Discrete Math - Day 26

Today we continued exploring relative primes and dove into Euler's totient function.

To begin class, I asked them if 49 was the smallest composite number that was relatively prime to 30? There was some discussion and one student explained that because 30 = 2 x 3 x 5, the next available prime number that could be used was 7 and that 7 x 7 would be the smallest composite number that could work.

To check for understanding, I asked students what would be a composite number that was relatively prime to 210. While not everyone got the idea, students came up with values like 169 = 13 x 13 and 121 = 11 x 11. For others, I asked them to look at their prime number sheet and see what prime numbers were available to use. The conclusion was that 121 was the smallest composite number relatively prime to 210 since 11 was the next available prime number to use.

The definition of relatively prime was extended to all positive integers so that positive integers n and m are relatively prime if gcd(n, m) = 1. This applies to both prime numbers and composite numbers. The class discussed the implication for two distinct prime numbers--they will be relatively prime. The class verified their understanding by identifying pairs of relatively prime numbers.

Euler's totient function was introduced. This function counts the number of integers that are less than or equal to n and are relatively prime to n. The totient function is designated using the Greek letter phi, that is φ(n).

For example, what does φ(4) equal? In this case, the integers ≤ 4 are {1, 2, 3, 4} and both gcd(1,4) and gcd(3,4) equal 1. Therefore, φ(4) = 2.

Students created a list of φ -values for all integers from 1 through 20. There was some confusion and hesitation at first but by answering their questions and verifying their first attempts they were off and running. The class then verified their results. A couple of values presented by one student were questioned and the class came to a consensus about what each value should be.

I asked the class to look at their results and think about any patterns that they made use of in creating their list of φ-values. Students mentioned that when working with even numbers they were able to immediately eliminate all the even values less than n. They also noticed that for any prime number p, φ(p) = p - 1.

Additionally, students saw that all the φ-values were even except for φ(1) = 1 and φ(2) = 1. One student conjectured that for n > 1, φ(2n) ≤ n and φ(2n + 1) > n. This presented the opportunity to discuss how mathematicians make observations like these and then try to determine the truth of these conjectures. I didn't want to dive into trying to prove the conjecture right at that moment. We will explore some additional properties of the totient function that will enable the class to more easily answer this question.

I asked the class to explore the multiplicative nature of the totient function, that is
when does φ(n x m) = φ(n) x φ(m) for positive integers n and m?

Students explored different examples using their list of values for the integers 1-20. In doing this they noticed several things:

• the multiplication doesn't work when n and m are both even
• the multiplication doesn't work when n = m, i.e. for square numbers
• the multiplication may or may not work for a given integer

One student decided to test the square number conjecture by finding φ(25) and comparing it to  φ(5) x φ(5), which demonstrated that they were not equal.

This last one was intriguing. Students found φ(18) = φ(9)φ(2) but that φ(18) ≠ φ(3)φ(6) and, similarly,
φ(24) = φ(3)φ(8) but that φ(24) ≠ φ(4)φ(6). The question is why? What is different in the values being multiplied that enables one product to be true while the other is not for the same integer?

Students pondered this for a while and someone conjectured that perhaps it was because in one instance the two integers had a common factor, causing the product to not work, while in the other instance the two integers did not have any common factors.

Putting this back in terms of relatively prime numbers the conjecture became φ(n x m) = φ(n) x φ(m) if gcd(n,m) = 1. The class explored this conjecture and decided it was true. The result was verified. The big question is why?

Students wrestled with this idea. As I walked around I asked them to consider what went into determining each φ-value. Students got to the idea that the common factors were somehow affecting the result. We looked explicitly at φ(24) = φ(3)φ(8) and φ(24) ≠ φ(4)φ(6). In the latter situation, the values 2 and 4 are eliminated from both lists as you go through and determine relative primes. In the first situation, this does not happen. I left it at this informal level as we will be exploring the totient function further.

To conclude the class, I had students do a brain dump in their notes of their thoughts about factoring, relative primes, and the totient function. I asked them to also record questions they had about what we've been doing. This process helps to students to start organizing their thoughts and to explicitly connect different pieces.

Tomorrow we'll explore additional ideas around the totient function that will lead to Euclid's algorithm.

Visit the class summary for a student's perspective and to view the lesson slides.

IPS - Day 26

Today we worked through a series of problems, determining the probability models and calculating expected values for the problems. There were good discussions about appropriate models and how to account for initial costs. By the end of working through and discussing the five problems we worked on everyone was feeling comfortable with creating probability models and calculating expected values using the models.

We then worked through a simulation to calculate the expected value from the simulation. This enabled students to compare the process for calculating expected values from a theoretical perspective and from a simulated perspective.

The next portfolio problem was assigned. This problem ties in creating probability models, verifying the model is valid, calculating an expected value, simulating the problem, and comparing the simulated result to the theoretical result.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 25

Today we finished the 39-Game Hitting Streak investigation.No issues came up in this discussion. Students understood what they did to calculate the probabilities and simulate the situation.

We then looked at the game show Deal or No Deal to investigate expected value and its use. The first task, which everyone readily got, was to determine the probability of picking the $1 million case. The next task turned out to be more difficult.

Students were asked to calculate the expected value of the brief case that was selected. There was a lot of discussion and confusion. I told the class they needed to create a probability model for the game. Again, this was a struggle. Finally, with some coaching and a simplified example, students began to create their probability model and calculate the expected value, which cam out to $131,477.54.

There was a discussion about what this value represented. I used the analogy of starting with the show's first airing and summing the values of every case that has been selected on the show and then calculating the average. The average value over all of the shows should be close to the $131,477.54. Another way of thinking of the value is to consider that if you were able to repeatedly play the game thousands upon thousands of times, the average value of the cases you pick would be close to the $131,477.54 value.

We then played the game. I like to keep track of the expected value at the end of each round and discuss this value versus the expected value. This allows students to think about the expected value versus the probability of actually having a winning case. At one point in the game there were 5 cases left. Three cases were valued at $50,000 or less, one had a value of $100,000 and one a value of $500,000. The deal was to take a guaranteed $53,000.

I explained that the probability of having a case less than this value was 3/5. The question of having a much larger expected value ($130,000) versus the chance of cashing in on the higher value comes into play. On the next turn the offer was $67,500 and there was only one case with a value higher than this. Again the probability of actually holding the case was only 1/4 while the expected value was $137,536. Again, in a situation like this, since you aren't able to play the game repeatedly, it would make more sense to accept the offered amount.

This is always a fun activity that students get into, especially as you go around the room and have different students select which case to open next. It also provides a nice example of relying too much upon making use of expected value exclusively when making decisions.

Next class will focus on practicing creating and using probability models.

View the class summary for a student's perspective and to view the lesson slide.

Discrete Math - Day 25

Today's class focused on the idea of relatively prime composite numbers.

Before diving into this topic I asked students to share conjectures they had written down from last class. There wasn't much forthcoming in this front with one notable exception. One student talked about the idea that while prime numbers are infinite, the spacing between them continues to increase and that as you continue going further out on the number line that the spread between the prime numbers becomes infinite. It was a nice informal way of thinking of limiting processes and the idea that finding ever larger primes will become next to impossible as the computational time required will exceed fixed time. Interesting ideas that really go beyond the scope of a high school discrete math class.

