How to Mathematically Make Millions in the State Lottery

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A group of students at MIT performed an independent analysis of the expected value of the state lottery for their research project. Their findings led them to win $3M, and most likely an A – without ever winning the jackpot.


How? This was because they concluded the expected value (average) of their total winnings, X, represented by the function, E(X), to be greater than the cost to play.

The expected value formula explains it all:
E\left(X\right)=\sum x\ P\left(x\right)
Where:

X=Average\ Prize\newline x=Prize\newline P(x)=Probability\ of\ winning\ x

Read: “The function E of X is the summation of x times the probability of x.” Really, x is a prize associated with a probability of winning said prize…

x=Prize\ for\ matching\ Number\ of\ matches\ of\ White\ Balls\ and\ Mega\ Ball.

Where the probability of winning each x is represented by the probability formula, P(x):

P\left(x\right)=\frac{n!}{r!\left(n-r\right)!}m

Where:

n=\ possible\ numbers\ in\ set\newline r=number\ of\ matching\ options\newline m=probability\ of\ Mega\ Ball

In the state lottery,

n=70, r=5, m=\frac{1}{25}.

For example, to match 5 of 5 White Balls and the Mega Ball, the probability equation, P(5,1) evaluates to the famous number: \frac{1}{302,575,350} This equates to the probability of winning the Mega Millions jackpot. There are 8 possibilities:

P(1+M),\ P(2+M),\ P(3),\ P(3+M),\ P(4),\ P(4+M),\ P(5),\ and\ P(5+M)

Where M is a matching Mega Ball number.


But the team started by James Harvey and Yuran Lu in ’05 at MIT, officially under the corporate identity Random Investment Strategies, didn’t hit the jackpot. How did they win so much?


The expected value formula only works for many plays. It stipulates that there is a linear relationship between the probability and number of plays! Meaning, the more plays, the higher the probability, until the probability of hitting the jackpot reaches 100% – which is when all combinations are played. Also, the average expected return is simply a summation of these products.


The big secret:


Random Investment Strategies only played when the expected value of the lottery was greater than the cost of a ticket! For example, say it’s $2.00 a ticket, like the state lottery, and you find E(X) = 2.38, then after many plays, a player should expect $2.38 for every $2.00 ticket.


According to the article “A Calculated Approach to Winning the Lottery” by Jay London, written for the MIT Alumni publication,

“… on February 7, 2005, Harvey placed $1,000 worth of bets … he won $3,000. [they] incorporated Random Strategies Investments … and began attracting investors.
The group eventually determined … they could garner an average profit of more than 15 percent … the group earned an estimated $3.5 million over seven years.”


Can we do better than the MIT students? Of course! They were limited by the cash prizes they played for in the game Cash WinFall, where the grand prize never exceeded $2M and I’m sure no one believed them. My formulation guarantees a win, every time, and everyone gets a large distribution. How?


As stated above, buying 302,575,350 tickets = t with every possible combination would guarantee a win. You’re also guaranteed many multiples of the smaller prizes. Let’s imagine you have to input all numbers by hand… if you can generate these tickets with a computer, then these next few parts (hiring workers) can be skipped.


By hand would mean generating your own set of numbers and copying those numbers onto the 5 ticket scantron. Where each ticket is a section on the scantron that allows for custom input of lottery numbers. The number of scantrons, S, we need to fill out, is represented by this formula: \frac{t}{5}

Then,

S=302,575,350\ tickets\ \times\frac{1\ scantron}{5\ tickets}=60,515,070\ scantrons

Suppose each field takes 30 seconds to fill in then it takes 150 seconds = s to fill out 5 fields on 1 scantron. This amounts to:

Ss=9,077,260,500\ seconds\ \times\frac{1\ day}{86400\ seconds}=105,060.89\ days\times\frac{1\ month}{30\ days}=3502\ months

…for one person to run the operation. This is where most critiques of this method end. None have asked: “How many people do we need for this task to take 1 week?” Here, I elaborate on this and describe a scenario that should prove profitable.
Really, since the formula is:

T=\frac{t(s)}{5p}

Where:

T\ =\ total\ time\newline t=number\ of\ tickets\newline p=number\ of\ people\newline s=150\ seconds

Then solving for p when T = 604,800 seconds (7 days) will tell us how many people must be involved. Therefore:

p=\frac{t(s)}{5T}

p\ people=\frac{302,575,350\ tickets\left(150\ sec\right)}{5\ tickets\left(604800\ sec\right)}\newline=15,009
Since we have 15,009 people working for us, we will pay them minimum wage and equity in the operation equal to \frac{p}{t} up to the first 73%, the next 2% goes to management, the last 25% goes to the mathematician that formulated the operation.


With the team being paid $15 an hour, working for 8 hours a day, for 7 days, for 1 week, we would need to run 3 shifts for optimum performance. This increases the number of people we need by 3 times to keep the same pace. 45,027 people working 3 shifts, getting paid for 56 hours of work, each. That means we need $37,822,680 to pay our staff.


First, we only play when the expected value is greater than the cost of a ticket. Now, consider the Mega Millions pot on July 29, 2022, of $1.28B. I ran my own independent analysis and found this; mind you, it’s a pro forma, or for sake of form, and not to be taken seriously or as investing advice.

Table 1: Expected Value Analysis of California State Lottery or July 29, 2022

Summation\ of\ xP(X)=\sum{xP(X)}=E(X)=2.82

My analysis says that the expected value was 2.82, a profit of 0.82, and an ROI of 41%. I ran this for analysis for 302M tickets and found this:

Table 2: Expected Value Analysis of Maximum Tickets

Accounting for the cost of tickets as part of the summation, we could have won $852.3M with this strategy. 61% would be removed due to taxes, leaving $332.4M

If every person fills in 6,720 tickets, then they each get a 1:6,720 split or a 0.01488% split of 73% of the pot – or $242.7M. This equates to a 100% probability of winning $36,110 each! Considering an average of 56 hours of work, this means the worker truly made $2,463 an hour and $5.37 per ticket.

The management team would split a 2% pot and the mathematician would garner 25% of the pot. Of the $332.4M, it’s $6.6M and $83.1M, respectively. After paying the workers, the mathematician leaves with $45.3M. That is:

Table 3: Projected Earnings For All Involved

Seriously, though, the lottery should be played for fun and I don’t think anyone should consider trying this strategy.

Works Cited:

London, Jay. “A Calculated Approach to Winning the Lottery.” MIT Alumni Association, 14 Aug. 2012, https://alum.mit.edu/slice/calculated-approach-winning-lottery.

OpenStax College. (2013). Statistics. OpenStax. https://openstax.org/books/statistics/pages/4-2-mean-or-expected-value-and-standard-deviation#:~:text=To%20find%20the%20expected%20value,%E2%88%91%20x%20P%20(%20x%20)%20.

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