match 5 of 6
1 in 39,000
12
$600,000
$50,000
match 4 of 6
1 in 800
587
$1.4m
$2,385
match 3 of 6
1 in 47
10,000
$600,000
$60
So the average ticket could be expected to bring home cash winnings of
$50,000 / 39,000 + $2385 / 800 + $60 / 47 = $5.53.
An investment where you make three and a half bucks of profit on a $2 investment is not one to pass up.*
Of course, if one lucky person hits the jackpot, the game turns back into a pumpkin for everybody else. But Cash WinFall was never popular enough to make that outcome likely. Out of forty-five roll-down days during the lifetime of the game, only once did a player match all six numbers and stop the roll-down in its tracks.*
Let’s be clear—this computation doesn’t mean that a $2 bet is sure to win you money. On the contrary, when you buy a Cash WinFall ticket on a roll-down day, your ticket is most likely a loser, just as it is on any other day. The expected value is not the value you expect! But on roll-down day, the prizes, in the unlikely event that you do win, are bigger—a lot bigger. The magic of expected value is that the average payout of a hundred, or a thousand, or ten thousand tickets is very likely to be close to $5.53. Any given ticket is probably worthless, but if you’ve got a thousand tickets, it’s essentially certain that you’ll make your money back and then some.
Who buys a thousand lottery tickets at a time?
Kids at MIT, that’s who.
The reason I can tell you the WinFall payoffs on February 7, 2005, down to the last dollar is because this figure is recorded in the exhaustive and, frankly, kind of thrilling account of the WinFall affair submitted to the state in July 2012 by Gregory W. Sullivan, the inspector general of the Commonwealth of Massachusetts. I think I’m safe in saying this is history’s only state fiscal oversight document which inspires the reader to wonder: Does someone have the movie rights to this?
And the reason it’s this particular day for which this data is recorded is that February 7 was the first roll-down day after James Harvey, an MIT senior working on an independent study project comparing the merits of various state lottery games, realized that Massachusetts had accidentally created an insanely profitable investment vehicle for anyone quantitatively savvy enough to notice it. Harvey got a group of friends together (at MIT, it’s not hard to get a group of friends together who can all compute expected value) and bought a thousand tickets. Just as you might expect, one of those 1-in-800 shots came through, and Harvey’s group took home one of those $2,000 prizes. They won a bunch of match-3s too; in all, they just about tripled their initial investment.
It won’t surprise you to hear that Harvey and his co-investors didn’t stop playing Cash WinFall. Or that he never did get around to finishing that independent study—at least not for course credit. In fact, his research project quickly developed into a thriving business. By summer, Harvey’s confederates were buying tens of thousands of tickets at a time—it was a member of his group who placed the mammoth order at the Cambridge Star Market. They called their team Random Strategies, though their approach was anything but scattershot; the name referred to Random Hall, the MIT dorm where Harvey had originally cooked up his plan to make money on WinFall.
And the MIT students weren’t alone. At least two more betting clubs formed up to take advantage of the WinFall windfall. Ying Zhang, a medical researcher in Boston with a PhD from Northeastern, formed the Doctor Zhang Lottery Club. It was the DZLC that accounted for the spike in sales in Quincy. Before long, the group was buying $300,000 worth of tickets for each roll-down. In 2006, Doctor Zhang quit doctoring to devote himself full-time to Cash WinFall.
Still another betting group was led by Gerald Selbee, a retiree in his seventies with a bachelor’s degree in math. Selbee lived in Michigan, the original home of WinFall; his group of thirty-two bettors, mostly made up of his relatives, played WinFall there for about two years until the game shut down in 2005. When Selbee found out the gravy train was getting back on the tracks out East, his course was clear; in August 2005, he and his wife Marjorie drove to Deerfield, in the western part of Massachusetts, and placed their first bet—sixty thousand tickets. They took home a little over $50,000 in pure profit. Selbee, with the benefit of his experience playing the game in Michigan, added an extra profit-making venture to his Cash WinFall tickets. Stores in Massachusetts got a 5% commission on lottery ticket sales. Selbee cut deals directly with one store, offering hundreds of thousands of dollars worth of business at a time in exchange for going halfsies on the 5% commission. That move alone made Selbee’s team thousands of dollars in extra profit every roll-down.
You don’t need an MIT degree to see how the influx of high-volume players affected the game. Remember: the reason the roll-down payoffs were so swollen was that a lot of money was being split among just a few winners. By 2007, a million or more tickets were being sold for each roll-down drawing, most of them to the three high-volume syndicates. The days of the $2,300 prize for matching four out of six numbers were long gone; if a million and a half people bought tickets, and one person in eight hundred matched 4, then you’d typically see almost two thousand match-4 winners. So each share of the $1.4 million kitty was now more like $800.