To start off the investigation of relative prime numbers I posted pairs of numbers and asked students to consider how they could group these pairs based upon ideas and properties they associate with prime numbers. The goal was for students to consider the multiplicative properties and factors of these numbers.

Many students grabbed their divisibility test sheet and started to consider primes that divided each number. Some students considered differences. One group in particular was stuck on the additive relationship of the pairs. I mentioned to this group that prime numbers were based on multiplicative properties and they then started to consider factors.

Most of the groupings looked at values divisible by 3 or values divisible by 5, etc. One student noticed that there were two pairs of numbers that did not have any common factors. Eureka! I was able to then highlight the idea that many of the pairs had two or more common factors but that for these two particular pairs the only common factor was 1.

I then defined relatively prime composite numbers using this idea. I provided the example of 15 and 16 being composite numbers that are relatively prime. I asked students what an equivalent definition would be using the greatest common divisor. This seemed to stump the class. I asked students what the greatest common divisor of 15 and 16 was? Several students responded that it was 1. I asked again how the greatest common divisor could be used to define relative prime composite numbers.

One student tentatively said that two composite numbers would be relatively prime if their greatest common divisor was equal to 1. This was precisely what is needed.

Students then used their Sieve of Eratosthenes to look for patterns for relatively prime composite numbers. This was a struggle for many students but a couple of patterns were identified. First, students noted that they could not have two relatively prime even numbers. Next, students noted that the two composite numbers either needed to be both odd numbers or that one had to be even and one odd.

Most students were still not comfortable with the idea of integers being relatively prime, so I asked each student to identify 10 pairs of relatively prime composite numbers and to group their pairs. It was interesting to see the tactics that students used. Some picked a value like 15 and started to find multiple values that would be relatively prime to 15. Others were randomly picking values and checking.

I then had students write down their responses to the following questions:

• What strategies did you use to find relatively prime numbers? 
• Which pair is the closest together? 
• Which pair is the farthest apart? 
• What made a pair easy or hard to find? 
• Are some numbers used more than others? 

As this was the end of the class I posed on last question: what is a composite number that is relatively prime to 30? It was interesting to hear students frantically trying values only to find they didn't work. I purposely picked 30 since the smallest relative prime composite number available would be 49. Finally, one student said that 49 would work.

For homework, I asked students

• to verify that 49 works,
• to determine if 49 is the smallest composite number that is relatively prime to 30, and
• to explain why 49 would have to be the smallest composite number, if it was the smallest

We'll start by looking at the responses to these questions and then generalize the idea of two values being relatively prime.

Visit the class summary for a student's perspective and to view the lesson slides.

IPS - Day 24

Today we continued looking at probability models and expected value. Students worked on the 39-Game Hitting Streak investigation. Students continued to struggle with how to simulate the given situations and how to calculate needed probabilities when there were four at-bats rather than three.

Overall there were many good discussions that took place between students and by the end of the class most were feeling more comfortable with the way to run trials and calculate results.

We'll compare and discuss their results next class.

Visit the class summary for a student's perspective and to view the lesson slide.

Discrete Math - Day 24

Today we continued looking at prime numbers and their distributions. I revisited the conjecture made by one student that the number of primes was infinite and the idea from another student about creating a formula of some sort to answer the question.

With this idea in mind I proceeded to present the idea of proof by contradiction. As an example, I used the idea that everyone agrees that when the temperature is 80 degrees or warmer then it is comfortable to wear shorts. Today the temperature is in the high 20s. Someone walks outside in shorts and concludes that it is not comfortable to wear shorts. What conclusion can we draw? Several students said that the temperature isn't above 80 degrees.

This is the idea with a proof by contradiction. We hypothesize the truth of a statement and using that assumption we follow a series of logical steps that leads to a contradiction. We are now in a position to prove that the number of primes is infinite.

This proof first appeared in Euclid's elements over 2,000 years ago and is a simple, yet beautiful illustration of proof by contradiction. First, assume there are a finite number of primes. We can list these primes as a set, albeit it might be a very large set. The primes in our set are {p₁, p₂, p₃, ... p_n}.

By the accepted properties of integers we may multiply these together and add 1 to the product to create another integer N. We have N = p₁ x p₂ x p₃ ... x p_n + 1.

Now, p_n <  p_n + 1 <  p₁ x p₂ x p₃ ... x p_n + 1 = N, so pn < N.

What happens if we divide N by p₁ or p₂ or any of the primes in our list? The class pondered this but had no response. I asked them to see what happens for 2 x 3 x 5 + 1. What happens if you divide this number by 2 or by 3 or by 5? Students saw that the prime numbers did not divide the number evenly. In fact, each time we divide by one of the primes in the list the remainder is 1.

This means that we have a number that is larger than any of our existing prime numbers and is not divisible by any of those primes. This contradicts the idea that p_n is the largest prime. Therefore there must not be a largest prime.

We then looked at creating a cumulative frequency graph of the number of primes. This graph roughly follows a logarithmic function. The asymptotic nature of the prime number distribution was proved by Gauss, but I did not get into that at this time. It is something that could be discussed now; I have a video, The Music of the Primes, that discusses this and which we will watch later.

I displayed one student's graph to help illustrate the nature of the distribution. Of course, students couldn't remember the name of the graph, although some recognized that it looked like a parabola flipped on its side or similar to a square root graph. I asked students to try to make connections between this idea and patterns that they see in their Sieve of Eratosthenes work. Again, not much was forthcoming; I asked students to continue to consider connections.

Next we answered some questions about the distribution of primes and length of contiguous composite numbers. Several students asked what a composite number was. I asked the class to help out and thankfully several were able to say that composite numbers have more than 2 factors.

We discussed the answers and they show that the number of integers needed to capture 10 prime numbers increases as the integer values go up. We also saw that the longest length of contiguous composite numbers goes up as the value of integers increase. These illustrate that while prime numbers are infinite, they become more and more spread out.

I then had students make conjectures about prime numbers and the patterns they were seeing. Several students wondered whether the prime numbers would eventually taper off to nothing. This was a wonderful opportunity to go back to the proof we did earlier. When students were reminded that we proved that the number of primes was infinite, they readily said, "Oh, that's right!" and then proceeded to modify their conjecture that the primes become more and more rare.

Next class we'll explore the idea of integers that are relatively prime. This will be followed by prime factorization and then the Euclidean Algorithm.

Visit the class summary for a student's perspective and to view the lesson slides.

IPS - Day 23

Today we continued looking at probability models and expected values. Many students did not attempt to complete the Shooting Free Throw worksheet. I let it be known, in no uncertain terms that when I give an assignment that I expect it to be completed.

Because of the issues with this I decided to skip over the simulation portion of shooting two free-throws and asked students to focus on looking at the problem from a theoretical perspective. In this case the struggles were with recognizing that scoring one point could occur in two different ways: a miss and a make or a make and a miss.

Students would ask if their result was correct and I would ask if their probabilities summed to one. When they realized they didn't it became an issue of why not? They then discovered that they hadn't accounted for all of the outcomes.

With the probability model constructed I asked students to calculate the expected value. I had to refer back to the One Girl Family Planning worksheet and its definition of expected value. Students calculated the expected value. We briefly discussed the meaning of expected value.

