# When Not to Expect What You’re Expecting

My wife and I played the Massachusetts PowerBall lottery last month. In one respect it was a good thing that we didn’t win: if we had, it would’ve made my job as a math popularizer that much harder. When a lottery-winner says that playing the lottery is a bad investment strategy, it comes across as hypocritical at best.

One way experts in statistics and probability try to explain why buying a lottery ticket every week is a bad retirement plan is by invoking the concept of expected value. To illustrate the idea, imagine for simplicity a scaled-down version of a lottery — a roulette game with a wheel that has 38 pockets. Suppose that you paid \$1 to bet that the ball will wind up in a particular pocket, and that the payoff if you guessed right will be \$36. Then in 37 possible worlds you win \$0 while in 1 possible world you win \$36. Your average payoff over those 38 equally likely parallel worlds is (37x\$0+1x\$36)/38, or about 95 cents, which is less than what you paid to play. If you like the ambiance of casinos (music, drinks, company) but hate the element of uncertainty, or if the idea of other yous in other worlds having different destinies freaks you out, you can fold all thirty-eight worlds into one by placing a bet on each pocket. Then you’ll pay \$38 and are sure to win \$36, for a net loss of \$2. (Seems like a bad idea, but maybe the croupier is cute and you’re hoping your unorthodox betting strategy will make a good conversation-starter.) State-run lotteries are a lot like this roulette game: if you were to bet on every “pocket” (ignoring the fact that there might not be enough money in the economy to enable you to do that), you’d lose big time. And it stands to reason that if buying all possible tickets is a bad idea, so is buying just one per week, or just one this time.

But getting back to the casino example: wouldn’t it be nice if the casino paid \$40 rather than \$36 for a winning bet? Then the strategy of betting on all 38 pockets would give you a profit of \$2 instead of a loss of \$2. A \$2 profit doesn’t sound like much, but the strategy scales up: bet \$1000 on every pocket and you’re guaranteed to make a \$2000 profit. (Don’t have \$38,000 lying around? Find some rich friends to front you the money and promise to split the \$2000 profit with them.) But how likely is it that a casino would offer such a good deal?

Amazingly, there is a casino that operates this way. It’s a floating casino that pops up from time to time, Brigadoon-like, in places like Virginia and Michigan and Massachusetts. It’s not a real casino, of course; it’s just various state lotteries, on particular days and weeks when the expected return on a ticket is greater than the cost of a ticket. In a case like that, how many tickets should you buy?

“How Not to Be Wrong”, by Jordan Ellenberg.

Before I tackle that question, I want to put in a plug for the book that taught me about what happened in Michigan and Massachusetts: Jordan Ellenberg’s “How Not to Be Wrong”. It’s a lot of fun, and has things to offer a wide spectrum of readers. Ellenberg discusses chance and other issues informatively and entertainingly, dispensing insights that will interest seasoned mathematicians and lay readers alike. You might also enjoy the Audible.com audio book, read by the author; I found it pleasant company while driving the car, walking the dog, washing the dishes, etc.

But let’s get back to the question about gambling. Should you buy lots and lots of tickets on days when the expected payback of a ticket exceeds the cost of a ticket? The kicker is the word “expected”. The expected value, as defined in probability theory, is an average over possible worlds, and averages can be misleading. Specifically, the average value of a quantity isn’t very informative in situations where the quantity varies a lot. If you buy a ticket that with probability 0.99 will prove to be worthless and with probability 0.01 will turn out to be worth \$1000, then the expected value of the ticket is \$10 (that’s 0.99×0 + 0.01×1000), but if you’re only buying the one ticket, that kind of calculation takes us into never-never land: you may win \$0 from your purchase, or you may win \$1000, but you will never-never win \$10. Paradoxically, you can expect, with 100% certainty, that whatever you win (\$1000 or \$0), it will not be the “expected” value (\$10).

A traditional rationale for the notion of expected value is the law of large numbers, which applies when you try the same experiment (or make the same wager) a large number of times. If you buy n of those \$1 tickets, then the average payoff of your n tickets (that is, the sum of their payoffs, divided by n) will tend to be close to \$10. But n needs to be fairly large before this effect sets in. In ordinary casinos (the kind that are profitable for the casino-owners), bets favor the house on average, and a large number of people make bets, so that the law of large numbers operates reliably in favor of the house. What about lotteries? A large number of people buy tickets, just as a large number of people play casino games, so when the odds are against the individual buyer (as they are against the individual bettor), the state can count on making money in the long run.

