Powerball Math

(The post below is from 2013)

For updated info on the 2016 drawing,
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This Wednesday evening’s Powerball jackpot is estimated to top $400 million.


Let’s use the interest in this compellingly large jackpot — one of the largest in Powerball history — to take a look at the math behind Powerball.

Just to make sure we all know how Powerball works, please watch this 1-minute video of the most recent Powerball drawing on Saturday, August 3, 2013:

Note: this is a loooong post — but it does show how we might approach math at TLC — and it’s an interesting look at the math involved in Powerball — but if you don’t have at least 10 minutes, you might want to come back to this post when you have more time. Also, at TLC, we would stop at various points along the way to make sure students are “getting it” along the way.

Okay, so if you were looking really carefully at that video, you counted 59 white balls in the container on the left and 35 red balls in the container on the right. What are the odds of choosing all five white balls correctly and also choosing the red Powerball? This is an important question because if you match all six on Wednesday night, you would win the jackpot — so this is literally a $400,000,000 question.

The lottery website says the odds of hitting the jackpot are a little more than 1 in 175 million:


That’s correct, but how does that math work? And how could we figure out the odds of other events happening?

This sort of math is called “statistics and probability”  — it’s part of the common core in 6th, 7th and 8th grade (why they repeat it so many times is beyond me — I think sixth graders are eminently capable of handling the same math as 8th graders, if it’s taught well).

The idea of probability is to figure out how likely it is that an event (or series of events) will happen. For example: “The odds of an average person living in the US being struck by lightning in a given year is between 1/750,000 and 1/500,000,” according to the Wikipedia article about Lightning strikes. Basically, you take the number of lightning strikes in the US in a given year and divide that number by the number of people in the US — see the link for more details.

With odds of 1-in-175-million-to-one, your odds of hitting the Powerball jackpot are more than 200 times less likely than your odds of getting hit by lightning.

But it’s at least 200 times more fun to win at Powerball than it is to get hit by lightning… so let’s use math to understand the odds of choosing five balls out of 59, plus the red Powerball.

If the Powerball people (let’s call them the PBPs) were picking just one white ball, I would have a 1-in-59 chance of winning. If they pick two balls, what would my odds be? Well, when there are two events, you multiply their likelihood of happening together. So if the odds of my flipping a coin “heads” once is 1-in-2, my odds of flipping two heads in a row is 1-in-4 (that’s 1/2 times 1/2). My odds of flipping three heads in a row is 1-in-8.

To see why, here’s a list of the possible flips with three coins:


Only one out of eight possibilities is three heads in a row.

The odds of flipping at least two heads when I flip a coin three times is 4-out-of-8, which is the same 50-50, or 1-out-of-2:


But the odds of flipping exactly two heads is 3-out-of-8 (the first one — H-H-H — would not count, because it’s not exactly two heads).

Okay, so let’s go from flipping coins to something a bit more complicated — Powerball.

We said above that the odds of getting one ball right is 1-in-59. But if I just get one white ball correct, nobody will give me anything — the prizes get interesting as you combine at least one white ball with the red Powerball.


And to win the Powerball jackpot, we need to get all five white balls, plus the red Powerball. You might think that the odds of getting the five white balls is 1-in-59 raised to the 5th power, just as the odds of flipping three heads in a row was 1-in-2 raised to the 3rd power (that’s 1/8).

[Note — if I were working with students, this is another spot where we would stop to make sure everyone is still “getting it”]

In the Powerball example, let’s say I get the first ball right (woo hoo!) Once I do that, the odds of getting the second ball right are actually 1-in-58, because now there are only 58 balls left (I got the first one right, remember?)

So the odds of picking five balls out of 59 would be 59 x 58 x 57 x 56 x 55, right?

Well, that ‘s what I thought — but that’s not quite right.  As I noted in my earlier math-related post, How big is the sun?, it’s actually a good thing to make mistakes when you’re doing math.

Let me write that again, because that’s kind of a big deal:

It’s actually a good thing to make mistakes when you’re doing math.

Taking the time to reflect — in writing — about our mistakes helps us solidify our understanding. That’s why we will blog about math at least twice a week at TLC Middle School.

In the sun problem, I confused radius with diameter — but now I know to be careful to distinguish between the two.

In this Powerball, I made the mistake of thinking that my odds of choosing all five balls in Powerball would be 1-in 59 x 58 x 57 x 56 x 55 (or 1-in-600,766,320, if you multiply it out).

But let’s say the winning numbers in Powerball are 10-20-30-40-50. The PBP do not care whether the numbers come out in the order of 10-20-30-40-50 or 50-40-30-20-10.  In fact, they could come out 10-30-20-40-50 and I would still win. 

So what I did in just multiplying 59 x 58 x 57 x 56 x 55 was to neglect to take into account the multiple ways that the PBP could select the winning numbers. It turns out there are 120 different ways to select five numbers. Imagine thinking of all the different ways to order these letters:

… all the way down to …

If this looks oddly like the H-T-H chart above, that’s because it is the same sort of thing. But we need to be precise in math, and in the H-T-H chart, we could have duplicates — you could get multiple heads or tails, because it’s the same coin. In the Powerball example, the odds change each time a ball is removed from the container, and we can’t have two balls with the same number (at least not in the white ball part of the program).

Mathematicians have figured out that the number of ways to order “n” objects is something called n!, or n-factorial.

Factorial — in its simplest form — means taking a whole number and multiplying it by the numbers below it until you get to two. I guess you could include one also, but that’s boring, since anything times 1 is itself.

As an example, 3! is 3 x 2, or 6.

5! would be 5 x 4 x 3 x 2, or 120.

Factorial, probabilities, combinations and permutations are all explained extremely well in this great resource called Combinations and Permutations from the fine folks at “Math is Fun.”