I then had students look at the simple dice game that I had used a couple of days ago. In this game, you roll the die and break even with a roll of 1, win $5 with a roll of 2 or 3, and lose $5 with a roll of 4, 5, or 6. I asked students to calculate the expected value for this game. I wanted to use this as a check to see if students knew the basic mechanics for calculating an expected value. I was pleased to see that a large percentage of the class was able to calculate the expected value of -$0.83 correctly.

I asked what the expected value meant. Students said it meant that you would lose money when playing the game. I tried to push their thinking further. I explained that the expected value was the amount I could expect to lose on average. For example, if I played one million times, I told students that on average I would have lost $0.83 cents. I then asked them how much total money would I have lost? This puzzled them for a few minutes. I said that after playing the game one million times my average loss was $0.83 cents. Students started working through and figured that I would lose $830,000 in total. I thought this was a good way to illustrate how the expected value represents an average of the results that will occur but the cumulative impact can be much different.

We then started to look at the 39-Game Hitting Streak investigation. Students struggled with how to verify a given probability. I told them to consider from a theoretical perspective the events that take place to reach the stated outcome.

The investigation asks students to consider the probability of hitting streaks of 1, 2, and 3 games. The tendency is to focus on the probability of getting at least one hit in a game and stopping. The issue is that a hitting streak of a given length means the streak ends. For a one-game hitting streak, a batter must get at least one hit in the first game and then get no hits in the second. For a two-game hitting streak, a batter must get a least one hit in games one and two and then not get a hit in game three. This pattern continues for whatever length hitting streak you may want to consider.

Students were just starting to understand the issue that the hitting streak must end when we got to the end of the class. We'll finish up this investigation on Monday and then take a look at some other situations involving expected values.

Visit the class summary for a student's perspective of the class and to view the lesson slide.

Discrete Math - Day 23

Today we continued to explore prime numbers and their properties. The class first verified that they had all the prime numbers less than 500 captured. We next discussed patterns that students found. Students talked about whole columns that could be eliminated and one student mentioned that the values along a diagonal were being eliminated, although she couldn't remember for which value this occurred.

The discussion then centered around how many prime numbers needed to be checked. Some students didn't stop until they were up in the seventies. One student said she stopped when the product of the next two primes exceeded 500. This meant that when she saw 23 x 29 > 500 she stopped. No one else had any thoughts about how far they had to go, so I asked the class how this student knew that she could stop looking for primes at this point.

There was a painfully long wait (over 2 minutes) of utter silence. I re-phrased the question and asked what was happening as you moved through the list of primes, crossing out multiples, that indicated you could stop at 23. Again there was no response.

I am not sure what is at the root of this issue. Is this a problem with student number sense? Is it an issue that students are not used to being active learners and just want an answer? Is it too early in the morning for them to be awake and thinking? It is something that I will have to monitor. I will say that students' natural abilities to be inquisitive and to reason are inhibited at this point.

I discussed what is happening as we proceed through the list. Each time, multiples of that prime are eliminated. When we get through 19, we have covered all numbers and their multiples that can be formed with primes valued at 19 and less. For 23, we already have 19 x 23 = 437. There are no other multiples of 23 that are not already crossed out, so there is no need to check further. A number like 31 has all of the lower multiples crossed off, such as 62, 93, 155, etc. Thus, one all the prime numbers less than square root of 500 have been checked there is no need to continue.

We then looked at some divisibility rules. Students by-and-large were not familiar with divisibility rules. For the most part students aren't interested in these rules because the can just pull out a calculator and check. Given that I have rearranged the course material, it may have been better to introduce these rules later when looking at modular arithmetic. I will move this piece for the next course.

As it was, most students did attempt to make use of the rules and work through trying to determine from the rules whether or not different values were prime. I discussed the divisibility rule for 7 as an illustration of how things work.

Finally we looked at the question of how many prime numbers exist. This was at the end of class but there was some interest in the philosophical nature of the question and the idea of the infinite. One student said they would go on forever because numbers themselves go on for ever. Another thought there might a fixed number. The question came up of couldn't there be some formula or equation we could use to determine the answer. This was a perfect set up for proving the infinitude of prime numbers.

There wasn't enough time in class to go through the proof but I was able to wrap up with the ideas that the students presented and that we would use these to prove the conjecture that there are infinitely many primes.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

Discrete Math - Day 22

The class started with taking a second look at induction. I used the sum of the first n odd integers to illustrate the induction process. It still didn't go well. I am going to look through the proof course on the Journal of Inquiry-Based Learning in Mathematics web site to get some ideas for next year.

With that the course transitions to the second unit, which covers Number Theory and Cryptography. Last year I went a bit overboard on the number theory portion. Although some of it was interesting, at least to me, the discrete course almost turned into a number theory course. This year I've removed some of the prime number investigations and will bring them in at the end of the unit to fill in space, if needed.

As a means of introduction I briefly discuss that cryptography is used in everyday communication devices and on web sites. Cryptography helps to keep our personal information secure. However, cryptography is based on prime numbers, so we need to spend some time better understanding the properties and characteristics of prime numbers. And so begins our journey into number theory.

Many students haven't thought about integers and prime numbers for a good number of years. And, although students have had two years of algebra, they don't necessarily recognize the underlying algebraic structures that are at play. This provides the opportunity to cover familiar ground from a more detailed mathematical perspective. The importance of precise definitions, the use of algebraic structures, and the development of important mathematical relationships can be developed using ideas and concepts with which students can readily relate.

To start of, I ask students to define an integer. The purpose of this isn't to get to a formal definition but to have students agree as to what an integer looks like. Students discussed that integers can't have fractional parts or decimals; are values like 1, 2, 3; and do change values in increments of one. I wrote out the following:
                      ...-2, -1, 0, 1, 2, ...

and repeated that these values have no fractional or decimal parts. I realize the latter is not technically true but for the purposes of understanding what an integer looks like it focuses students on the important idea.

Next I asked students to consider properties of integers that they know. Students talked about integers being positive and negative, that there was a zero value, and that they didn't have fractions.

I asked students about operations and one student said you could add two integers. The class giggled because they thought it was obvious, not realizing this was a key algebraic operation that can be performed over integers. I pressed students to consider other operations. They responded you could multiply but not divide, another key algebraic property. After a few more ideas we took a look at the properties of integers under addition and multiplication.

The properties looked at were:

Addition

• for any integers a and b, a + b is also an integer
• for any integer a and value zero, a + 0 = a
• for any integers a, b, and c, (a + b) + c = a + (b + c)
• for any integer a, there exists integer b such that a + b = 0
• for any integers a and b, a + b = b + a

Multiplication

• for any integers a and b, a x b is also an integer
• for any integer a and value one, a x 1 = a
• for any integers a, b, and c, (a x b) x c = a x (b x c)
• for any integers a and b, a x b = b x a

I present these in this manner because it sets up considering the similarity and differences of operations under addition and multiplication and mirrors how algebraic rings are defined.