It’s pretty easy to figure out how much a big player stood to gain from Cash WinFall—the trick is to look at it from the point of view of the lottery itself. If it’s roll-down day, the state has (at least!) $2 million of accumulated jackpot money it’s got to get rid of. Let’s say a million and a half people buy tickets for the roll-down. That’s $3 million more in revenue, of which 40%, or $1.2 million, goes into the state’s coffers, and the other $1.8 million gets plowed into the jackpot fund, all of which is to be disbursed to bettors before the day is through. So the state takes in $3 million that day and hands out $3.8 million:* $2 million from the money already in the jackpot fund and $1.8 million from that day’s ticket receipts. On any given day, whatever the state makes, the players, on average, lose, and vice versa. So this day is a good day to play; ticket buyers, in the aggregate, took $800,000 from the state.
If players buy 3.5 million tickets, it’s a different story; now the Lottery takes $2.8 million as its share and pays out the remaining $4.2 million. On top of the $2 million already in the kitty, that amounts to $6.2 million, less than the $7 million of revenue the state took in. In other words, despite the generosity of the roll-down, the lottery has gotten so popular that the state still ends up making money at the expense of the players.
This makes the state very, very happy.
The break-even point comes when the 40% share of the roll-down day revenue exactly matches the $2 million already in the pot (that is, the money contributed by the players who were unsophisticated or risk-loving enough to play WinFall without a roll-down). That’s $5 million, or 2.5 million tickets. More sales than that, and WinFall is a bad bet. But any fewer—and over the life span of the WinFall game, it always was fewer—and WinFall offers players a way to make some money.
What we’re really using here is a wonderful, while at the same time commonsensical, fact called additivity of expected value. Suppose I own a McDonald’s franchise and a coffee shop, and the McDonald’s has an expected annual profit of $100,000, while the coffee shop’s expected net is $50,000. The money might go up and down from year to year, of course; the expected value means that, in the long run, the average amount of money the McDonald’s makes will be about $100,000 a year, and the average am
ount from the coffee shop $50,000.
Additivity says that, on average, my total take from Big Macs and mochaccinos together is going to average out to $150,000, the sum of the expected profits from each of my two businesses.
In other words:
ADDITIVITY: The expected value of the sum of two things is the sum of the expected value of the first thing with the expected value of the second thing.
Mathematicians like to sum up that reasoning in a formula, just as we summed up the commutativity of addition (“this many rows of that many holes is the same thing as that many columns of this many holes) by the formula a × b = b × a. In this case, if X and Y are two numbers whose values we’re uncertain about, and E(X) is short for “the expected value of X,” then additivity just says
E(X+Y) = E(X) + E(Y).
—
Here’s what this has to do with the lottery. The value of all the tickets in a given drawing is the amount of money handed out by the state. And that value isn’t subject to uncertainty at all;* it’s just the amount of roll-down money, $3.8 million in the first example above. The expected value of a sure $3.8 million is, well, just what you expect—$3.8 million.
In that example, there were 1 million players on roll-down day. Additivity tells you that the sum of the expected values of all 1.5 million lottery tickets is the expected value of the total value of all the tickets, or $3.8 million. But each ticket (at least before you know what the winning numbers are) is worth the same. So you’re summing 1.5 million copies of the same number and getting $3.8 million; that number must be $2.53. Your expected profit on your $2 ticket is 53 cents, more than 25% of your wager, a handsome profit on what’s supposed to be a sucker’s bet.
The principle of additivity is so intuitively appealing that it’s easy to think it’s obvious. But, just like the pricing of life annuities, it’s not obvious! To see that, substitute other notions in place of expected value and watch everything go haywire. Consider:
The most likely value of the sum of a bunch of things is the sum of the most likely values of each of the things.
That’s totally wrong. Suppose I choose randomly which of my three children to give the family fortune to. The most likely value of each child’s share is zero, because there’s a two in three chance I’m disinheriting them. But the most likely value of the sum of those three allotments—in fact, its only possible value—is the amount of my whole estate.
BUFFON’S NEEDLE, BUFFON’S NOODLE, BUFFON’S CIRCLE
We have to interrupt the story of the college nerds versus the lottery for a minute, because once we’re talking about additivity of expected value I can’t not tell you about one of the most beautiful proofs I know, which is based on the very same idea.
It starts with the game of franc-carreau, which, like the Genoese lottery, reminds you that people in olden times would gamble on just about anything. All you need for franc-carreau is a coin and a floor with square tiles. You throw the coin on the floor and make a bet: will it land wholly within one tile, or end up touching one of the cracks? (“Franc-carreau” translates roughly as “squarely within the square”—the coin used for this game was not a franc, which wasn’t in circulation at the time, but the ecu.)
Georges-Louis LeClerc, Comte de Buffon, was a provincial aristocrat from Burgundy who developed academic ambitions early on. He went to law school, perhaps with the aim of following his father into the magistracy, but as soon as he finished his degree he threw aside legal matters in favor of science. By 1733, at the age of twenty-seven, he was ready to stand for membership in the Royal Academy of Sciences in Paris.
Buffon would later gain fame as a naturalist, writing a massive, forty-four-volume Natural History that laid out his proposal for a theory intended to explain the origin of life as universally and parsimoniously as Newton’s theory had explained motion and force. But as a young man, influenced by a brief meeting and long exchange of letters with the Swiss mathematician Gabriel Cramer,* Buffon’s interests lay in pure mathematics, and it was as a mathematician that he offered himself to the Royal Academy.
