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We get to play with more toys. Yay!
Each group of students shares one plastic spinner with a clear base.
We can put a spinner onto a circle like the one below to make a game.
The first circle is a very boring game. Half the time a player wins $1. Half the time a player loses $1.
What we want to focus on is that the game is fair. This means that if someone played it a whole lot—long enough for any rare streaks of good or bad luck to cancel out—then overall they would not gain or lose money.
1. How can we draw lines to make a spinner game fair if winning has twice the gain of losing?
1. Pretend the circle is a pie, and we are cutting the pie into slices.
You could think that the +2 needs to happen half as often as the −1.
Or you could think that the −1 needs to happen twice as often as the +2.
Either way, we should give the −1 two pie slices, but only give the +2 one pie slice.
That is a total of 2 + 1 = 3 pie slices. So we cut the pie into thirds. The result looks like this:
2. How can we draw lines to make a spinner game fair if winning has three times the gain of losing?
2. Pretend the circle is a pie, and we are cutting the pie into slices.
You could think that the +3 needs to happen one-third as often as the −1.
Or you could think that the −1 needs to happen three times as often as the +3.
Either way, we should give the −1 three pie slices, but only give the +3 one pie slice.
That is a total of 3 + 1 = 4 pie slices. So we cut the pie into quarters. The result looks like this:
3. How about this spinner game with three possible outcomes? Can it be made fair?
3. Pretend the circle is a pie, and we are cutting the pie into slices.
If we give the +1 and +2 each one slice of pie, the slice for the +2 counts double. We effectively have three slices of pie for winning +1.
This means we need three slices of pie for the −1.
The two winning numbers each get one slice. The −1 needs three slices.
That is a total of 2 + 3 = 5 pie slices. So we cut the pie into fifths. The result looks like this:
The probability of a situation happening is the ratio of desirable outcomes to total outcomes. (This ratio is often changed into percent format.)
A classic example of probability is rolling two dice and adding their values.
4. When rolling two dice, what is the probability of the sum being seven?
4. Looking at the green boxes on the chart, we see that six out of thirty-six possibilities have a sum of seven. So the probability is 6⁄36, which we should reduce to 1⁄6 or change to about 16.7%.
5. When rolling two dice, what is the probability of the sum being ten?
5. Looking at the pink boxes on the chart, we see that three out of thirty-six possibilities have a sum of seven. So the probability is 3⁄36, which we should reduce to 1⁄12 or change to about 8.3%.
Imagine there is a gumball machine with equal amounts of three colors of gumballs: red, green, and blue. The table below shows all twenty-seven possibilities for getting three gumballs.
6. If you get three gumballs, how likely is it to get at least one blue gumball?
6. Nineteen of the twenty-seven possibilities have at least one blue gumball. So the probability is 19/27 or about 70%.
The odds of a situation happening is the ratio of desirable outcomes to undesirable outcomes. (This ratio is often reduced, but never changed into percent format.)
7. When rolling two dice, what are the odds of the sum being seven?
7. Looking at the green boxes on the chart, we see that six out of thirty-six possibilities have a sum of seven, and thirty do not. So the odds are 6 to 30, which could be reduced to 1 to 5.
8. When rolling two dice, what are the odds of the sum being ten?
8. Looking at the pink boxes on the chart, we see that three out of thirty-six possibilities have a sum of ten, and thirty-three do not. So the odds are 3 to 33, which could be reduced to 1 to 11.
9. If you get three gumballs, what are the odds of getting at least one blue gumball?
9. The odds of getting at least one blue gumball are 19 to 8, which can be reduced to 2 to 1.
The weighted average of a group of situations measures the "average result" of that group.
To find an expected value, use a table. Each possible outcome is a row. Work across with multiplication: the value for that outcome times its percent probability. Then add those products.
15. When rolling two dice, what is the weighted average?
15. This answer requires making a table, as below. The answer is not surprising. Most people already know that the "average value" when rolling two dice is seven. The expected value table confirms that common knowledge is precise instead of rounded: the expected value is indeed seven exactly, not slightly more or less. (The original Google spreadsheet is here.)
16. Every time little Billy is taken to the grocery store he takes three pennies for this gumball machine. He wants a blue gumball and will spend up to three cents trying to get one. What is the average number of pennies he spends?
16. Here is a sample spreadsheet that shows Billy spends an average of about 2.1¢ each trip to the grocery store.
For many students the most commonly used weighted average table is finding their overall grade in a class.
17. A student has earned the grades below. What is the student's overall grade in the class?
17. Here is a sample spreadsheet that shows the overall grade is 81.9 in the class.
The weighted average is sometimes called the expected value. It does make sense to say "the expected value of the sum of two dice is 7". It almost makes sense to say "the expected value of one of Billy's trips to buy a gumball is 2.1¢." But it does not make sense to call the overall class grade an "expected value" because that situation does not involve condensing mutually exclusive outcomes into an average outcome.
18. At a carnival, a booth offers a dice game in which you roll two dice and find the sum. If the sum is even the booth operator will pay you $1. If the sum is odd you will pay him $1. Is this a fair game?
19. Your little brother thinks that ten is a very big number. He wants to play a dice game about the number ten. He proposes a game where you each start with a pile of candies, and he finds the sum of two dice several times. Whenever the sum is less than ten, he gives you one candy. Whenever the sum is ten or greater, you give him more than one candy—but he is not sure how many is fair. Help your brother finish inventing his game by using an expected value table to find how many candies must you give him when he "wins" so that the game has an expected value of zero.
20. You are going to be the instructor of a Biology class and need to write the syllabus. You want attendance to be worth 10%, the final exam to be worth 30%, and the rest divided between nine lab reports and two midterms with each midterm worth as much as three lab reports. How much is each lab report worth? How much is each midterm worth?
18. Yes, the game is fair. The expected value table has two rows, and the sum of the products is zero. Overall you will gain and lose the same amounts, and end up at zero.
Result Value Probability Product even +1 0.5 +1 × 0.5 = +0.5 odd −1 0.5 −1 × 0.5 = −0.5
19. We need to make an expected value table. We will do it from your little brother's point of view, so giving you a candy is negative but receiving candy is positive.
Result Value Probability Product less than ten −1 30⁄36 − 30⁄36 ten or more +y 6⁄36 (6 × y)⁄36
We want the game to be fair, which means the sum of the products should be zero. Therefore the numerators of the two products should match (but one is negative and one is positive). So 30 = 6 × y. Each time your brother wins you should give him 5 candies.
20. Be careful, this is not an expected value problem! All we are trying to do is have the graded values add up to 100%. The nine lab reports and two midterms need to sum to 100% − 10% − 30% = 60%. Since each midterm should be worth three lab reports, that means we have a total of 9 + 3 + 3 = 15 lab report equivalent things to together make up that 60%. So each is worth 60% ÷ 15 = 4%. In other words, lab reports are worth 4%. Midterms are worth three times that, so midterms are worth 12%.
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LCC Math 25 Packet Problems
Likelihood problems are in the packet on pages:
• R-6 to R-12