This is an opportunity to discuss the ideas of closure, identities, associative property, commutative property, and the existence of an inverse (for integers there is an additive inverse but not a multiplicative inverse). I also relate this algebraic structure to other entities that possess the same basic properties: matrices and polynomials. Students have seen and worked with these but probably never connected the operations to a common algebraic structure, in this case what we call an algebraic ring.

I also point out that the commutative property is actually a special property that does not apply to all situations, for example matrix multiplication.

Students at this point realize there is a lot more going on in their known math world than they had previously considered.

I next asked students to define a composite number. Again, I wasn't looking for a formal definition at this point, only a working definition that we could use. Students came up with two ideas:

1. A composite number is the product of two other integers other than one and itself.
2. A composite number has factors other than 1 and itself.

The class was satisfied that these definitions fit the idea of composite number, so we moved to the next definition, that of a prime number. Here, students decided that a prime was only divisible by 1 or itself or that its only factors were 1 and itself.

This was perfect because it allowed me to bring up the question of whether or not 1 is a prime number. Students said that 1 was not. I told them that was correct and that 1 was the multiplicative unit. I asked students to consider the definition of a prime number and if 1 fit into the definition. They agreed that it did.

The question posed was how to define a prime number so that 1 would not be included in the definition. After some discussion, students decided that defining a prime as an integer with exactly two factors fit the bill.

This allowed us to now re-visit the definition of a composite number. Students concluded that a composite number would be a number with 3 or more factors. This left open the ability to characterize the multiplicative unit as a value with exactly 1 factor.

With these definitions out of the way it was now time to define what factor meant, since both the prime and composite number definitions required factors. Students decided that a factor b of a means that b divides a. I used this as an opportunity to introduce the symbol b | a to designate that b divides a.

Of course, this brought up the issue of what it means to say that b divides a. Students really struggled with this. It is a simple concrete idea that is so ingrained that students have trouble thinking about the idea more abstractly. After some questioning and discussion, students started to think about this in terms of factors. I warned them to be careful because we didn't want to get into circular definitions. After some more discussion we said that b divides a if there is another integer c such that bc = a.

I didn't emphasize that all of this is made possible through the properties of integers that we had discussed. This is something that I'll have to bring up tomorrow.

With these definitions in place we proceeded to look at a three questions about prime numbers:

• Is 221 prime? If not, what are its factors?
• What’s the largest divisor you need to check to be sure that 397 is prime? How do you know it’s the largest?
• Is 8171 prime, or not? How do you know for sure?

I allowed students to work on these on their own, no smartphone searches allowed, and then to discuss in their groups. Students wanted to know the easy way to get to these answers. I let them explore.

After trial and error, students found that 221 = 13 x 17. It's factors are 1, 13, 17, and 221. Students initially said the factors were just 13 and 17 but it's important to count all the factors.

Most students thought they would need to check values up to 200. We talked about this and I listed out the following primes: 2, 3, 5, 7, 11, 13, 17, 19, 23. After talking about this I claimed that once I that the first 8 primes did not divide into 397 that there was no need to check any further. We discussed this further and students were reluctant to accept that this would happen. I provided a simpler example, using 10 as the number to examine. This helped to illustrate what was happening.

For the last problem, we didn't verify if 8171 was prime or not but discussed how many primes would need to be checked. Students found they would need to check primes up to 90. I then asked students if they knew all the primes less than 100. Of course, they didn't and I didn't expect them to know.

This allowed me to introduce the Sieve of Eratosthenes. I provided a worksheet to use the sieve on the first 500 positive integers. I illustrated how the sieve works and asked students to work through the sieve as homework.

Tomorrow we'll use the results to examine the distribution of prime numbers.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 22

Today we continued our exploration of expected value. Students finished working on and discussing the One Girl Family Planning investigation from yesterday. There were some good conversations and the class discussion that followed was a healthy examination of how to find expected value.

I first checked with students about how they simulated the situation. I like to do this so students can hear about different ways to represent the given scenario and to identify any potential issues. Students presented using 0 or 1, 0-4 and 5-9, and flipping a coin to represent the situation. These all work and are easy to implement.

One girl presented her simulation by saying she picked individual random digits and whatever digit came up that was the size of the household. I asked the class what they thought of this representation.There were puzzled looks and some seemed unsure of what to think. I asked what the probability of a 9 child household would be using this technique and students responded 1 out of 9. I then asked them what the theoretical probability of having a 9 child household was. The response was it was a tiny number. I again asked if the model being discussed would work and the class replied that it wouldn't because the probabilities for the events didn't match.

This was a good opportunity to emphasize that the probability of occurrence should be accurately reflected by the random digit assignment. It is also important to think about how to apply this assignment to simulate the given scenario.

By this time, students were starting to understand the mechanics of finding the average number of children per household. Calculating the theoretical expected number was confusing to some students. I had asked these students to consider what they had done when calculating the average number of children per household. This helped students to think about what they had done and why.

During the class discussion, one student discussed how she had multiplied out number of children, an assumed household count of 500, and probability for each family size, added these values together, and then divided by 500 to get the average children per household. Another student presented essentially the same calculations but didn't bother to use the 500 value.

The class wanted to know which way was right. I told them they needed to consider what they were trying to find, the information they had to use, and what they had been doing. Some students took to this and had healthy discussions, performing various calculations to cross check results and techniques.

The final discussion centered on the algebraic equivalency of the two approaches. As one student put it, if you multiply each value by 500 and then divide the result by 500, you are cancelling out the 500 values.

The investigation concluded with providing the definition of expected value:

We then proceeded to work on the Shooting Free-Throw investigation. This investigation followed a similar tract to the One Girl Family Planning investigation. Students needed to simulate the situation, examine their results, calculate theoretical values, and compare results.

The first issue students face is how to simulate the scenario. Most students recognize that a 60% free-throw success rate could be simulated by using the digits 0-5 for a make and 6-9 for a miss. Some students, however, wanted to assign a 50-50 probability to making and missing. The thinking is that there are two outcomes and the inclination is that they should have equal probability. I simply ask these students what is the probability that a free-throw is made and they quickly realize they can't use a 50-50 split on their random numbers.

Students struggle with how to account for the one-and-one situation in the problem. The tendency is to say there are 3 outcomes: 0 points made, 1 point made, and 2 points made. The issue is that these events are not all equally likely. Referring students back to the One Girl Family Planning problem and asking them how the simulated a household having 3 children helped them to realize they needed to generate successive random numbers.

From here the next hurdle is calculating the expected point value. I asked students to refer back to the expected value definition. This helped students move beyond what to do and to begin recognizing how an expected value can be calculated.

Students also had questions about how to calculate the theoretical probability associated with the situation. Students are not good at representing the situation using a diagram or table. I encouraged students to try to make a visual representation of the problem. Many started to use tree diagrams, which helped them piece things together.

I asked students to complete the investigation as homework and that we'll discuss the results next class.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 21

Today we started looking at the connection between simulations, probability models, and expected value.

We started class looking at a roulette table. Gambling games and Las Vegas catch students' interest and it is valuable for students to learn that large, ornate hotels and casinos are paid for by the public's misconception of expected value.

I use the roulette table as an opener but, for now, it is only on an informal basis that we talk about what should be expected to happen if the game were played repeatedly. In this discussion, students felt like they would come close to breaking even.

