible numbers.

So, I sat with my calculator (i.e. computer) and started figuring things out with a little help from the web…

## Calculating Chances and Combinations

The formula for calculating chances looks like this:

$$C(n,r) = \frac{n!}{r!(n-r)!}$$

This might be called the r-combination or “n choose r” or the binomial coefficient.

The formula gives us the number of ways a sample of r elements can be obtained from a larger set of n distinguishable objects, where order does not matter and repetitions are not allowed.

### Using all 49 numbers

\begin{align*}C(n,r) &= \frac{49!}{6!(49-6)!} \cr &=13,983,816\end{align*}

There are 13,983,816 possible combinations. You would have one chance in 13,983,816 of guessing this number.

### Using only Odd Numbers

There are 25 odd numbers between 1 and 49, so there are 177,100 possible combinations.

\begin{align*}C(n_\text{odd},r) &= \frac{25!}{6!(25-6)!} \cr &=177,100\end{align*}

13,983,816 (total possible combinations) - 177,100 (possible combinations with odd numbers) = 13,806,716 losing combinations! That doesn’t bode well.

### Using only Even Numbers

There are 24 even numbers, so, 134,596 possible combinations.

\begin{align*}C(n_\text{even},r) &= \frac{24!}{6!(24-6)!} \cr &=134,596\end{align*}

That leaves us 13,849,220 losing combinations. Again, not great.

### Using 3 Odds and 3 Evens

According to How to Win the Lotto 6/49 According To Math, you can choose 3 odd and 3 even numbers to up your chances.

The author states there are 4,655,200 ways to win.

While I can see that that number is the product of these r-combinations:

\begin{align*} \frac{25!}{3!(25-3)!} \times \frac{24!}{3!(24-3)!} = 4,655,200\end{align*}

I am unsure of the logic here. I believe it, but I admit that I lost some interest at this point.

And… my interest went in another direction.

## How long would it take to win?

(and what would it cost?!)

Lotto 6/49 has two draws per week. A single “line” or chance costs $3.00 CAD. Let’s imagine I will play one line per draw and play the same numbers for both of those draws. And, let’s also imagine that the payout never changes and is based on the payout from 29 January 2022: Match Prize 6 Mains$7,102,956.40
5 Mains + Bonus $60,724.30 5 Mains$2,595.10
4 Mains $87.10 3 Mains$10.00
2 Mains + Bonus $5.00 2 Mains Free Play (=$3.00)

I slapped together some sloppy Python, which I am only in the process of learning, so be kind:

import random
import babel.numbers

# 6/49 ASK FOR 6 NUMBERS BETWEEN 1 AND 49 (Inclusive)

YOURNUMBERS = []
YOURBONUS = 0

print("Choose 6 numbers between 1 and 49")

x = 1

while x != 7:
CHOICE = input(f"Number {x}/6: ")
if int(CHOICE) <= 49 and CHOICE not in YOURNUMBERS:
YOURNUMBERS.append(CHOICE)
x += 1
else:
pass

# Get a Bonus

while True:
RANDOM49 = random.randint(1, 49)
if RANDOM49 not in YOURNUMBERS:
YOURBONUS = RANDOM49
break
else:
continue

# Payout Calculations

WINNING_BONUS = 0
PURCHASES = 0
WINNINGS = 0.0
WINNING_NUMBERS = []
MATCHES = []

while len(MATCHES) != 6:
MATCHES.clear()
WINNING_NUMBERS.clear()

y = 1

while y != 7:
CHOICE = random.randint(1, 49)
if str(CHOICE) not in WINNING_NUMBERS:
WINNING_NUMBERS.append(str(CHOICE))
y += 1

while True:
RANDOM49 = random.randint(1, 49)
if RANDOM49 not in WINNING_NUMBERS:
WINNING_BONUS = RANDOM49
break
else:
continue

for num in WINNING_NUMBERS:
if num in YOURNUMBERS:
MATCHES.append(num)

if len(MATCHES) == 0:
WINNINGS += 0

if len(MATCHES) == 1:
WINNINGS += 0

if len(MATCHES) == 2 and YOURBONUS == WINNING_BONUS:
WINNINGS += 5

if len(MATCHES) == 2 and YOURBONUS != WINNING_BONUS:
WINNINGS += 3

if len(MATCHES) == 3:
WINNINGS += 10

if len(MATCHES) == 4:
WINNINGS += 87.10

if len(MATCHES) == 5 and YOURBONUS == WINNING_BONUS:
WINNINGS += 2_595.10

if len(MATCHES) == 5 and YOURBONUS != WINNING_BONUS:
WINNINGS += 60_724.30

if len(MATCHES) == 6:
WINNINGS += 7_102_956.40

PURCHASES += 1

# print(PLAYS)

TOTALCOST = babel.numbers.format_currency(PURCHASES * 3, "CAD", locale="en_CA")
NET = babel.numbers.format_currency(WINNINGS - PURCHASES * 3, "CAD", locale="en_CA")

WINNING_NUMBERS.sort()
YOURNUMBERS.sort()

print(
f"""
My Numbers {' - '.join(YOURNUMBERS)} + {YOURBONUS}
Winning Numbers {' - '.join(WINNING_NUMBERS)} + {WINNING_BONUS}
You would have to buy {PURCHASES:,} tickets before winning the grand prize.
The cost of playing would be {TOTALCOST}.
You would GROSS {GROSS}.
Your NET winnings would be {NET}."""
)


And went about playing with it.

As I am writing, I just received this output:

Choose 6 numbers between 1 and 49
Number 1/6: 2
Number 2/6: 20
Number 3/6: 42
Number 4/6: 7
Number 5/6: 33
Number 6/6: 49

My Numbers 2 - 20 - 33 - 42 - 49 - 7 + 36
Winning Numbers 2 - 20 - 33 - 42 - 49 - 7 + 25
You would have to buy 15,731,789 tickets before winning the grand prize.
The cost of playing would be $47,195,367.00. You would GROSS$35,228,605.40.

## The Next Distraction

Adding KaTeX to my blog… which I will write about in a future post.