We roll two dice and try to guess what the sum will be.
Question: Are some totals more likely than others? Predict what sums appear more often, and how often they will appear.
Answer: Each combination of numbers where we list which of the dice had which number is equally likely, but some totals are more likely than others. Here's a chart with all the possible rolls. The numbers in bold along the top row show the value of the first dice. The numbers in bold along the left column show the value of the second dice.
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
You can see that there are 36 different possible combinations. To find out how likely it is to get a particular total, count how many times it appears in the chart, and divide by the total number of combinations.
Probability of what you want = (How many ways you can get what you want) /
(How many ways you can get anything)
For example, there are 5 ways (the orange squares) to roll a sum of 8 out of the 36 possible combinations, so the probability of the two dice summing to 8 is 5/ 36.
We are given three cards in a bag, one which has a sticker on each side, one with a sticker on one side and blank on the other side, and one that is blank on both sides. We pull a card out of the bag and look at one side. It has a sticker on it.
Question:What's the chance of having a sticker on the back of the card?
Answer: 2/3. One way to see this is to think of the cars as (sticker1, sticker2), (blank1, blank2), (sticker3, blank3). Then out of the three stickers, only one of them (sticker3) is on the card whose sides are different. So, if we see a sticker, there is a 1/3 chance that it came from the (sticker3, blank3) card, and a 2/3 chance that it came from the (sticker1, sticker2) card.
Have you ever wondered how the bank decides how much money to offer a contestant on the game show ''Deal or No Deal''? If you were a contestant, would you know when to take the money and run? Hopefully, you will when we're done with this activity.
We're going to investigate some of the math behind the game show ''Deal or No Deal''. We'll do so by playing the game. Before we begin, let's remember the rules:
Some notes to the teachers/ activity leaders: One teacher will be with each team, letting them talk strategy, but not letting them get too far off topic. Let them do what they want to do, but make them give a mathematical justification for it (ie, to the Bank: 'Why are you offering that much?' To the Contestant: 'Why did (didn't) you think that was a good enough offer?' Also, nudge them towards understanding the concept of expected value. However, it is very importantto not call it expected value. Instead, call it the average. The objective of this game is to get the students to use math in everyday situations. They know how to compute the average; by introducing new terminology like expected value, we run the risk of some disengaging, and not doing any math at all.
You have been given the opportunity to be on a game show with a chance to win a million dollars.
Game Show Host: Behind one of these three doors is one million dollars. Behind the other two doors are two goats. Which door do you think has the million dollars?
You: I'll try door number 1
Game Show Host: Let me show you what's behind door number 2....It's a goat! Do you stick with door number 1, or do you want to switch and try door number 3?
Question: What do you do? Do you stay with your first choice, or do you switch?
Answer: When you choose door number 1, there was a 1/3 chance that it was the correct door. There was a 2/3 chance that the million dollars was behind door number 2 or door number 3. These probabilities haven't changed. But, now you know that the million dollars is not behind door number 2. This means that there is a 2/3 chance that the million dollars is behind door number 3. You should switch!
Sarah is a student at Dewitt Middle School in Ithaca. She has two sisters, who are much older (they live in their own houses). Her oldest sister, Jennifer, lives up by the Pyramid Mall, and her other sister, Kristen, lives down near the Commons. Every day after she is done with her activities at school, Sarah visits one of her sisters. Sarah is very busy with many after-school activities so sometimes she leaves school by 3:30, but sometimes it is as late as 6 or even 8. Her schedule is so complicated that the time she leaves school each day is pretty much random. She also visits her sisters on the weekends at random times.
She visits her sisters by walking down to the bus stop for TCAT bus #30 (of course, she walks with several friends to be safe). Not wanting to favor one sister over the other, Sarah decides to take the bus whichever direction comes first. If the bus is going North, she'll visit Jennifer; if it's going South, she'll visit Kristen.
Here's a picture of the bus route:
After a few months of doing this, Kristen was getting upset with Sarah. Kristen complained, "It seems like you're always visiting Jennifer, not me. Do you like her better?" Sarah was sad about this, since she thought her way of choosing which sister to visit was fair. The bus runs in a loop, so half the time it should be going North and half the time it should be going South. Sarah decided to take some data to see if Kristen was right. Sarah got a notebook and decided to put a mark by Jennifer's name whenever she visited her. If she visited Kristen, Sarah would put a mark by Kristen's name. After several weeks of doing this, her data was surprising! It was true that she was visiting Jennifer more often. In fact, Sarah visited Jennifer almost four times as often as she visited Kristen!
Question: Why is this happening?
Answer: While Sarah's way of choosing seems fair, it in fact isn't. Since Sarah's bus stop is significantly closer to Jennifer's house than to Kristen's house, she visits Jennifer more often. As a bus travels around the route, it spends the great majority of its time (approximately 80% of its time) travelling South from Sarah's stop, down to the Common, and back North to Sarah's bus stop. If Sarah arrives at her bus stop during any of that time, she'll catch the bus as it heads North towards the mall. By comparison, in order for Sarah to catch the bus to Kristen's house, she must show up at her bus stop during the relatively small amount of time (20% of the total time on the route) that the bus spends travelling North from Sarah's bus stop, around the mall, and back to her bus stop. So, she has a 4/5 (80%) chance of visiting Jennifer, but only a 1/5 (20%) chance of visiting Kristen. Not very fair!
If Sarah's bus stop was in the middle of the route, then her way of choosing would be fair. In that case, the bus would spend have of its time between the bus stop and Jennifer's house, and half of its time between the bus stop and Kristen's house.
This workshop was originally created for Cornell's Expanding Your Horizons conference 2004 by Melanie Pivarski, Kristin Camenga, Jonathan Needleman, and Mia Minnes.