Next was a discussion of how probability models could help in looking at what is expected. I used a simple example that helps illustrate the issues. In this example you are playing a game that costs $5 to play. I you roll a one you receive $5. If you roll a 2 or 3 you receive $10, and if you roll a 4, 5, or 6 your receive nothing. The question is what do you expect to happen if you play this game repeatedly?

I let students think about the situation and then discuss it among themselves. First, I was pleased that students readily accounted for the cost and related that the actual outcomes for the situation were $0, $5, and -$5.

The discussions tended to focus on the values of the outcomes and not to account for the frequency of each outcome. About two thirds of the class felt they would come out a head or even. The others thought they would lose. Those thinking they would win focused on the winning of $10 and that this would offset the loss of $5.

Those that thought they would lose focused on the frequency of occurrence. In this case, half the time they would lose $5. For the other half, they would not win something part of the time and therefore they would not win as much as the lost. We set up a probability model to help illustrate the issue. The class agreed that this argument made sense.

Dice Game Probability Model

Outcome	$0	$5	-$5
Probability	1/6	2/3	1/2

We didn't get into calculating the expected value at this point. Rather, I told the class that not only could we determine if we would win or lose but we could also determine how much we could expect to lose.

With that we dove into the first of a series of investigations drawn from Navigating Through Probability, Grades 9-12. The first investigation we looked at was One Girl Family Planning. This activity builds on previous investigations done in class.

Students were comfortable with the simulation piece. The issues came in with calculating average children per household. The tendency is to either calculate average households per group or average children per group. It takes some prodding to get students to understand they need to determine the pool of children available and how many households these children are being divided up by.

The next hurdle comes with calculating theoretical probabilities. Again, some questioning helped students realize how this should be calculated. They were also able to calculate their expected values by using a quantity of 500 or 1000 to get a theoretical mean.

I was able to have a couple of students explore the idea of whether they needed the value of 500, say, to calculate the mean. These students thought that it was necessary. I asked them to describe how they performed their calculations. They basically said they multiplied all the values by 500 and at the end divided by 500. I asked them if they were multiplying everything and dividing everything by 500 if the 500 was necessary. The students started to realize that maybe the 500 value wasn't necessary. I asked them to check to see what would happen if they didn't use the 500 when calculating the mean.

The rest of the class will finish off the investigation at home and we'll discuss at the start of next class.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

Discrete Math - Day 21

Today was a rough lesson, which I had anticipated. We are coming off a four day weekend and the topic for today was mathematical induction. Probably not the best topic to tackle first thing off a four-day break.

Most students did not have proofs covered in geometry, so this is their first look at the idea of proof. This is changing, as geometry classes are again covering proofs but that isn't the case for the students in this class.

The issue is relevant because students are blindly solving problems without any justification or checking as to the reasonableness of their solutions. I used one problem that many students tackled on their portfolio problem as an illustration. Every one who tried this problem got it wrong and in the same way.

The problem was if you take the first 8 letters of the alphabet and require that the letter sequence ABC always appears, how many different arrangements of the letters can be made. Students stated the solution was ₈C₃ because the had 8 letters and they needed to keep the 3 letters (ABC) together. There was no checking, verification, or other justification. While this is a nice conjecture, it needs to be affirmed as to its reasonableness. It basically needs to be proved that the solution always works.

I reiterated that larger problems should be broken down to smaller entities so that the reasonableness of solutions can be checked. In this case, making a table of the number of additional letters and the number of arrangements quickly shows that _nC_r does not work.

I moved from this to the concept of proof. I referenced back to triangular numbers and the closed form of the formula. The first question posed was to get students thinking about how to demonstrate a formula always works. As expected, students thought you could plug in values to test the solution. I then asked how many values is enough? Do you need 10, 100, 1000? How do you know if the formula will still work on the 1001st term?

To get students thinking about what it takes, I then asked them what they would need to see in order to be convinced. Students struggled with this. There was some thinking about seeing a pattern that clearly repeated.

I conveyed the basic tenets of mathematical proofs:

• Proofs use accepted truths (called axioms)
• Proofs are a series of statements that are logically connected
• Proofs demonstrate that the proposition will always be true under the stated conditions

The last tenet emphasizes that the context of statement truth needs to be considered. For quadratic equations, all real solutions fall on the x-axis. However, this statement is not true over the field of complex numbers. The truth of the statement is within the confines of the stated conditions.

I asked students to consider why proofs are important. There were some good assessment of the need for proofs:

• They verify the validity of a statement
• They create a foundation to build upon
• They provide understanding of a problem

I then introduced mathematical induction as one way of proving statements or equations.

I presented the steps to conduct a proof by induction:

• Demonstrate a statement is true for a starting value.
• Demonstrate that from any given starting point (k) you can reach the next step (k+1)

and asked students why this would demonstrate that a formula always works?

Students were wrestling with the ideas and had no clear understanding of the logical basis for what was happening. I used the analogies of climbing a ladder and dominoes falling to try and make a concrete connection between the concept of induction and something that students could grasp.

I tried starting things off with a simple example. I think a different example needs to be used here, perhaps using the sum of the first n integers or the sum of the first n odd integers? For this class I used every positive even number can be written as 2n for n > 0. I started with a value of n=1. In this case 2 x 1 = 2, which is even. I then assumed that 2k was even. The next even number is 2k + 2 (having discussed the recursive formula that E_n = E_n-1 + 2. Then 2k + 2 = 2(k + 1) so the formula works for the next even number and by induction will be true for all positive integers.

Students were confused and, not surprisingly, many seemed indifferent to the idea. I asked them to try using the idea of induction to prove the triangular number result. Most got stuck and had to be led through the process. Some students managed to figure out how to demonstrate the result for n=1, but didn't know where to go from there.

We'll try working through one or two more before diving into number theory and cryptography, where proofs will show up again.

Although it was a struggle, I think it is worthwhile to convey the ideas of proof and to have students work through the reasoning and logic that proofs require. We'll continue to look at proofs throughout the semester. As a first lesson on proof, I think students started to understand this was an important, and perhaps difficult, task.

View the class summary for a student perspective and to see the lesson slides.

Discrete Math - Day 20

Today we practiced working with Bayes theorem. The first task was helping students to decode text to understand what events to use and what probabilities to determine.

I presented a structure for students to consider and asked them to work through identifying the components they would need to use to answer the question. As anticipated, there was a lot of confusion and misunderstanding they arose when attempting this. This led to a good class discussion about how to identify components.

The next task was actually calculating the necessary probabilities. This involved taking weighted averages, which again caused problems with students. There were good questions about why you simply couldn't add two values together and divide by two, whether the probability was for an overall rate or just for a specific event, etc.

Students then tackled another problem. Although there continued to be confusion on identifying events of interest and how to calculate an accumulated probability, more students were feeling comfortable with how to proceed and how to resolve these issues. By the end of this problem, the majority of students were feeling more confident about working through a Bayesian probability problem. They admitted they didn't like doing it but they felt they could do it.

I briefly touched upon what they just worked on and how this relates to spam filters, explicitly connecting events and probabilities in the Bayes formula to the scenario.

Working through and discussing these two problems took over 60 minutes. Since there was only about 20 minutes left, I decided to pass out the take home quiz and give them time to work on it in class. The quiz covered counting concepts such as combinations, permutations, and pigeonhole problems.