The paper Buffon presented was an ingenious juxtaposition of two mathematical fields that had been thought of as separate: geometry and probability. Its subject wasn’t a grand question about the mechanics of the planets in their orbits or the economies of the great nations, but rather the humble game of franc-carreau. What was the probability, Buffon* asked, that the franc would land entirely within a single tile? And how large should the floor tiles be to make the game a fair bet for both players?
Here’s how Buffon did it. If the coin has radius r and the square tile has a side of length L, then the coin touches a crack exactly when its center lands inside a smaller square, whose side has length L − 2r:
The smaller square has area (L − 2r)2, while the bigger square has area L2; so if you’re betting on the coin landing “squarely in the square,” your chance of winning is the fraction (L − 2r)2 / L2. For the game to be fair, this chance needs to be 1/2; which means that
(L − 2r)2 / L2 = 1/2
Buffon solved this equation (and so can you, if that’s the kind of thing you’re into), finding that franc-carreau was a fair game just when the side of the carreau was 4 + 2√2 times the radius of the coin, a ratio of just under seven. This was conceptually interesting, in that the combination of probabilistic reasoning with geometric figures was novel; but it wasn’t difficult, and Buffon knew it wouldn’t be enough to get him into the academy. So he pressed forward:
“But if instead of throwing in the air a round piece, as an ecu, one would throw a piece of another shape, as a squared Spanish pistole, or a needle, a stick, etc., the problem demands a little more geometry.”
This was an understatement; the problem of the needle is the one for which Buffon’s name is remembered in mathematical circles even today. Let me explain it more precisely than Buffon did:
Buffon’s Needle Problem: Suppose you have a hardwood floor made of long, skinny slats, and you happen to have in your possession a needle exactly as long as the slats are wide. Throw the needle on the floor. What’s the chance that the needle crosses one of the cracks separating the slats?
Here’s why this problem is so touchy. When you throw the ecu on the floor, it doesn’t matter which direction Louis XV’s face ends up pointing. A circle looks the same from every angle; the chance that it crosses a crack doesn’t depend on its orientation.
Buffon’s needle is a different story. A needle oriented nearly parallel to the slats is very unlikely to cross a crack:
but if the needle lands crosswise to the slats, it’s almost certain to do so:
The franc-carreau is highly symmetric—in technical terms, we say it is invariant under rotation. In the needle problem, that symmetry has been broken. And that makes the problem much harder; we need to keep track of not just where the center of the needle falls, but also what direction it’s pointing.
In the two extreme cases, the chance the needle crosses a crack is 0 (if the needle is parallel to the slat) or 1 (if the needle and the crack are perpendicular). So you might split the difference and guess that the needle touches a crack exactly half the time.
But that’s wrong; in fact, the needle crosses a crack substantially more often than it lands wholly within a single slat. Buffon’s needle problem has a beautifully unexpected answer: the probability is 2 / π, or about 64%. Why π, when there’s no circle in sight? Buffon found his answer using a somewhat intricate argument involving the area under a curve called the cycloid. Computing this area requires a bit of calculus; nothing a modern-day sophomore math major couldn’t handle, but not exactly enlightening.
But there’s another solution, discovered by Joseph-Émile Barbier more than a century after Buffon’s entry into the Royal Academy. No formal calculus is needed; in fact, you don’t need computation of any kind. The argument, while a little involved, uses no more than arithmeti
c and basic geometric intuition. And the crucial point is, of all things, the additivity of expected value!
The first step is to rephrase Buffon’s problem in terms of expected value. We can ask: What is the expected number of cracks the needle crosses? The number Buffon aimed to compute was the probability p that the thrown-down needle crosses a crack. Thus there is a probability of 1 − p that the needle doesn’t cross any cracks. But if the needle crosses a crack, it crosses exactly one.* So expected number of crossings is obtained the same way we always compute expected value: by summing up each possible number of crossings, multiplied by the probability of observing that number. In this case the only possibilities are 0 (observed with probability 1 − p) and 1 (observed with probability p) so we add up
(1 − p) × 0 = 0
and
p × 1 = p
and get p. So the expected number of crossings is simply p, the same number Buffon computed. We seem to have made no progress. How can we figure out the mystery number?
When you’re faced with a math problem you don’t know how to do, you’ve got two basic options. You can make the problem easier, or you can make it harder.
Making it easier sounds better—you replace the problem with a simpler one, solve that, and then hope that the understanding gained by solving the easier problem gives you some insight about the actual problem you’re trying to solve. This is what mathematicians do every time we model a complex real-world system by a smooth, pristine mathematical mechanism. Sometimes this approach is very successful; if you’re tracking the path of a heavy projectile, you can do pretty well by ignoring air resistance and thinking of the moving body as subject only to a constant force of gravity. Other times, your simplification is so simple that it eliminates the interesting features of the problem, as in the old joke about the physicist tasked with optimizing dairy production: he begins, with great confidence, “Consider a spherical cow . . .”
How Not to Be Wrong : The Power of Mathematical Thinking (9780698163843) Page 21
