Lotteries: Finding Expected Values of Games of Chance

Lotteries: Finding Expected Values of Games of Chance
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Lesson Transcript
Instructor: Cathryn Jackson

Cat has taught a variety of subjects, including communications, mathematics, and technology. Cat has a master's degree in education and is currently working on her Ph.D.

Most of us won't have a problem with winning the lottery. But is it a realistic goal? Do you really have a chance? In this lesson, learn how luck and probability collide when finding expected values in games of chance.

Understanding Lotteries and Expected Values

Megan is buying a lottery ticket with her friends. The ticket asks her and her friends to pick 4 numbers between 1 and 10. But her friends don't feel as if this is a good deal. They want to choose another lottery game, one where they pick 6 numbers from 1 to 48. They feel as if they have a better chance of winning. Each week they buy 1 ticket. Which ticket should they start buying?

It's a common statistical problem: What are my chances of winning the lottery? In this lesson, you will learn about the different types of lotteries and how to find the expected value for each game.

First, let's discuss expected value, which is the number of successful outcomes expected in an experiment. In other words, what are your chances of winning? The formula for expected value is n * P. The n represents the number of trials, and the P represents the probability of success on an individual trial. However, in this case we don't know the probability of success, but we do know the number of successful outcomes: 1. Meaning that there is one winning ticket. Therefore, we will need to write our problem like this: n * P = 1 and rearrange our formula to find our probability which is this: P=1/n.

If each ticket only contained 1 number, and 500 people bought a lottery ticket, our problem would look like this: 1/500 or 0.2%. Yikes! That's a pretty small chance.

Unfortunately, lotteries are a little more complicated than this. Let's look at two different types of lotteries.

Pick 4 Lotteries

Remember the first game that Megan and her friends played, where she had to pick four numbers between 1 and 10? The order of the numbers does not matter in this game, and each number can only be chosen once.

We can use a combination formula to solve this particular probability problem. The numbers that Megan and her friends pick doesn't matter. The probability will always be the same.

Before we calculate the different combinations of numbers possible, you will need to understand the concept of combinations. Let's say that you were drawing cards from a regular deck of cards. What are the odds of you drawing a 6 of hearts, a 4 of clubs, a 7 of diamonds, and a king of spades? This is similar to our problem with Megan and her lottery ticket. First, you probably know that drawing a 6 of hearts is a 1 out of 52 chance. This is because there is one six of hearts and there are 52 cards total. However, once you draw that first card, the next probability will be 1 out of 51, because you never replaced that first card. Since it doesn't matter if you draw the 6 of hearts first, second, third, or fourth, you have different probabilities for drawing a six of hearts from the deck at any given time. You also have to account for each of the other cards in this scenario: the 4 of clubs, 7 of diamonds, and a king of spades.

It can get very confusing with so many possibilities. To solve this problem, we will need to use the combination formula. The combination formula is a probability formula that uses factorials to find the number of possible combinations of all the outcomes in the experiment. The combination formula looks like this:


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You may notice that this formula uses an exclamation point, also known as a factorial in mathematics. Probability and statistics problems don't often use factorials, except when it comes to combinations. For large numbers, you will need to use a graphing calculator to find the factorial values. For more information about factorials, check out our other lessons!

Since we are using smaller numbers, we can solve these factorials by hand.

In this scenario, n equals the number of possible choices, or the amount of numbers that Megan and her friends can choose which would be n=10. x equals the number of trials. So Megan and her friends have x=4, or four trials - four opportunities to guess at a winning number. Now our combination formula looks like this:


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The 10 over 4 at the beginning of this formula is read as 10 choose 4. In other words, you have 10 numbers and you are choosing 4 of them. So what are all of the possible combinations of numbers? We need to find the different combinations of numbers that are possible first, before we can figure out the probability of a winning combination.

The top of the formula shows 10 factorial. Meaning we want all of the different combinations out of choosing from the numbers 1 through 10. However, we need to look at the different combinations after you've selected one number - there are nine out of the ten left to choose. This is just like when we had 51 cards to choose from after selecting our first card. That's why the first part of the formula is 10 factorial over 10 minus 4 factorial. This will give us the combinations of numbers when selecting 4 numbers out of 10 and no numbers can repeat.

I've taken the first part of this formula and broken it down for you. If order was important in this particular lottery, then we would just use this part of the formula, and our answer would be 1 out of 5,040. However, because the order of the numbers does not matter, we need to add the second part of the formula.

I've added in the second part of the formula, 4 factorial, that accounts for the different order combination the numbers could appear. For example, if the winning numbers were 7, 3, 1, and 10, then those numbers could appear in different orders, such as 10, 1, 3, and 7 or 3, 1, 7, and 10. Once you multiply your numbers, you will have the probability of Megan winning this lottery, which is 1 out of 210. This is the same as the expected value we discussed earlier: P=1/n, or your probability of winning P is equal to 1/210.

You may be wondering why you have a better chance of winning when order doesn't matter. If you think about it, when order matters, then it narrows your chances of winning significantly. Let's look back at our example if the winning numbers were 7, 3, 1, and 10. If order mattered, then a ticket with 3, 1, 7, and 10 would be different from a ticket with the numbers 10, 1, 3, and 7. Remember, this would make the number of possible combinations equal to 5,040. However, when order doesn't matter, all of these tickets would be the same winning ticket and would decrease the number of possible combinations.

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