View the class summary for a student's perspective of the class and to see the lesson slides.

IPS - Day 20

Today was focused on looking at probability models and simulating results. The What is the Probability of a Hit investigation from Navigating Through Probability Grades 9-12 provided the source material for today.

This task provides application of much of the basics of probability that has been covered. To complete the investigation, students need to consider independence, mutual exclusivity, sample spaces, unequal probability, and simulation.

I was pleased how students were able to list out the possible outcomes for the sample space. This typically is an issue, but not this time. Calculating the probabilities for each outcome was more problematic. About one-third of the class realized they had independent events and that to have an outcome over three events that the probabilities for each would be multiplied.

The rest of the class simply wanted to count the number of successful outcomes and divide by the total number of outcomes. Since event probabilities are not equal, this approach leads to an incorrect result. After some questioning and guidance, most students understood why they needed to multiply the probabilities.

The next issue came in calculating the outcome of at least a certain number of events. Some students wanted to multiply probabilities, some wanted to count outcomes, and some did not include all of the outcomes they should. After discussion with classmates and questioning, they figured out what they needed to include and how to calculate the results.

The rest of the investigation went smoothly. Students said they were feeling more comfortable with understanding and using probability.

The remainder of the class was spent going through the probability quiz. Students realized they were over-thinking or misunderstood what was being asked. I passed out a second quiz that students will complete over the weekend.

View the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 19

Today we continued exploring conditional probability, independence, and the relationship between the two. We tackled another investigation from Navigating Through Probability. In this case the Independent or Not Independent task.

This task looks at identifying an association between students with asthma and students living within a smokers' household. Overall the class went well. Students were being thoughtful about what they were doing and were asking relevant questions that would push them toward a solution. Students definitely appear to becoming more comfortable with the notation, although several still struggled with how to describe the meaning of P(A | B).

The biggest hang-up can when explaining the meaning (the implication) of when P(A | B) and P(A) are not equal. Several latched onto the idea that the events are not independent but were hard pressed to explain why this result indicates independence.

With a little questioning and guidance, students started to see that the perception of likelihood of living in a smokers household changed when they knew a person had asthma versus not knowing anything about the individual. This influence of one event (having asthma) on the probability of a second event (living with smokers) is why the two events are not independent.

We'll continue to explore this through a couple of more examples before moving into probability models and expected values.

Visit the class summary for a student perspective of the class and to view the lesson slides.

Discrete Math - Day 19

Today was focused on providing practice in working through probability problems. After a brief review of representing probabilities, how conditional probabilities look in these representations, and Bayes theorem, students worked through 10 problems.

I use a subset of questions from Basic Ideas in Probability on education.com. These problems provide a nice mixture of conditional probabilities, independence issues, and Bayes theorem.

Students typically struggle with wording of questions, trying to decipher what exactly is the sample space and what is the event of interest. Overall the class did well on these, although there are still issues with using Bayes theorem.

I think we'll spend some more time working through a couple more problems involving Bayes theorem and take a look at mathematical induction. This latter topic fits into the question of how do you know that a formula will always work. We'll revisit some of the work on figurate numbers to answer this question.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 18

Today we completed the Abby's Kennel investigation. Although it took a couple of days, the discussion today about the relationship between conditional probabilities, multiplying probabilities, and independence was healthy. It appears there is a better understanding of the interplay of these concepts that I have seen in the past.

After discussing the calculations and their meaning, we looked at a few more examples of conditional probability. These were to assess student understanding. Overall, the class seemed to grasp what was happening when a condition was placed on a probability.

I briefly touched on ways that we have been able to represent probabilities and mentioned that conditional probabilities can be looked at from the perspective of any of these representations. I will cover this more explicitly next class before students work through practice problems.

We then worked through a simply presentation of tabular data that represented infant hearing screening tests compared to actual hearing loss. Several questions were posed and the most difficult aspect of these questions is deciphering precisely what is the condition and what is the desired outcome. Again, there was healthy discussion about the meanings of the problems and the results.

Students wrapped up the class by summarizing in their notes what they saw as the connection between conditional probability and independence.

We'll work through problems next class and then take a closer look at the idea of independence.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

Discrete Math - Day 18

Today's focus was on conditional probability, independence and Bayes Theorem.

We first looked at some simple examples and students were asked what common characteristics they saw across the problems. Besides making connections to ordinary probability they observed that constraints were being placed on the problem through the given portion of the problem statement.

Conditional probability was defined informally, notation was given, P(A | B), and several examples were provided. The class worked through these and the results were discussed. These problems are intended to build conceptual understanding so students can easily see how their sample space is being altered by the given in the problem.

Students practiced on several problems. These problems set up the ability to explore the idea of independence. Two definitions of independence were given:

     P(A and B) = P(A) x P(B)     or     P(A | B) = P(A)

These are equivalent definitions, using one it can be shown that the other must hold.

Students then were asked to use these definitions to show independence in two of the problems just worked on. They then used the general multiplication rule P(A and B) = P(A) x P(B | A) to calculate the probabilities both way to determine independence. This was followed by calculating probabilities that were not independent.

Although these problems are somewhat simplified, the purpose of having students work through these is so they can connect the formulas, calculations, and results. These problems are building toward Bayes theorem. There needs to be a certain level of comfort with how conditional probabilities work before diving into Bayes theorem.

To get to Bayes theorem, the equivalent forms of P(A and B) are set equal to each other

     P(A and B) = P(B) x P(A | B) = P(A) x P(B | A)

Then, algebraically, we get

     P(A | B) = P(A) x P(B | A) / P(B)

This effectively reverses the conditional statement, i.e. we start with knowing the probability of B given the event A and we can reverse the relationship to now determine the probability of A given the event B.

A simple example using to boxes containing red and green balls can be used to illustrate the theorem.
We are given a green ball and ask the question, "What is the probability the ball was drawn from the first box?"

In this example, event A is being drawn from the first box and event B is a green ball being drawn. We need to use the fact that box 1 contains 2 green and 5 red balls and box 2 contains 4 green and 5 red balls. We calculate the following:

     P(a ball being drawn from box 1) = P(A) = 7 / 16
     P(a green ball | the ball is drawn from box 1) = P(B | A) = 2/7
     P(a green ball) = 6/16

Bayes theorem states:

     P(a ball drawn from box 1 | a green ball) = 7/16 x 2/7 / 6/16 = 2/6 = 1/3

Students are a bit intimidated the first couple of times working through a problem like this but do build confidence and comfort the more problems they tackle.

I then referenced that email spam filters are based on Bayes theorem. An email is received and based on characteristics that are present in the email a probability is attached as to how likely the email is spam:

     P(spam | characteristics present in the email)

Bayes theorem is used in forensics and investigations as well. This brings back the issues raised in the video we watched on the first day. There was not time to discuss this aspect in today's class but it will make a nice connection as we proceed ahead.

We concluded with students recording their thoughts about conditional probabilities and independence. The next class will focus on working through some problems to gain procedural fluency.

Visit the class summary for a student's perspective and to view the lesson slides.

IPS - Day 17

Today was a bit odd. This was the Friday of Spirit Week and there was a spirit assembly before this class. Being the last period of the day for most people, they were anything but in a mindset to focus on mathematics.

