Probability and Expected Value in Lotteries | Free Discrete Mathematics Lesson (4 College Credits) | UPI Study

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Probability and Expected Value in Lotteries

Discrete Mathematics · Chapter 7: Probability in Games, Gambling, and Strategic DecisionSection 7.5 · lesson 44 of 878 min · free with a free account

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In this lesson

Lottery odds by combination, a ticket's expected value, and the house edge.

  • Count lottery outcomes with combinations
  • Compute the odds of every match count
  • Find a ticket's expected and net value
  • Define house edge and return to player

Transcript

Everything said in the video, section by section (1,179 words). Read along, or search it.

Probability in Games, Gambling, and Strategic Decision

Probability and Expected Value in Lotteries

This section applies counting and expected value to lotteries. We'll count the ways to win with combinations, turn those counts into a ticket's expected value, and then introduce one genuinely new idea built on top of them. The house edge.

The Opening Problem

A 3,000,000-Dollar Headline

Here's an illustrative example. A state lottery game called Lucky Six asks players to pick six numbers from one through forty-two, for two dollars a ticket. This week, every ticket terminal in the state advertises a three million dollar jackpot. That number is real, and someone will eventually match all six. But it says nothing about what a single ticket is worth on average. Or about how much of every dollar wagered the lottery expects to keep for itself. Answering both takes tools already built. Combinations from section five point seven, and expected value from chapter six.

7.5.1 Counting the Ways to Win

5,245,786 Possible Tickets

A ticket is a choice of six numbers from forty-two, and the order they're written in doesn't matter. So the number of distinct tickets is forty-two choose six. Canceling the shared factorials leaves forty-two times forty-one, down to thirty-seven, over six factorial. That's three billion, seven hundred seventy-seven million, roughly, over seven hundred twenty. Which gives five million, two hundred forty-five thousand, seven hundred eighty-six. Those combinations are all equally likely. So a single ticket matches the jackpot with probability one in five million, two hundred forty-five thousand, seven hundred eighty-six.

Partial Matches

Right Numbers, Times Wrong Numbers

Lucky Six also pays smaller prizes for partial matches. Each of those needs two combinations, multiplied. The number of ways to choose correctly among the six winning numbers, times the number of ways to choose the rest from the thirty-six losing numbers. Then divide by the total. For exactly three matches, there are six choose three, twenty ways, to pick the three correct numbers. And thirty-six choose three, seven thousand one hundred forty ways, to pick the other three from the numbers that weren't drawn. That's one hundred forty-two thousand eight hundred favorable tickets. A probability of about zero point zero two seven, or about one in thirty-seven tickets.

Table 7.5.1

Every Possible Match Count

Here's every possible match count. Matching none of the numbers happens about thirty-seven percent of the time, and matching exactly one about forty-three percent. Two matches, about seventeen percent. Three matches is about one in thirty-seven. Four is about one in five hundred fifty-five. Five is about one in twenty-four thousand. And the jackpot is one in over five million. Every ticket lands in exactly one row, so the seven probabilities sum to one.

7.5.2 Expected Value of a Ticket

Weight Each Prize by Its Probability

Chapter six defined expected value as each value times its probability, summed. Here it applies to a new random variable, the payout of one ticket. Lucky Six pays five dollars for three matches, fifty for four, two thousand for five, and this drawing's three million for all six. Fewer than three pays nothing. Multiply each prize by its probability. Three matches contributes about thirteen and a half cents. Four matches, about nine cents. Five matches, about eight cents. And the jackpot contributes about fifty-seven cents. Adding the four contributions, the expected payout of one ticket is about eighty-eight cents.

Net Expected Value

About 1.12 Dollars Lost Per Ticket

A two dollar ticket returns about eighty-eight cents on average. So its net expected value, the payout minus the price, is about minus one dollar twelve. The negative sign is the point. On average, across many tickets, a player loses about a dollar twelve for every two dollars spent. That's not special to this game. A negative net expected value is characteristic of lottery tickets in general. A ticket has to fund its own prizes, plus other costs, before anything is returned to players.

Figure 7.5.2

The Jackpot Dominates

This chart shows how each prize tier contributes to the eighty-eight cent expected payout. The jackpot tier dominates, despite its minuscule probability. Its prize is large enough to outweigh the other three tiers combined.

