Playing the Lottery, part 2

In my previous post, I talked about a calculation called “expected value”, which helps measure just how fair or unfair a given game is. I also talked about “the gambler’s downfall”, which basically means that the player is much more likely to run out of money before the house does. In this post, I’ll talk about five ways the state lottery tries to trick you into thinking that the game is better than it really is.

#1 The prize might be divided among several winners. They want you to think about the size of the jackpot and ignore the fact that several winners may end up splitting the jackpot. A $72 million jackpot sounds bigger than a $24 million jackpot, but that’s just an illusion. The $72 million jackpot is much more likely to be split three ways, so each gets $24 million.

#2 They lie about the value of the jackpot. I’m not talking about taxes; that’s a whole other subject. Imagine a game where, if I win, you have to pay me right now, but if you win, I take 30 years to pay you. How fair does that sound? When they tell you that the prize is $24 million, that’s a deception. The truth is that they are essentially offering you a gift certificate which is only worth $14 million. You can trade it for $14 million in cash, or you can trade it for an annuity that pays $800K per year for the next 30 years. The problem here is the difference between Present Value and Future Value. $24 million is the Future Value, spread out over 30 years. But I don’t care what it will be worth in the future. What matters is what it’s worth right now. The Present Value is only $14 million, not 24. There exact ratio of Present Value to Future Value depends on interest rates, but right now PV is roughly 60% of FV over 30 years.

#3 They use huge numbers in order to confuse people. Most people can understand small numbers like $50 and $1,400 but they have trouble understanding just how big is a million, or a billion. The lottery takes advantage of this by offering what seems like a large prize and burying in the fine print the fact that the odds against you are even more astronomical. Sometimes it’s 14 million to 1 against, sometimes it’s 292 million to 1 against. Your brain sees both those numbers as just “really big”, even though the second one is twenty times higher.

#4 They make the game complicated. This has the double whammy of making it more fun (because it feels like you have some control) yet it also makes it harder to understand the odds. Even if you’re one of the rare people who learned Pascal’s formula for expected value, they are betting you won’t be able to apply it to such a complicated game. It has been said that the lottery is a tax on people who are bad at math. The truth is that even people who are relatively good at math have trouble understanding the lottery. Luckily, you have me to help you.

#5 The exact parameters of the game aren’t known until after it’s over. In order to figure out how much you might win, you need to know how many tickets will be sold. But that’s not known at the time you buy your ticket. And they are constantly adjusting their estimate of what the jackpot will be. In fact, the size of the jackpot also depends on how many tickets get sold.

Let’s try to estimate what the expected value of the lottery really is. First, it’s not guaranteed that someone will win. It’s very possible that there won’t be any winning tickets. The more tickets get sold, the greater the chance that someone will win, but that also increases the chance that the jackpot will be split. And remember that the advertised number isn’t the actual jackpot. Unfortunately, we’ll have to make educated guesses for some of the numbers. Suppose they advertise that the jackpot this week will be $18 million and we think there will be 30 million tickets sold. Suppose this is a standard $1 “pick six numbers from 1 to 49” lottery, with 13,983,816 unique combinations of numbers. Let’s call that last number n; you’ll see why in a minute. All things being equal, any ticket has 1/n chance of winning. But, assuming that someone wins, the best guess for how many winning tickets there will be is 30 million divided by n. The jackpot will be divided by this number, which means we’ll actually multiply the jackpot times n over 30 million. And lastly, remember that the actual jackpot is only about 60% of the advertised jackpot. And we need to multiply all this by the probability that someone will win. Given 30 million tickets, I’ll estimate that to be 75%.

EV = (75%)x(1/n)x(60%)x($18 million)x(n/30 million)-$1.00

Notice there’s an n in the numerator and denominator, so the n’s cancel. Same goes for the “million”. That just leaves…

EV = (75%)x(60%)x($18)x(1/30)-$1 = $.27-$1.00 = -$.73

This is a really bad expected value. It’s negative (of course) meaning the odds are tilted against the players. On average, each ticket costs $1.00 and loses $.73. That’s a huge profit for the house.

Well, let’s suppose that no one wins the jackpot this week and it rolls over to another week. Now they’ll another 30 million tickets and the advertised jackpot is $36 million.

EV = (75%)x(60%)x($36m)x(1/30m)-$1 = $.54-$1.00 = -$.46

This is better, but it’s still a large profit for the house, and that’s on top of all the profit they made last week when there were no winners at all.