I started the class by rearranging seats partially because it was about time to do this and partially to try to get students more focused on what was going on in class.

We continued our work on the Abby's Kennel investigation. Today's work focused on the definition of independence and on conditional probabilities. The investigation defines the independence of two events, A and B by comparing P(A and B) to P(A)P(B). If these two values are equal the events are independent.

The conditional probability P(A | B) is defined and there are comparisons made between the situations of P(A | B) and P(A | C), where event A is calculated given A and then given B. When these two values are equal, it is another indication of independence.

Of course, some students didn't get very far and some rushed through the questions without really considering why they were doing certain calculations and what their results meant. I had to have them go back and explain more about what was happening and what it means.

I'll use these series of questions as the next portfolio problem as conditional probabilities and independence play a key role in what we'll be doing throughout the semester.

Visit the class summary to view a student's perspective of the class.

Discrete Math - Day 17

Today we got back to discrete probability. I posted 3 problems based on cards games. Students were to find the probability for each event.

This was another struggle for students. The tendency is for students to want to add things together. For finding the probability of being dealt a "blackjack" they wanted to add 4/52 + 16/52. I kept reminding them to look back at how we were counting outcomes on other problems. The same could be said for the other two problems.

I finally had to step in to help get things re-focused for the class. I first asked how many ways could I be dealt two cards from a deck of 52? After some thought I got responses that there were ₅₂C₂ ways. I told them this was the value of the denominator; it reflected all the possible outcomes that could occur.

I then asked what the numerator should be. There was a lot of mumbling but no responses. I focused on the first card and asked how many different aces could I be dealt. The response was four. I then asked how many ways could I be dealt a face card or 10. The response was 16. I then asked how many ways could the 4 aces be mixed and matched with the 16 face cards/10s. Some students responded there were
4 x 16 = 64. I related this scenario to mixing and matching pants and tops.

I then asked students what if we were dealt a face card/10 first, how many ways would be match this situation with an ace. Students readily saw there were 16 x 4 = 64 ways for this to happen. The question is whether the desired outcomes were 64 or 128. This can be related back to the basketball teams problem. Once we have an ace and a 10 in our hand, does it matter whether the ace or the 10  was dealt first. The resultant probability is

     P("blackjack") = 64 / ₅₂C₂ = 64 / 1326 = .048265

So, approximately 1 in 20 hands results in a blackjack.

For being dealt a flush, students now realized the denominator would need to be ₅₂C₅. The issue was how many flush hands could be dealt. I asked the class how many cards do we have to draw from to form a flush. Students realized there were 13 cards to draw from and we were picking 5 of these, irrespective of the order. So, we have ₁₃C₅ ways to draw 5 cards of the same suit. But this is only one suit and we have four. Therefore, there are 4 x ₁₃C₅ = 4 x 1287 = 5148 possible flush hands. Our probability is

     P(flush) = 4 x ₁₃C₅ / ₅₂C₅ = .001981

or about .2%.

The last problem is one that confuses students because they are now primed to look at hands dealt and ways to get a single card from 13. I like to present this problem third for just that reason. Students can approach it from this perspective and they will reach the correct solution but a little thought will simplify things greatly.

In the game of spades, with four players, each player receives 13 cards. The queen of spades can only end up in one of those 4 hands, so the probability of getting the queen of spades dealt to you is 1/4. It is instructive to allow students to work through this problem the long way and come to the realization that the result is 1/4.

The long way first looks at how many ways a hand of 13 cards can be dealt, which is ₅₂C₁₃. The next issue to address is how many ways can the queen of spades be dealt in a hand of 13 cards. This means we have 12 cards that can be selected from 51 cards, since we are assuming our hand would have to contain the queen of spades. This means we now have ₅₁C₁₂ options. Finally,

P(holding queen of spades) = ₅₁C₁₂ / ₅₂C₁₃ = .25

There is usually a moment of shocked surprise at this result. Then students realize that this makes perfect sense.

Class concluded with students recording which probability rules are confusing to them and things they can do to help remember what to do.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 16

Today we started class reviewing probability problems and then taking a quiz. The work ethic of some students hasn't been what it should be, so I was expecting some students would struggle with the problems, which was the case.

After the quiz, we started investigating conditional probability. I use the Abby's Kennel investigation from NCTM's Navigating Through Probability Grades 9-12. This investigation allows students to build a conceptual understanding of conditional probability and to develop procedural knowledge.

We had time for students to complete their data collection portion of the lesson. We'll finish working on this investigation next class.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

Discrete Math - Day 16

We took a break from discrete probability to focus on reading problem statements and discerning key elements to help identify possible solution routes.

Students were given 21 problems and asked to group the problems based on what they were seeing in the problem statements. There were some good discussions on what the problems contained, although a couple of groups looked at more superficial elements such as problems using letter or problems using numbers.

This is okay as part of this exercise is to reflect later on about the appropriateness of the groupings. I then had group members rotate through the other tables. One group member remained at their table to present their problem groupings and answer any questions about how or why problems were classified in such a manner.

Those circulating were to focus on how the problem groupings were made, to look through the problems actually in each group, and to ask questions they had about the problem groupings. All the tables were visited in this manner.

When students reached their original tables they shared what they saw and learned with the table-mate who stayed behind. If needed, they could re-group their problems based upon what they had seen others do.

Next, students selected three problems, each from a different problem grouping, and solved them. They were supposed to consider how they solved the problem versus how the problem was classified. Were the solution methods similar or different, were the groupings consistent with the solution method, etc.

We discussed a few problems that students had questions about. At this point I assigned their next portfolio problem. The assignment is described below:

• Discuss how these two problems were classified and what you see as similarities and differences in the problem statement
• Solve each problem
• Discuss the similarities and differences in the solution
• Comment on the appropriateness of the classification you made

Afterward, we refocused on basic probability rules. We discussed examples of the different probability properties and rules. Class closed with students recording the rules in their notebook along with examples to help them understand the meaning of the rules.

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 15

Today was a bit of a scramble. I had to host a literacy lab at the last minute during the previous period and the debrief of the lesson took place during this class. As a result, I had a substitute cover today.

I have been wanting to give students additional practice with calculating probabilities in preparation for a quiz on probabilities. I used this period as that time, preparing several different worksheets of problems that reviewed many of the probability properties that we have been covering.

Students worked through the problems and they will have a quiz tomorrow. There are some outstanding questions about some of the answers that I'll go through with them before the quiz.

We'll then dive into an investigation of conditional probability.

Visit the class summary for a student's perspective of the lesson.

Discrete Math - Day 15

Today we started to look more closely at discrete probability. I re-visited the idea that we are counting desirable outcomes and total possible outcome. We then divide the desirable count by the total count to determine the probability.

To start things off I asked students to calculate the probability of being dealt a four of a kind (four cards of the same rank) in poker. This boils down to drawing 5 cards and having four of the same rank.

Students were basically stumped on this problem. It took a lot of guiding questions, asking students to look back at what we had done before, think about connections to the flag problems we had worked on, etc.

Even the denominator, which we had determined yesterday, was somewhat of a struggle. Students were thinking it would be ₅₂C₅, but they had no confidence that their thinking was correct.

