# How many tickets will be sold before Wednesday? …and other burning Powerball questions

*This is my second post on Powerball. Here’s the first, which looks at expected value.*

Before Saturday night’s then-record Powerball drawing, a reader asked me to calculate the odds that more than one person would hit the jackpot. That would require us to know how many tickets might be sold; unfortunately, I was traveling (see photo at right), and couldn’t run the stats to make a prediction.

Lucky for us, no one won, so we get another few days to play with even bigger numbers.

According to LottoReport.com, which keeps sales and jackpot data on Powerball and other games, there were 440,321,172 tickets sold for Saturday’s ~~$700~~ ~~$800~~ ~~$900~~ $947.9 million drawing. On Saturday afternoon, I took a quick glance at the previous data and made a guess of 350-400 million based on what I thought was a pattern in the numbers. I thought I was going to be high.

That number – 440,321,172! – is incredible. Every single person in the United States could have had one ticket and there would’ve still been more than 100 million left over.

I know what many of you are thinking: the odds of hitting the jackpot are 1 in 292,201,338 million, so someone probably should have won, right? Well, not necessarily.

## How come no one hit the jackpot?

The probability that any one ticket will fail to hit the jackpot is 292,201,337 / 292,201,338, or 99.9999996577702%. To calculate the probability that *all* of the 440,321,172 tickets would lose, we have to multiply that probability by itself 440,321,172 times. Put another way:

0.999999996577702^{440,321,172} ≈ 0.2216

There was a 22.16% chance that no one would win. That’s actually pretty surprising, even to me, but that’s math for you! (I’m assuming all numbers were chosen at random, but even if 1,000,000 people played the *Lost* numbers, it wouldn’t make that big of a difference.)

Given those sales, here were the odds that we would have a certain number of winners:

# winners | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

probability | 22.2% | 33.4% | 25.2% | 12.6% | 4.8% | 1.4% | 0.4% | 0.1% |

Feel free to try to calculate them yourself, if you’d like. As for me, I hate dwelling on the past, so let’s look ahead to Wednesday’s drawing for $1.3 billion. (That’s with a *B*, followed by an *illion*.)

## How many tickets will be in play?

From LottoReport.com, I took data for Powerball from January 2013 through the present and plugged them into Excel. I was interested to see whether the size of the jackpot (the independent variable) affects the number of tickets purchased (the dependent variable).

The graph of those 316 lottery drawings looks like this:

It’d be great if we could connect these points with a straight line; indeed, if you were to draw a line from left to right, you’d start it out flattish, climbing slowly. As the jackpots get higher, however, the line starts to curl upward.

*Note: the next few paragraphs are rather technical, so if you are unfamiliar with regressions or don’t want to get lost in the statistical weeds, feel free to skim until the next blue text.*

Excel is powerful enough to calculate power regressions by itself, but I wanted to be sure the results were statistically significant. I took the natural logarithm of the values *J* (jackpot, in millions) and *T* (number of tickets sold, in millions), then used StatPlus to compare them. I got the following, all values *p* < 0.001:

- All 316 drawings: ln(
*T*) = -0.2211 + 0.6808 ln(*J*) - Highest 50 jackpots: ln(
*T*) = -6.6924 + 1.865 ln(*J*) - Highest 25 jackpots: ln(
*T*) = -8.4909 + 2.1631 ln(*J*)

(N.B. on the cut-offs in terms of jackpots: 50th was $188 million; 25th was $255 million.)

Saturday’s drawing was for $947.9 million and had 440.3 million tickets sold. Plugging *J* = 947.9 into each formula, I got the following outcomes for *T*:

- All 316 drawings: 85.2 million tickets (error: -80.6%)
- Highest 50 jackpots: 441.8 million tickets (error: 0.3%)
- Highest 25 jackpots: 564.3 million tickets (error: 28.1%)

That Highest 50 number is really, really close. To check it further, I took Saturday’s jackpot out of the data and re-ran the regression; the new formula predicted Saturday’s drawing within 0.4%.

Good enough for me; we’re not trying to be entirely scientific here. (There are a few issues with this approach, of course, which I discuss below.)

