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January 8, 2016

If you play Powerball, play it smart – here’s how

Be sure to check out my follow-up post, which looks for ways to predict how many tickets will be sold, among other statistical fun!

My college stats professor, Dr. Paul Sommers, has a simple set of rules when it comes to playing the lottery:

1. Don’t play the lottery.

It’s an easy guide that I normally follow, but like many people, I will probably buy some Powerball tickets for Saturday’s draw. The lure of the record $800 million jackpot is hard to ignore.

A store in Bergen Beach, Brooklyn, awaiting the onslaught for the $640 million MegaMillions jackpot in March 2012. (It’s also the last time I bought a lottery ticket.)

Professor Sommers is guided by the principle of expected value, a topic I’ve explained in my Guide to Game Theory. The concept is simple: if I buy a huge number of tickets, do I expect my average winnings to be greater than the cost of the tickets?

You can also think about it this way: if I had enough money to play every single possible combination of numbers, would it be profitable?

This is one of the rare times when the answer is yes – except there’s some pesky fine print we need to deal with. (And then there’s the problem of taxes, which almost renders this discussion moot.)

The expected value of Saturday’s drawing

I’m going to put very high-level details in this section; if you want to learn more about the calculations, keep reading after you’ve finished.

In Powerball, the break-even point of expected value (EV) is when the lump-sum jackpot is about $491 million. As of this moment, the lump-sum jackpot is advertised at $496 million, and I expect it will grow significantly over the next 24 hours.

(Note: the $800 million prize that’s advertised is paid out in 30 equal annual installments, which decline in real value over time due to inflation, interest rates, and other factors. In my calculations, I’ll always use the lump-sum prize, which is an estimate of the current value of those future payments.)

White Red Prize Combinations Odds (1 in…) EV
0 0 $0 190,612,800 1.53 $0.0000
1 0 $0 79,422,000 3.68 $0.0000
2 0 $0 10,416,000 28.05 $0.0000
0 1 $4 7,624,512 38.32 $0.1044
1 1 $4 3,176,880 91.98 $0.0435
2 1 $7 416,640 701 $0.0100
3 0 $7 504,000 580 $0.0121
3 1 $100 20,160 14,494 $0.0069
4 0 $100 8,000 36,525 $0.0027
4 1 $50,000 320 913,129 $0.0548
5 0 $1,000,000 25 11,688,054 $0.0856
5 1 $496,000,000 1 292,201,338 $1.6975
Totals 24.87 $2.0173

Our expected value calculation shows that a $2 ticket expects to return a little under $2.02, meaning, on average, players will make a small amount of money.

But there’s a big problem with this assumption, and you probably know what it is: when there’s more than one jackpot winner, the lucky players split the top prize.

If two tickets share the jackpot, they’ll each take $248 million. The unfortunate thing for our purposes is it drops the overall expected value to around $1.17, meaning the average player should expect to lose 83 cents for each ticket purchased.

This past Wednesday, when the jackpot was ‘just’ $528.5 million, over 175 million tickets were sold. How many people do you think will play this time around? And how many do you think will hit the jackpot? Odds are good it will be more than one.

Should you choose Power Play?

If a lottery is offering you something, chances are it’s bad for you. That is true for Power Play.

Unless you’re in California, “Power Play” is a $1 add-on option that multiplies your non-jackpot winnings by a randomly selected number. The multipliers are 2x, 3x, 4x, and 5x; if the jackpot is under $150 million, 10x is also a possibility.

The expected value of the multiplier is 2.767 when the 10x is in play, and 2.595 when it’s not. On average, Power Play will return about 50 cents or 46 cents, respectively, making it a big loser when you’ve paid a dollar to play.

Power Play is only a “good” deal – compared with the regular game, that is – when the lump-sum jackpot is under $175 million.

Your numbers don’t matter

It doesn’t matter whether you pick your own numbers or choose the “QuickPick” option, so save yourself some time and pick the latter. And unlike what some sites have claimed, past performance of certain numbers has no bearing on future results. (This is called the Gambler’s Fallacy; I’ve written about it before.)


Calculating your odds and payouts

If you’ve been reading The Final Wager for a while, you know I don’t talk about statistics without showing you how to do the math. Keep reading if you’re interested in learning more about expected value and combinations.

Total possible combinations

The odds of hitting the red ball are easy to calculate: there are 26, so the probability you’re correct is 1/26. (The red ball is officially known as the Powerball, but I’ll refer to its color to avoid confusion with the name of the game itself.)

For the white balls, since the order in which they’re chosen doesn’t matter, we’re looking at a combination. (If you’re unfamiliar with this concept, check out my overview. In fact, I included an example on a simple lottery for just this reason!)

Out of 69 possible white balls, the machine will select 5. We can calculate the total combinations, therefore, as follows:

69C5 = 69! = 69! = 69 x 68 x 67 x 66 x 65 x 64!
5! x (69-5)! 5! x 64! 120 x 64!

That comes out to 11,238,513 combinations of white balls.

Multiply that by 26 red balls and you get 292,201,338 combinations – which, if you play every single combination at $2 each, will set you back

$584,402,676

If you have this much money lying around, please send me an email.

Combinations of n white balls

Finding all possible combinations is fairly straightforward. Since our prize depends on how many balls we correctly match, however, we want to find out how many different ways we can “hit” exactly n white balls.

Let’s start simple. How many combinations will “hit” all 5 white balls? That easy: just one. I’ve got A, B, C, D, and E on my ticket, and the machine spits out A, B, C, D, and E in some order. (Knock on wood.)