Let’s take it one more step. Suppose that once again there are no winners and it rolls over again. This week they advertise the jackpot to be $72 million. Why such a big jump? Because they expect to sell more tickets! This week there will be 60 million tickets instead of just 30 million. Now it’s 90% certain that someone will win.

EV = (90%)x(60%)x($72m)x(1/60m)-$1 = $.65-$1.00 = -$.35

The house still expects to make a 35% profit this week, on top of all the profit they made last week and the week before. The only way that your EV becomes positive is if there’s a rollover followed by a week where very few people buy any tickets. But they’ve convinced everyone to buy the tickets because $72,000,000 sounds great.

Now, let’s use the real-world numbers from last week’s Powerball Lottery. The jackpot was advertised as $1.5 billion and they sold 371 million tickets.

EV = (85%)x(60%)x($1586m)x(1/371m)-$2 = $2.18 – $2.00 = $.18

Amazingly, we’ve actually found a game with a positive expected value, which means it favors the players (slightly). Players could expect a 9% return on their investment. This leads to another question. If you could buy one of every single ticket, would it be worth it? First, consider that n=292,201,338 for Powerball, so you’d need to buy that many tickets. And you’d have to hire an army of 120,000 people to help you buy them, which would cost around $50 million to pay their salaries for one week, bringing your total investment to $634 million. You’d be increasing the number of tickets sold to 663 million, and the jackpot would go up another 200 million or so. Also, you’d be guaranteeing that there would be at least one winner.

EV = (100%)x(100%)x(60%)x($1786m)x(292m/663m)-$634m = $471m-$634m = -$163 million

So that would be a really bad business plan. It’s not smart to invest $634 million when you expect to lose $163 million of it. The EV changed when you altered the game by buying so many tickets.

Anyway, what actually happened last week is there were 3 winners, so each of them got $328 million (Present Value), which isn’t nearly as big as $1.586 billion but it’s still huge. But is it really all that great? Will it make you happy? Will it solve your problems? That’s the topic for part 3.


Playing the Lottery, part 1

There was a lot of talk last week about the huge jackpot in the Powerball Lottery. Many people were wondering… is it true that this Lottery was somehow better than other Lotteries? As a former math teacher, I find such questions interesting. I’m going to tackle this in three parts. Part 1 will be about how to judge the fairness of gambling games in general. Part 2 will talk about strategies the Lottery uses to try to trick you. Part 3 will discuss the complicated question of how winning and losing affects your happiness.

First, let’s look at a simple way to judge whether games are fair.

The simplest game I can think of is a coin flip. I flip the coin, you call it heads or tails. If you guessed right, you win. Otherwise, I win. But what, exactly, do we “win”? Suppose we each put up a dollar and the winner gets to keep all the money. I think you’ll agree that this is a completely fair game. Neither of us has an advantage over the other.

But that’s not generally how gambling works. The person who sets up the game (called “the house”) can adjust the game to give themselves an advantage. As the saying goes, “the house always wins.” But there are varying degrees of just how lopsided the game might be.

Let’s consider a slightly more complicated game. I’ll roll a six-sided die underneath a cup and invite you to guess what number is showing on the die. But you have to put down $3 before you guess. If you guess the number, I’ll give you $12. How fair is this game? Fortunately, a very smart man called Blaise Pascal came up with a way of calculating this, 400 years ago.

Imagine yourself playing the game many times (or many copies of yourself playing the game together). On the average, do you come out ahead? Consider the six-sided die game described above. Clearly, you’ll lose this game most of the time. But it might be worth it to play, if the amount you win is sufficiently large compared to the amount you lose. Imagine playing the game six times and winning just once. You lose five times. All together, you’ve won 1x$1=$12 and you’ve lost 5x$3=$15. So you’re down by $3 after six games. That’s an average loss of 50 cents each time you played. This gives us a measure of how fair the game is. Your “expected value” is -.50 , which is roughly 18% of what it costs you to play the game. From the house’s point of view, they expect to make an average of 18% gross profit from each game.

This isn’t precisely how Blaise Pascal did it. He suggested that you should consider all the possible outcomes, calculate the probability of each, and multiply each probability times the win/loss associated with that outcome, and add up those values. Let’s try it his way.

You probability of winning is 1/6. Your probability of losing is 5/6. Multiply 1/6 times the $12 win and you get $2. Multiply 5/6 times the -$3 (negative because it’s a loss) and you get -$2.50. Add together $2 and -$2.50 and the result is -$.50, which is the same expected value we calculated above.