But what about the short run, specifically in those windfall weeks when the expected value of a ticket exceeds its price? Many people who write about lotteries (such as the recent PowerBall lottery) act as if there’s a big qualitative difference between those once-in-a-blue-moon days when expected winnings exceed ticket cost versus all those other days, but caution is in order. Unless you spend thousands of dollars on tickets, the number of tickets you buy will still count as a “small” number as far as the law of large numbers is concerned, so expected value is of questionable significance. On those once-in-a-blue-moon days, if you buy one ticket, or even a hundred, all you can really say is that your chance of winning a big prize is slightly higher than usual.

There is a loophole here, which you may have noticed when I wrote “unless you spend thousands of dollars on tickets”. Why not spend thousands of dollars on lottery tickets when expected value is in the buyer’s favor? The main reason is risk; there’s a sizable chance that you’ll blow your investment with nothing to show for it. But what if you bought so many tickets that success was assured?

In fact, that’s just what one ticket-buying cartel in Virginia did in the 1990s (as described in an entertaining Planet Money podcast). The cartel’s answer to the question “How many tickets should we buy?” was, correctly: All of them! They bought all the possible combinations that the lottery allowed, thereby assuring themselves the jackpot (or at least a share of it). Virginia took a while to respond to this heist, and the cartel kept creatively finding new loopholes, but eventually Virginia succeeded in making it unworkable for cartels to game their lottery.

What about other states? Your usual state lottery, unlike Virginia’s, uses more number combinations than any cartel could amass, but despite this obstacle, there may still be a way to beat it every now and then using the law of large numbers. That’s because many lotteries don’t have just a single big jackpot prize; they award smaller prizes to people who manage to guess several of the jackpot numbers. Some savvy folks exploited this to game the Michigan lottery a few years later. They bought many thousands of tickets, and those tickets repaid the investment by reaping many small prizes. Because the small prizes were one-in-a-thousand shots rather than one-in-a-million shots, the law of large numbers kicked in for the cartel, virtually guaranteeing a certain number of the small prizes; and because those small prizes were more valuable than usual that week, the cartel was virtually guaranteed a profit.

Ellenberg points that the the victim here is not the state running the lottery: the state still makes a profit from running the game, and the extra hype and publicity generated by once-in-a-blue-moon days helps fill state coffers. What the cartels are doing is creating unlicensed virtual casinos that piggy-back off state lotteries: by buying so many tickets that the law of large numbers applies to the tickets they buy, the cartels are able to enjoy the reduction of risk that casino-owners count on. The real victims were all the other people who poured money into the system.

A few years after the Virginia and Michigan escapades, some MIT students came up with a variant of the mass-purchasing gimmick that reduced the element of risk even below what the law of large numbers would usually involve. When the jackpot went unclaimed for a long time, the payout for matching 4 of the 6 jackpot numbers in the Cash WinFall game was \$2,385, and the chance of winning such a prize was a whopping 1/800 (approximately). That’s an expected payout of \$2.98 per ticket, well in excess of the \$2 ticket price. So the MIT kids apparently figured out a scheme for filling out their tickets in such a way that each of the 163,185 four-number combinations got represented equally often (or nearly so).

The MIT students who did this haven’t revealed exactly what they did, but Ellenberg describes one particular route that the students might have followed in designing their scheme. It’s based on finite projective geometry, and it serves as a nice example of the power of mathematical abstraction. Recall the poem “Paradox” by C. R. Wylie that that I quoted in Why This Blog?. In describing mathematics, it refers to “the power that this game, played with the thrice attenuated shades of things, has over their originals”. Euclidean geometry is already one step removed from the real world of stubby pencil-points and wobbly lines; projective geometry, with its inclusion of ideal points “at infinity”, is at a second remove; and finite projective geometry, which replaces our rich, varied and seemingly infinite universe with a finite set of points and lines, is more abstract still. Yet somehow such constructs, deployed at the right time, can make the right people a lot of money.