At TLC, we’d assign students to click on that link and read over the material there (it’s long, but extremely well written — it even includes 10 practice problems at the end) and then we’d discuss it when we met the next day.

If you’re getting into this post, please do consult that resource.

For Powerball math purposes, the key formula from that resource is this one:


With Powerball, we have 59 things (white balls) and we are choosing 5 of them. So “n” is 59 and “r” is 5.

When I multiplied 59 x 58 x 57 x 56 x 55, I was just doing n!/(n-r)! — I forgot about the different ways the five balls could be ordered, which is represented here by 5!

So if we plug all this in, we get 59! divided by 5! x 54!

59! is the same as 59 x 58 x 57 x 56 x 55 times 54!, so the 54!s cancel out [again, working with students, I’d make sure they “got” this] and we’re left with 59 x 58 x 57 x 56 x 55 divided by 5!, or 120.

So that 600 million number I had earlier needs to be divided by 120, which gives us roughly 5 million.

Wait — the odds of winning the Powerball jackpot are 1-in-175-million. What’s up?

Well, we need to multiply the odds of getting all five white balls (1-in-5-million) by the odds of also choosing the red Powerball (1-in-35) and that works out to roughly 1-in-175 million.

So now here’s an application question: is Powerball a good bet with an expected jackpot of $400 million?

Generally speaking, it costs $2 to get a ticket, and each ticket has a 1-in-175 million chance of winning.  Is that a good bet?  Well, it depends on what the jackpot is… a ticket is worth the investment only when the expected monetary value (EMV) of the ticket exceeds the purchase price of the ticket — in this case $2.

With an expected jackpot of $400 million, a 1-in-175-million chance times a $400 million payoff is worth 1/175,000,000 times $400,000,000, which works out to about $2.2857 — we will round up to call that $2.29.

So right now a ticket is worth $2.29 (really a little more, since you can also win smaller prizes, but we are simplifying to just consider the jackpot)

We can see that when the jackpot hits $350,000,000, the EMV is $2.  As soon as the jackpot goes over $350 million, it actually makes sense — mathematically speaking — to buy a ticket.

If we are being precise, it’s 175,223,510 times $2 that we care about — and that works out to $350,447, 020.  But basically, when the jackpot hits 351 million, it’s time to buy tickets.

But there are some practical problems. As we noted earlier, you are 200 times more likely to be hit by lightning than you are to win Powerball, so you probably won’t win, unless you actually bought 175 million tickets.

And even if you bought a ticket a second — it takes longer than that, and you would need breaks — it would take you 175 million seconds to buy all the tickets and ensure that you get the winning ticket.  That’s 2.9 million minutes, or 48,000 hours, or 2025 days…

There’s a lot of math to explore in Powerball.

And at TLC, we would also stop to consider the morality of state-sponsored lotteries. Who are more likely to buy lottery tickets? Rich people or poor people? A book written by a pair of Duke professors, Selling Hope, calls lotteries an “insecurity tax” and argues that such lotteries are regressive, as opposed to progressive.

But lotteries fund all sorts of worthwhile projects.

According to the North Carolina Education Lottery website, lottery funds have contributed $2.9 billion to education programs since it started:

nc lotto

Should North Carolina fund its education programs partially through its share of the lottery? Currently, 43 states currently participate in Powerball — is that a good thing or a bad thing? That could be an interesting conversation — and one that we’d have at TLC, assuming students were interested.


About Steve Goldberg

I teach students at Research Triangle High School (RTHS) about US History. RTHS is a public charter school in Durham, NC, whose mission is to incubate, prove and scale innovative models of teaching and learning. The blog posts here reflect my own personal views and not those of my employer.
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5 Responses to Powerball Math

  1. John Burk says:

    I think this is a great lesson—I really like how you extend it to consider the morality of how lotteries are used to fund education. Also, I love that you are so open in discussing the mistake you made in not initially recognizing that the balls can be placed in any order. I think of the mathematical practices I’d like to see students most learn in school is how to find and analyze their own mistakes.

  2. John Burk says:

    I came across this article which explores the issues you raise in a bit more depth and might be of interest to you and your students.

    • Steve Goldberg says:

      Hi John.

      Thanks so much! That article is a great example of how, in most math models, we have to over-simplify.

      The author of the article you posted above does the opposite, and I think that would be very cool for students to see — though I’m not sure I would have most middle school students read the whole article — I’d just get them to consider each of the complications he raises 🙂

  3. Great post Steve. It is a really tricky topic, with lots of places for errors. Here is a way to solve the problem without the formula:

    Since we select five numbers, we actually have a 5/59 chance of correctly choosing the first number. Assuming we get a hit on the first number, we now have a 4/58 chance of correctly choosing the second number. Continuing the pattern, 3/57, 2/56, 1/55. Taking the product of those five probabilities is equivalent to the formula you used. (I always hated memorizing formulas, preferring to derive them myself when needed.)

    One more thing…there really isn’t repetition in the prob and stats strand across the middle grades Common Core. I particularly like the seventh grade standards in this area because they show how the study of statistics (learning about a population when we don’t know the true probability) is based on the study of probability (learning how empirical probability approximates true theoretical probability). Here is a great source for understanding the progression of ideas across the middle school in probability and statistics:



    Eric (from London)

    • Steve Goldberg says:

      Thanks, Eric. I like the way you set the problem up with 5/59, 4/58, etc…

      What I am wondering about the common core’s approach to probability and statistics is whether you could present the whole thing (6th 7th and 8th grade material) to a sixth grader so that he/she would understand how the strands connect. Why do you need a whole year (or two) to “get” the idea of moving from univariate to bivariate data?

      I hope you enjoy London!

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