To calculate the other combinations, let’s think about things differently. Each of the 69 white balls will fall into one of two groups: “winners” (population: 5) and “losers” (population: 64).

Let’s say I pick n winners. The remainder – 5 minus n – must be losers.

Winning combinations Losing combinations
5Cn 64C(5-n)

When we multiply those two values together, we see how many total combinations there are where you match n winning white balls.

For example, when n = 2, we get

Winning combinations:

5C2 = 5! = 5! = 5 x 4 x 3! = 10
2! x (5-2)! 2! x 3! 2 x 3!

Losing combinations:

64C3 = 64! = 64! = 64 x 63 x 62 x 61! = 41,664
3! x (64-3)! 3! x 61! 6 x 61!

…which means there are 416,640 different ways you can hit exactly two white balls.

Here is the full table of combinations.

Balls correct Combinations
White (of 69) Red (of 26) White Red Total
0 0 7,624,512 25 190,612,800
1 1 7,624,512
1 0 3,176,880 25 79,422,000
1 1 3,176,880
2 0 416,640 25 10,416,000
1 1 416,640
3 0 20,160 25 504,000
1 1 20,160
4 0 320 25 8,000
1 1 320
5 0 1 25 25
1 1 1
Total 11,238,513 26 292,201,338

Next, we’ll plug in the prize values to see how much we’ll expect to win. For the jackpot, I’m going to use the $496 million “lump sum” payment the winner can select.

Balls correct
White (of 69) Red (of 26) Combinations Prize ($) Total ($)
0 0 190,612,800 0 0
0 1 7,624,512 4 30,498,048
1 0 79,422,000 0 0
1 1 3,176,880 4 12,707,520
2 0 10,416,000 0 0
2 1 416,640 7 2,916,480
3 0 504,000 7 3,528,000
3 1 20,160 100 2,016,000
4 0 8,000 100 800,000
4 1 320 50,000 16,000,000
5 0 25 1,000,000 25,000,000
5 1 1 496,000,000 496,000,000
Total 589,466,048

In the best-case scenario, you can expect to win

$589,466,048

of your “investment” in tomorrow’s Powerball jackpot, for a profit of just over $5 million.

Ah, but there’s that old catch: if you buy every single combination then have to share the jackpot, you’ll actually lose about $240 million – and it’ll be worse if more than two players correctly pick all six balls.

And then, once again, there’s taxes – maybe you can write off all of your non-winning tickets as ‘losses’, but perhaps it’s better you talk to a professional.

Click to continue to part two of this Powerball series.

11 Comments
  1. Chuck C permalink

    What I would like to know is how many combinations were still left unchosen.

    • I’d love to see how frequently each white number is chosen. I’m guessing 12 and below are highest, then 31 and below, then 69 (just because), then the rest

  2. Doug B permalink

    Regarding “your numbers don’t matter”, don’t you think that, because some percentage of people play their birthdays, their family’s birthdays, etc., that your odds of having to split the jackpot with someone are higher if you play low numbers (<=12, <=31), so you would slightly increase your expected value by playing higher numbers? Granted, we don't know by how much, but it would seem that it's probably a little bit.

    • Yes, probably increased odds that you’ll split on lower numbers. But you’d have to hit all six numbers first, which is where the real problem lies.

  3. Kelly permalink

    Don’t forget about taxes too (although the hypothetical person who bought a huge enough number of tickets to make deducting them worthwhile could use that to offset the taxable amount).

    • Yes, taxes are really only an issue *after* you win. I suppose there could be an upside for deducting after a big loss but I’m guessing most readers won’t find that useful.

  4. Doesn’t the best-case scenario also include 25 wins of $1 million and all the combinations of the smaller wins as well? Probably doesn’t push it past the tipping point, but still to be considered.

    Also, the last time there was some giant lottery fever, I remember reading something about the bigger problem for the best-case scenario was being capable of purchasing every combination (between filling out and scanning all 100mil+ the forms necessary to get every combination, it would take more than 4 days of manpower?)

    • Yes, the total for the best-case scenario includes all of the non-jackpot prizes.

      I highly doubt Powerball has a way to feed numbers directly into the central computer, so I’m guessing people would have to do it the old-fashioned way. I recently learned this happened in 1992: an Australian gambling group bought 5 million of the 7 million combinations in the Virginia lottery, winning a $27 million jackpot. (Imagine if they had missed!)

      • Yes, that’s the story I read as well. As you say, they were able to orchestrate purchasing only 5 million tickets (out of 7) … you’d need a lot more tickets for this one, but you’d also have a lot more stores to work with. I just don’t understand how they manage to “hire” all the people necessary to fill out the forms and do all the purchasing without having to split the jackpot too many ways.

        -shrug- Large numbers are fun.

      • I’ve been thinking about that a bit today – I assume they printed out as many cards as they could in advance and gave them to their package-carriers, who would take them to the stores to be scanned. Too many possibilities for mistakes (and way too time-/$-consuming) if done manually.

        Particularly impressive as this was in the early ’90s, when the sign of technological luxury (as I recall from my early days) was having a fax machine in your car :)

  5. I was living in Fredericksburg VA at this time and I had organized a small ticket pool for my co-workers and myself.
    You are correct that the group from Australian group used preprinted tickets. They also monopolized the machines at many locations to the exclusion of many irate “regular folks”. Following this incident, the VA lottery changed their rules so that players could not use pre-printed slips and that players buying “large blocks” of tickets could not monopolize the machines. Also, that lotto drawing was held once a week, so the time they had to do all was a little longer than the twice a week cycle you have today.

What do you think?