A finite projective plane (the ““Fano plane”“), consisting of seven points and seven lines. The “points” are 1, 2, 3, 4, 5, 6, 7; the seven “lines” are the triples {1,2,3}, {1,4,5}, {2,4,6}, {3,4,7}, {1,6,7}, {3,5,6}, {2,5,7}. Each two-element combination occurs exactly once among the seven triples. If there were a pick-three-numbers-from-1-to-7 lottery that awarded a small prize for guessing two of the day’s three lucky numbers, and you bought seven tickets corresponding to the seven lines in the Fano plane, you’d be certain to win either the jackpot or exactly three small prizes. See Ellenberg’s book for details.

If you’re not using a scheme like this for playing the lottery, my advice for blue-moon betting is to buy just one ticket. Then you’re still buying the dream of a bright fortune-filled future, without having lost much money in the mortgage-and-tuition-filled present. Sure, for twice the money you could double your probability of winning, but that chance is so small that doubling it doesn’t budge it very far away from zero.

One numerical measurement of how well or badly the expected value of a quantity provides meaningful information is a number called the “variance”. Some other time I’ll talk about variance on a more technical level, at which time I’ll address common questions. But I want to keep things really accessible today. So it’s enough to understand that a quantity has high variance if it tends to vary a lot from its expected value, low variance if it tends to vary a little, and zero variance if it’s always equal to its expected value.

When you add a bunch of random quantities together, the variance might go up or it might go down, depending on whether the quantities are “positively correlated” or “negatively correlated”. (Loosely speaking, two quantities X and Y are positively correlated if, when one of them is large, the other tends to be large as well, and vice versa; they’re negatively correlated if, when one of them is large, the other tends to be small, and vice versa.) We can see this dichotomy at work in our roulette example. If you bet on just on pocket, there’s some unpredictability to what you’ll win, and if you place 38 bets on that same pocket, the difference between your best case scenario and your worst case scenario only gets wider. But if you place one bet on each pocket, then your bets, taken together, exhibit no variance at all: exactly one of your bets will win, and the rest will lose. In the former situation (38 bets on the same pocket), your bets are positively correlated with one another; in the latter situation (38 bets on different pockets), your bets are negatively correlated with one another, and indeed, the negative correlation causes their individual variances to cancel out. This sort of negative correlation is exactly what the Massachusetts cartel achieved.

There’s a lot more to say about lotteries than I have time to get into here. For starters, there’s the issue of whether the value of the jackpot is what it seems to be. I don’t just mean that you need to take into account the payout schedule, and income taxes, and the costs of managing your newfound wealth, and the chance that you’ll have to share the jackpot with other winners; there’s also the fact that gaining a billion dollars probably won’t make you a thousand times happier than gaining a mere million will, and this nonlinear relationship between money and happiness (or, in more technical language, between money and “utility”) makes the whole idea of the expected value of a lottery ticket even more suspect. To find out more, read Ellenberg’s book!

One pre-reader of this essay, who is both a mathematician and a poker player, pointed out a flaw in my thinking in the passage where I wrote “Unless you spend thousands of dollars on tickets, the number of tickets you buy will still count as a small number as far as the law of large numbers is concerned, so expected value is of questionable significance.” My mistake lay in thinking of playing one particular lottery as an isolated event in a person’s life. Throughout your life you’ll have many chances to take chances. If the gambles are small ones, so that risk isn’t an issue, then your best policy is to take precisely those gambles that have positive expected value. Even if all the gambles are one-shot deals different in character from one another, as long as they are statistically independent of each other, a version of the law of large numbers will still apply.

Also, one can quibble with my earlier assertion “And it stands to reason that if buying all possible tickets is a bad idea, so is buying just one per week, or just one this time”; Ellenberg himself would probably fault it for being unduly glib, though not actually wrong. One of the main themes of Ellenberg’s book is that lots of thing in life aren’t linear; it’s not true in every situation that if a little bit of something is good, twice as much is necessarily better, let alone exactly twice as good. But some things do behave linearly, and expected value (as Ellenberg himself stresses) is one of them.

Here are three puzzles (of the popular “hat puzzle” variety) that may at first seem far removed from the subject matter of this essay, but which, properly considered, are about designing schemes that make the variance of some quantity as small (or as large) as possible.