We’ll use the formula without Saturday’s results, which is:

ln(*T*) = -6.698 + 1.866 ln(*J*)

If *J* = 1,300, then ln(*T*) = 6.681; this model therefore predicts we’ll see 797.4 million tickets head out the door – and that’s not taking into account the near-certainty the prize will rise before the draw.

If that number is indeed correct, there will still be a 6.5% chance that no one wins, and soon we’ll be talking about using the Powerball jackpot to pay off the national debt. The chance that two or more tickets will split the big prize is now over 75%, however, up from around 44% on Saturday.

**HEY YOU** – if you were skimming, you can stop skimming now.

Here are the various probabilities of *n* winners on Wednesday, assuming 797.4 million tickets are out there:

# winners | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

probability | 6.5% | 17.8% | 24.3% | 22.1% | 15.1% | 8.2% | 3.7% | 1.5% |

Be sure to factor those probabilities into your expected value calculations, if you’re making any.

## Why are you hesitant about your model?

This is just a fun exercise so I can talk about statistics; I wouldn’t make any scientific claims based on these data.

First and foremost, we were *already* deep into uncharted territory. Before Saturday, the jackpot had never reached $600 million; it’s hard to extrapolate with just one data point connecting the previous record to a number more than twice as large, even if that in-between data point fit almost perfectly into the model.

Second is the reason we’re in this mess in the first place. As I mentioned previously, Powerball recently changed its format to make it much, much more improbable to hit the jackpot. The reduced odds of winning (from “basically zero” to a smaller “basically zero”) *might* cause some people to not play, but as one of my friends told me yesterday, she’s buying tickets simply for the ability to *dream* about winning the jackpot.

My cut-off points were arbitrary; maybe I should figure out at what jackpot value ‘regular people’ (i.e., people who don’t play habitually) start to take notice, and use only the data points above. If I use $500 million as my cut-off, for example, the line of best fit suggests ‘only’ 655 million tickets will be sold.

If the jackpot keeps growing, the model suggests the number of tickets will continue to rise at a faster clip. But my guess is almost everyone who’s planning to buy a ticket did so on Saturday, and they’re not necessarily going to increase their purchases just because the jackpot climbs even higher. Given some of the videos I’ve seen of people waiting in line, some lottery sellers might already be at capacity.

One final reason: my model doesn’t take into account data from 2012, when Powerball switched to its $2 format. Either I didn’t see I could click to access older data, or I was just lazy; you pick. Unfortunately, that means I missed a drawing where 281 million tickets were purchased for a $587 million jackpot. That would have flattened my curve quite a bit, and I’ll probably go back and run those numbers again later.

## What are the odds I didn’t hit that number at all?

One statistics question for you to try, if you like, before we part ways.

I did end up buying some tickets for Saturday’s drawing – 45 of them, in fact. You see, Saturday was my grandfather’s 90th birthday, and I thought *maybe* it would be good luck if I bought $90 worth of lottery tickets. (I am normally an overly rational person, but I have my moments of weakness.)

After the drawing, I went through my long list of numbers to see what (or if) I had won. I had a few lines that matched 32, the first ball up. But try as I might, I couldn’t find a single entry that had the second ball up, 16.

Question: What were the odds that a winning white ball – *any* winning white ball, not just the number 16 – would not appear *anywhere* in my 45 picks? My calculations are below. (Remember, there are 69 white balls in all, 5 of which will be selected by the machine.)

## Go get ’em, tiger – and remember, skip the Power Play

If you’re playing again on Wednesday, good luck! Just please, *please* be sure you can afford it. If you can’t, and you’re still tempted to play, I urge you to reach out to one of the many organizations that can help you make the right decision.

Here’s the answer to the above question:

Correct response: | Show |
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I really want to know how many ways 1 2 3 4 5 PB6 would split. My guess is a several

hundred.

I recall reading about a big lottery where the first three numbers were 1 2 3. They interviewed the winner and she said she almost threw her ticket away when if spit out the numbers like that. I have also had people insist that you shouldn’t pick consecutive numbers because they won’t hit. Further proof the lottery is a tax on people who are bad at math.