Note: when the house has the advantage (which is nearly always), the expected value will be negative. If you ever find a game with a positive expected value, that means the player has the advantage over the house. Such games are very rare.

Let’s compare that game to another one. In this next game, I’ll write down a single digit, anywhere from 0 to 9. You have to guess the number to win. I sell you a ticket for $2 and you write your guess on the ticket. If you guessed right, you win $17. Let’s calculate the expected value for this game. There are 10 choices so your probability of winning is 1/1o. Multiply that by $17 to get $1.70. Then multiply 9/10 by -$2 and get -$1.80. Add together $1.70 and -$1.80 to get -$.10, which is the expected value. But wait. There’s a mistake there. Did you catch it? I said I’ll sell you a ticket for $2 and you write your guess on the ticket and maybe you win $17… but you don’t get your $2 back! So even when you win, you really only came out ahead $15, not $17. So really we should have added together (1/10)x($15)+(9/10)x(-$2) = -$.30.

So the six-sided die costs $3 to play and you expect to lose $.50 (about 18%) but the 0-9 game costs $2 to play and you expect to lose $.30 (which is only 15% of $2). So you lose 18% of your money in the first game but only 15% of your money in the second game. Hence, the 0-9 game is a better bet for you. Of course, better for you means worse for the house.

Generally speaking, if the house takes less than 10%, that’s pretty good for the players. If the house takes more than 20%, that’s bad for the players. If the house takes more than 50%, that’s terrible for the players.

So now let’s look at a game where the house takes more than 50%. It’s a charity fundraiser. They walk around the room selling tickets for $1. At the end of the night, they count how much has been collected, put 1/4 of the money into a big glass jar and then randomly draw a number to see who wins. If your ticket matches the number, you win what’s in the jar. Suppose they sell 600 tickets, taking in $600. $150 of it ends up in the jar. So if your ticket wins you come out $149 ahead. The chance that your ticket wins is 1/600. The expected value is (1/600)x($149)+(599/600)x(-$1) = 149/600 – 599/600 = -450/600 = -.75 which means that, for each $1 ticket sold, the house keeps 75 cents. That’s great for the house, but bad for the players.

Let’s go back to the 0-9 guessing game for a minute. We calculated that, on average, you’ll lose 15% of your money. Imagine that you start with $100 and you use it to buy 50 tickets. On average, 5 of those tickets will be winners and 45 will be losers, so you end up with $85. What happens next? Suppose you buy 42 more tickets. You’d expect 4 of them to win and 38 to lose. Now you have $69. I think you can see where this is going. If you keep using your winnings to buy more tickets, eventually you’ll end up with no money left at all. The point is that recycling your winnings back into more tickets causes you to lose even more money than what the expected value says.

Of course, there’s a tiny tiny chance that you’ll have a lucky streak, winning game after game, until the house runs out of money (called “breaking the bank”). But honestly, who do you think will run out of money first, you or the house? This is called “the gambler’s downfall”.

Let’s talk about roulette. You pick a number from 1 to 36 and hope that the ball will land on that number. The “odds” are 35 to 1, but there are 38 spaces on the wheel. So if you bet $10 you have a 1/38 chance to win $350 (and keep your $10 too), plus a 37/38 chance to lose your $10. The expected value is (1/38)x($350)+(37/38)x(-$10) = -10/19, approximately -53 cents. That’s barely 5% of your bet, so this is a good game from the player’s point of view. But think about what would happen if you recycle your winnings. Suppose you walk in with $500 and you say “I’m going to bet $10 on my lucky number fifty times”. We don’t know precisely what will happen, but you’ll probably win once, maybe twice, and walk out with either $360 or $720. Not bad. But if you take those winnings and keep on betting it, you will eventually run out of money and walk out with nothing.

On the other hand, Suppose you walk in with $500 and say “I’m going to bet $10 on my lucky number until I run out of money or until I win, at which point I’ll stop”, then you have almost a 50-50 chance to walk out with more money than you came in with. You might even walk out with $850 if you win on the very first spin. As Kenny Rogers sang, “Know when to walk away.”

From the house’s point of view, it’s still a win for them. Suppose it takes you forty bets until you eventually win one and quit. Imagine you and 1,000 other people all making forty $10 bets. According to the expected value, the house predicts an average profit of 53 cents times forty bets times 1,000 people, which comes to $21,200. On a good day, the house might make $30,000 profit at that roulette table. On a bad day, they might only profit $10,000. But the house always wins.

That’s all for Part 1. In Part 2, we’ll discuss state-run Lotteries and how they trick you into thinking that the game is better than it actually is.