Puzzle #1: A jail contains 100 prisoners. (We’ll assume that all the prisoners are male to make the pronouns simpler; we’ll let the jailer be female.) The prisoners have been blindfolded, and hats have been placed on their heads. Each hat is either red or blue in accordance with the outcome of a coin flip, but these flips were done by the jailer and were not revealed to the prisoners. The jailer tells the prisoners that in a few minutes, the blindfolds will be removed so that each prisoner can see everyone’s hat but his own. At that point, without communicating with anyone else about what he sees, each prisoner will be required to write on a slip of paper a prediction of the color of his hat; if at least half of the prisoners are right, they will all be spared, but if more than half are wrong, they will all be executed. With a great show of magnanimity, the jailer lets the blindfolded prisoners converse and strategize, secretly confident that they will derive no benefit from their conversation, since none of them has any information about anyone’s hat color yet. Is there some strategy that will guarantee the prisoners’ survival?

This is not a think-outside-the-box puzzle (unless the “box” is the jailer’s error of thinking that collective strategizing is useless, or the error of restricting oneself to strategies that treat all the prisoners the same way). As an example of a strategy that doesn’t solve the problem, but comes close, consider the strategy suggested by pre-reader David Jacobi: The prisoners agree that when the blindfolds are removed, each prisoner will look at the other 99 prisoners, see which hat-color the majority of those 99 prisoners have on their heads, and then guess that his hat will be that color too. As long as there is an initial imbalance between blue and red, you can check that this strategy will lead to a majority of the prisoners guessing correctly (and hence to all of the prisoners being spared). Unfortunately, if 50 of the prisoners have red hats and 50 have blue hats (as will happen about 8% of the time), all 100 prisoners will guess wrong, and all will be killed.

If you think that this is the best you can do, try thinking harder. If it helps, replace the number 100 by the number 2 to get started.

Puzzle #2: This is similar to puzzle #1, except that this time, the jailer says that she will execute everyone unless ALL guesses are correct. Since each of the 100 prisoners has a 1/2 chance of guessing correctly, it would seem that the chance that all 100 will guess correctly would be 1/2 to the power of 100, which is minuscule: surely the prisoners are doomed. But this analysis assumes that the prisoners act independently of one another, which they are not required to do; remember, they may agree to a strategy in advance. Is there a strategy that gives the group a 50% chance of survival?

(The solution to puzzle #1 that appears below may be useful for solving puzzle #2, and vice versa.)

Puzzle #3: Now there are just three prisoners. The jailer says that a prisoner may opt to pass (that is, to decline to guess), but she requires that at least one prisoner guess the color of his hat, and she says that unless every guess that is made is correct, all three prisoners will die. It is true that each guess that is made has a 50% chance of being right and a 50% chance of being wrong. Nonetheless, there is a strategy that, with 75% probability, will lead to survival for the prisoners. Can you find such a strategy?

There is a magic trick related to puzzle #2. Some magicians (five, say) are blindfolded, and a member of the audience puts a dot on each one’s forehead, using either a red pen or a blue pen, in any fashion she likes, before the blindfolds are removed. Then, in succession, the magicians are asked to predict the colors of the dots (learning as they go whether their guesses were right or wrong). Miraculously, every prediction that is made is correct, with the possible exception of the first prediction! The five magicians assure the audience that the trick is a purely mathematical one, and that they are not communicating in any way after their blindfolds are removed, except for the fact that each of them hears the predictions that are made (and learns which predictions are right). So how do they do it?

Next month (March 17): Believe It, Then Don’t.

Thanks to Jordan Ellenberg, Sandi Gubin, Dan Halbert, David Jacobi, Brent Meeker, James Tanton and Peter Winkler.

ENDNOTES

Puzzle #1: Since each prisoner’s guess has a 50% chance of being correct, the expected number of correct guesses among the 100 prisoners is 50. On the other hand, the only way the prisoners can survive is if at least 50 guesses are right. So if the prisoners are to guarantee their survival, they must find a scheme that ensures that no matter what happens, 50 of them will guess right and 50 of them will guess wrong. (Do you see why? If there were situations in which more than 50 of them guessed correctly, there would also have to be counterbalancing situations in which fewer than 50 of them guessed correctly; otherwise the expected number of correct guesses couldn’t be exactly 50. And every situation in which fewer than 50 of the prisoners guess correctly is a situation in which all the prisoners die.)

While this partial analysis of the problem doesn’t give a strategy for the prisoners, it does point us in the right direction, by constraining the sorts of strategies we might want to consider.