7.5.3 House Edge

Expected Loss as a Business Model

Now the new idea. The house edge is the percentage of each dollar wagered, that the operator expects to keep over the long run. It's minus the net expected value, divided by the ticket price. Its complement, the return to player, is the percentage expected to come back to players as prizes. For Lucky Six, the house edge is one dollar twelve over two dollars, about fifty-six percent. And the return to player is about forty-four percent. So for every two dollar ticket, the lottery expects to keep about a dollar twelve, once every prize tier is paid out across many drawings. Lotteries are widely understood to carry a much higher house edge than most casino table games. A large share of each ticket goes to prize funding, retailer and administrative costs, and in many places, a designated public purpose.

7.5.4 More Tickets?

10 Tickets, 10 Times the Loss

Does buying more tickets change the math? Suppose a player buys ten different tickets for one drawing. By linearity of expectation, the total expected net value is ten times the net value of one ticket. That's about minus eleven dollars twenty. Ten times the spending, ten times the expected loss. No better, and no worse, per dollar wagered. The probability side does improve, and it's worth being precise. Two different tickets can't both match the single winning draw. So jackpot wins on distinct tickets are mutually exclusive. That means the addition rule applies exactly. Ten tickets give ten over five million two hundred forty-five thousand, and so on. About one point nine in a million. Ten times more likely than one ticket, which is a real improvement. But still under two in a million.

2 Ideas, Often Confused

Chance of a Win Versus Value per Dollar

A friend argues that with ten tickets, their expected value gets better, because they have ten times the chance of winning the jackpot. That's partly true, and the mix-up is worth naming. The probability of winning something, including the jackpot, does multiply by ten. But expected value per dollar wagered doesn't improve at all. Each ticket carries the same negative value, so the total scales by ten in the unfavorable direction. A higher chance of some win isn't a better average return. The first improves with more tickets. The second is fixed by the odds and prizes, and spending more can't change it. And both numbers are real. The three million dollar headline is true for one combination, on one drawing. The minus one dollar twelve is true for every ticket, every drawing.

Takeaways

Count, Weight, Subtract

To summarize. The number of equally likely lottery combinations is n choose r, and the chance of any single specific outcome, like the jackpot, is one over that count. Partial matches multiply two combinations, the right numbers from the winners and the rest from the losers, then divide by the total. A ticket's expected payout weights each prize by its probability. Subtracting the price gives a net value that's characteristically negative. And house edge and return to player describe the same fact, from two directions. Here's something to think about. A player only buys tickets in games with bigger jackpots, reasoning that a fifty million dollar game must have better expected value than a three million dollar one. When could that reasoning be wrong? What would you need to know about ticket price, total combinations, and the whole prize structure, before comparing the two?

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FAQ

Questions students ask us

About Probability and Expected Value in Lotteries

How do you count lottery outcomes and find the jackpot odds?
Count the tickets with combinations, \binom{n}{r}, because the order of the numbers doesn't matter. The jackpot probability is 1/\binom{n}{r}. For a game that picks 6 numbers from 42, \binom{42}{6}=5{,}245{,}786, so one ticket wins the jackpot with probability 1/5{,}245{,}786. All combinations are equally likely.
How do you find the probability of matching exactly some of the winning numbers?
Multiply two combinations, then divide by the total number of tickets. One counts ways to pick the right numbers from the winners, and the other counts ways to pick the rest from the losers. For exactly 3 matches in the 6-of-42 game, \binom{6}{3}\binom{36}{3}/\binom{42}{6}=20\cdot 7140/5{,}245{,}786\approx 0.027, about 1 in 37.
How do you calculate the expected value and net value of a lottery ticket?
Multiply each prize by its probability and add the products to get the expected payout. Then subtract the ticket price to get the net expected value. In the 6-of-42 game, the 3-, 4-, 5-, and 6-match prizes contribute about 0.88 dollars in total. On a 2-dollar ticket, the net value is about 0.88-2=-1.12 dollars.
What are house edge and return to player in a lottery?
House edge is the share of each dollar wagered that the operator expects to keep long term. It equals -(\text{net EV})/\text{price}. Return to player is the complement, 1-\text{house edge}, the share paid back as prizes. With a net value of -1.12 dollars on a 2-dollar ticket, the house edge is about 56\% and the return to player is about 44\%.

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