Is Powerball a Mug's Game?

A mathematician's guide to the news.
Aug. 31 2001 8:30 PM

Is Powerball a Mug's Game?

It all depends on when you play—and what value you put on a dollar.

Illustration by Robert Neubecker

Were you stupid not to play?

I don't have to ask myself; I played. My father and his Ph.D. in statistics put me in for a 20 percent share of his four tickets. But I got enough razzing from friends and neighbors that I thought it was worth explaining why, from a mathematician's point of view, last Saturday's drawing wasn't necessarily dumb.

The question to ask is: What is the expected value of a lottery ticket? If the expected value is more than a dollar, and the ticket costs a dollar, you should buy a ticket. If the expected value is less than a dollar, you should keep your money.

"Expected value" doesn't just mean "what do you expect?" After all, you probably expect the ticket to be worth nothing. Yet people don't think lottery tickets are worthless; if they did, they wouldn't buy them. "Expected value" as I mean it here is a mathematical definition that assigns a fixed value to an object whose true value is subject to uncertainty.

Suppose an object might be worth either V1 or V2 dollars, and suppose the probability is P1 that it is worth V1, and P2 that it is worth V2. Then the expected value is defined to be

P1 x V1 + P2 x V2.

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For instance, suppose you place a bet on a horse that has a 1/10 chance of winning, and the bet pays $100. Then the probability is (1/10) that your ticket will be worth $100 and (9/10) that your ticket will be worth nothing. So, the expected value of the ticket is

(1/10) x $100 + (9/10) x 0 = $10.

Why is $10 a good definition of the value of the ticket? Because if you spent a week at the track and bought, say, 250 such tickets, you'd probably end up winning about 25 times; you'd make $2,500, or $10 per ticket. So, if you were paying more than $10 for each ticket, you'd be a loser; less, and you'd be a winner.

So, what's the expected value of a Powerball ticket? Here's a wrong argument I heard a lot. People who knew the jackpot odds figured: "I've got a 1 in 80 million chance at $280 million, so the expected value is

(1/80 million) x $280 million + (79,999,999/80 million) x 0 = $3.50.

That's a good bet!"

The problem with that argument is that we weren't playing for the $280 million. We were playing for our share of the $280 million, thanks to the possibility of multiple winners. If I win the Powerball, the chance is pretty good that somebody else is going to win, too. Already my jackpot's down to $140 million. And the more people who play, the more the prize will tend to divvy up. If, as happened in real life this week, four people win, you're looking at just $70 million.

So how many people could you expect to share the prize with? I worked this out (requires Adobe Acrobat). The chance of no winner is about 8 percent; of one winner, 21 percent; of two, 26 percent; of three, 21 percent; of four, 13 percent; of five or more, 9 percent. So Saturday's fourfold victory was a bit of a surprise but not a real shocker. A Powerball enthusiast suggested to me that people are especially fond of picking lucky 21 on the red ball and that this could explain the large number of winners. But in fact, Powerball records show that the number of people who picked 21 on the red ball was only half a percentage point greater than would have been expected by chance. That's not enough to make a serious difference.

On average, the winner could expect a 37 percent share of the jackpot, or $103 million. A 1 in 80 million chance of $103 million should still have an expected value of (1/80 million) times $103 million, or $1.29. Still a good deal—especially because we haven't even thrown in the chance of winning a lesser prize.

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