Here’s the strategy I like best: when the prisoners confer, they divide themselves into two equal-sized groups. Prisoners in group A will predict their own hat colors based on the assumption that an even number of the hundred prisoners are wearing red hats; prisoners in group B will predict their own hat colors based on the assumption that an odd number of the hundred prisoners are wearing red hats. (For instance, if a prisoner in group A looks around at his 99 fellow prisoners and sees an odd number of red hats, he will predict that his own hat is red, since that will make the total number of red hats even.) Depending on the actual number of red hats worn by the 100 prisoners, either everyone in group A will guess correctly and everyone in group B will guess incorrectly, or vice versa. Either way, half of the prisoners will be right, so everyone will be spared.

Puzzle #2: When the prisoners confer, they agree that they will all predict their own hat colors based on the assumption that an even number of prisoners are wearing red hats. If this assumption is correct, all prisoners will guess correctly (and all will be spared); if the assumption is incorrect, all prisoners will guess incorrectly (and all will be killed). The two outcomes are equally likely, so the prisoners survive with probability 50%. No scheme can do better than this, since each prisoner has a 50% chance of being wrong. (In particular, the first prisoner that the jailer asks has a 50% chance of being wrong right off the bat.)

It’s illuminating to compare the strategy for puzzle #2 with the strategy for puzzle #1. In both puzzles, each individual guess has a 50% chance of being correct, so that the expected number of correct guesses is 50, but there the resemblance ends. In puzzle #1, the number of correct guesses is always exactly 50 (the variance is as small as possible). In puzzle #2, the number of correct guesses is never 50; it’s either 0 or 100 (the variance is as large as possible).

Puzzle #3: This variant is due to Todd Ebert. A prisoner who sees two hats of two different colors should pass; a prisoner who sees two hats of the same color should predict that his own hat is of the opposite color from what he sees. 25% of the time, when all three hats are the same color, this strategy will cause all three prisoners to guess wrong; but the rest of the time, the odd man out will correctly guess his own hat color, and the other two will keep silent, resulting in survival for all three. In the eight different possible worlds, there are six life-saving correct guesses and six fatal wrong guesses, but they aren’t evenly spread among the eight possible worlds; two of the worlds contain all the wrong guesses (three per world) while the other six worlds contain all the correct guesses (one per world). That’s why, even though the incorrect guesses are just as numerous as the correct guesses, it’s the latter that prevail in six-eighths of the parallel worlds.

If you find puzzle #3 intriguing, and want to try something along similar lines but harder, try it with 7 or 15 prisoners instead of 3. (When the number of prisoners is 21, there is an elegant way to ensure that the probability of survival is 1(1/2)n , described in Ezra Brown and James Tanton’s article; when the number of prisoners is not one less than a power of two, no general formula is known for the optimal probability of survival.)

The magic trick that I described is based on puzzle #2, but with a twist. Once the audience-member who drew the dots reveals whether the first magician’s prediction was correct or not, the other four magicians know whether the total number of red dots on their foreheads is even or odd, and hence by looking at each other’s foreheads can predict with complete accuracy the colors of their own dots. In a variant of the trick (feasible if the magicians are good psychologists), the audience-member is not asked whether the first prediction was right, or the second, etc. But she is likely to have some sort of noticeable reaction (a “tell”) that reveals to the magicians whether the first prediction announced was right or wrong, and they can base their answers on that.

REFERENCES

Ezra Brown and James Tanton, “A Dozen Hat Problems”. (If you have trouble downloading Ezra Brown’s private copy, try mine.)

Jordan Ellenberg, How Not to Be Wrong. See the chapter “What to Expect When You’re Expecting to Win the Lottery”.

(I bought my copy of “How Not To Be Wrong” at Porter Square Books in Cambridge, MA. You can buy your own copy at your local independent bookstore — if you don’t know of any, visit the Indie Store Finder — or order one from AbeBooks; AbeBooks is now owned by Amazon, alas, but it’s the closest thing you’ll find to an on-line syndicate of independent booksellers. If you buy something I’ve recommended from a local independent bookstore, please consider posting a comment to that effect, so that over time I can compile a database of bookstores that carry the sorts of books that I tend to recommend on this blog. Ditto for independent sellers of mathematical games, puzzles, and toys, such as Eureka Puzzles in Brookline, MA.)

The Planet Money podcast on lotteries (including the one in Virginia).

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