The headline is misleading. It should be “when to play the lottery such that your expected return is positive”. Given that your odds of winning are so low, you’d have to play for a very long time to actually get a return. So don’t.
Anyway: The week I wrote this post, MegaMillions paid out a record jackpot of $640 million (annuitized value). Because jackpots roll over when they aren’t claimed, it actually becomes possible for the expected value to be positive. Lotteries can’t allow this in the long term (or else they wouldn’t make money) but when the jackpots roll over, you’re essentially making money off early players. Same idea behind progressive slot machines.
The odds of winning Mega Millions are ~1:175million (source). Tickets cost $1. So if if the jackpot is over $175 million your expected return is positive, right?
WRONG! Because if multiple people win, they split the jackpot. This is where the question gets tricky, you don’t have a way of knowing how many people are buying tickets. Rather, how many tickets are being bought. Sales in this latest record-setter were close to $700 million, well over twice what they would be if every person in the 300 million US bought a single ticket.
Shown above are the ticket sales (in dollars) vs. jackpot size (all numbers in millions)[1.5]. Since the cost is $1/ticket, that’s also the number of tickets sold . For your expected return to be positive, the jackpot has to be larger than max(1, # of winners) x $175 million.
Since the record jackpot is $640 million, I’m only going to consider numbers that high. Below is a table of predicted number of players for a set of jackpot sizes, and the expected number of winners given independent ticket choices (a reasonably good assumption unless somebody buys all the possibilities).
|Jackpot Size (millon dollars)||Expected # of Players (millions)||Expected # of winners|
We see that the expected number of winners doesn’t even reach 1 until around 180 million tickets, which only occurs in a jackpot of $350 million. So for most games it’s unlikely for anybody to win.
So an average, everything looks okay. The expected number of winners is < 1 up to about $375 million. At $500 million, where your expected return from a solo win would be a jaw-dropping $1.85, you’ likely end up sharing your winnings with 1 other person, bringing your winnings down to $250/$175 -$1 = $0.43.
There are a few problems at this high level, though. First, one can see from the graph that the regression is much less accurate. However, this works in a players favor since my equation overestimates the number of players.
A worse problem is the variance, which I have ignored. Since there’s no such thing as 0.3 of a winner, one should consider the probability of 0, 1, 2, 3 winners. These will follow a binomial distribution, with the number of trials being the number of players and the probability of success being p = 1/175million. The variance is Np(1-p) ~= Np = mean. Standard deviation = sqrt(variance) ~= sqrt(mean). When 4 people win on average, there is a (roughly) 70% chance of between 2 and 6 people winning. 6 people splitting a $600 million jackpot would kill your winnings.
So it’s risky. One could conceivably buy all 175 million tickets when the jackpot reached a high level, and the probabilistic margin for profit would exist. But it would not be a guarantee.
 Like any physicist, I’ll be rounding numbers for simplicity. The real number is (56 choose 5) x 46, or 175,711,536.
[1.5] Data is from Jan. 2011 – Mar. 2012. Lots of states joined in 2010, an increasing sales base disrupts the regression.
 There is a “megaplier” option which costs more, increases certain payouts, but doesn’t affect the jackpot payout. I believe those sales figures have been excluded from this data.