After this, students were thinking that there would be 13 choices for the rank of the 4 of a kind. Then they were stuck again. I tried to get students to think of the other problems we worked on and what was happening with the cards. Finally a few students started considering the five cards in the hand and realized they could focus on the fifth, non-matching card. They determined there were 12 ranks this card could take on and then 4 suits. I had these four students go around to the other groups to discuss their idea. Afterward, I went around to the groups and asked them to communicate what they had learned. It appeared this was effective in helping the other students to make sense of the problem.

We finally arrived at a result of P(4 of a kind) = 13 x 12 x 4 / ₅₂C₅

I  then had students review what probability that they knew. It was limited to very vague concepts and a few considerations as to what valid probability values look like and must sum to.

I went through a few basic rules and then asked students, for homework, to produce examples of these rules.

The rules covered were:

1. 0<= P(A) <=1     all probabilities fall between 0 and 1, inclusive
2. Sum of all probabilities equals 1
3. P(A) + P(not A) = 1     probability of an event and its compliment are 1 (this can be used to calculate P(A) by using 1 - P(not A))
4. P(A or B) = P(A) + P(B)     for mutually exclusive (non-overlapping) events
5. P(A and B) = P(A) x P(B)     for independent events (when one event does not influence the outcome of the second event)

Visit the class summary for a student's perspective of the class and to view the lesson slides.

IPS - Day 14

Today we continued exploring calculating probabilities and running simulations. We looked at the birthday problem with the intent of determining the probability.

The first task was to simulate the situation. I had students consider the scenario and then we discussed their ideas. The class was in general agreement that they could generate values from 0-364 or 1-365, do this 40 times and look for duplicates in their lists. The point of confusion came about in what data was to be collected from this simulation.

The question of interest was "What is the probability that at least two people have the same birthday in a room of 40 people?" In performing a simulation, the focus should then be on either the room does or doesn't contain at least one match. In fact, we don't care if there are multiple matches, just as long as there is one.

Students, on the other hand, wanted to focus on the number of matches that they had. Despite discussing the issue and explicitly directing them to record the number of trials that had matches, many students continued to count total matches in their 40 trials. It took some individual discussions to get everyone reporting the same data.

We generated 104 simulations and had 90 occurrences of two or more matches. This translated to a probability of 86.5%. I did wonder about one group's data as it showed a much smaller match rate than everyone else (9 out of 15). This one group accounted for almost half of the misses and although I did not analyze the results statistically to affirm consistency, the data appears suspect on the surface.

The 86.5% result is on par with the theoretical result of approximately 90%. In order for students to see this they were asked to calculate the theoretical probability. This calculation takes a lot of thought and can be quite involved. I like to present the challenge, not with the expectation that students will be able to determine how to calculate the probability but to help them make connections as to how basic probability rules can be applied in complex situations.

Students were stuck as to how to proceed. I asked two students to the front of the class and asked what is the probability that their two birthday's match? Students were quick to respond 1/365. I then asked another student to join the first two. I asked what is the probability that at least two of these students have the same birthday? This caused students to pause.

Say we have students A, B, and C. We then must consider that AB match, AC match, BC match, and ABC match. This starts to become a more complicated endeavor. If we add a fourth person, D, then there are even more combinations of birthday matches to consider. Continuing to 40 people seems beyond possible to calculate.

I have told the class before, and this problem is a great example, that when calculating probabilities and the calculations start to get messy and unmanageable, they should look at calculating the complement of probability.

Students are still not comfortable with the idea that P(A) + P(A^C) = 1. In this case P(A^C) is the probability that no one has the same birthday. This probability is a much more straightforward calculation than P(A), the probability that at least 2 people have the same birthday.

We proceeded sequential, the probability that the first two people do not have the same birthday is 364/365. Another person is added to the room. The probability this person's birthday does not match either birthday is 363/365. We continue, each time adding a person and subtracting an available day from the numerator. The result is
               P(A^C) = 364/365 x 363/365 x 362/365 x 361/365 x ... x 325/365

And the P(A) is simply 1 - P(A^C).

This calculation still takes some work. I showed students how to perform this calculation on their calculator using list operations and list math functions. Typically students do not know more than simple basics on how to use their graphing calculators. I want to show them that they can perform complex operations with just a little effort. At the same time, I probably need to rethink how to best go through this. I am thinking a step-by-step investigation worksheet may work better. Students would probably be more willing to follow the steps and answer checking questions along the way. There's always something new to add for the next semester.

I concluded with having students summarize their work and writing down two questions they still have about probability and simulations.

Visit the class summary for a student's perspective and to view the lesson slides.

Discrete Math - Day 14

Today we continued to work on the grouping problem from last class. Students did not have any new insights. I figured something like this would happen and took this opportunity to re-iterate that looking at smaller problems helps to get a sense of the problem dynamics.

I had three people stand at the front of the room. I asked the class how many ways could I break these three people into three different groups? After gaining clarification that reordering doesn't matter, students determined that there was only one way to form three groups.

I had another student join the other three up front. We now had four students and we wanted to form one group of 2 and two groups of 1. The question was how many ways could this be done? To emphasize what was going on, I had the students rearrange and pair themselves in different ways.

Students started to perform mental calculations and were thinking it was 6 different ways. Then some students questioned that result because the felt the number of people in the two groups of 1 were not being accounted for. There were a lot of questions and uncertainty coming out of the discussion.

I stepped in and asked students how many people were we choosing from to form the first group? Students responded that there were four. I then asked how many of these four were we choosing to form the first group. The response was 2. At this point students started to ask whether this meant that they could use ₄C₂. I asked students to calculate this value and they found it matched their count of 6.

I then asked what would happen with the two groups of 1? There again was some confusion but students finally resolved that those two groups were now fixed since there were only two people left to fill the two empty slots.

We then moved to five people and first looked at one group of 2, one group of 2, and one group of 1. Students started to pick up that there would be ₅C₂, ₃C₂, and 1 ways of selecting each group. There were then discussions about whether to multiply or add these values.

I related this to having 4 shirts/blouses and 3 pairs of pants. Students knew that they would have 12 outfits. When asked why they responded that was what they were told to do. At last someone said that every time they selected a pair of pants they would have four options for their top. This made sense to the class.

I then asked the class what that meant for the group situation. There was still some confusion. I asked how this problem was related to the pant-top situation. For the first group of 2, this group could be matched to other groups formed. I again used the students at the front of class to illustrate what was going on.

I asked students to sit down and now looked at forming groups with 6 people. The first option was one group of 3, one group of 2, and one group of 1. Students started to see that there would be ₆C₃, ₃C₂, and 1 option.

We then tackled three groups of 2. Here students thought there would be ₆C₂, ₄C₂, and 1. Someone asked if the last group should be calculated as ₂C₂, which is technically what is happening. I agreed that that best represents the situation. When students calculated the result was still one.

In the course of this discussion students noticed that ₅C₃ and ₅C₂ came out to the same value and wondered why. We took a more in-depth look at how permutations and combinations are calculated. I also related back to the connection between combinations and Pascal's triangle with the idea that the symmetry in the triangle will manifest itself as symmetry in values. Students could see how the formula for calculating combinations would lead to this symmetry.

For those of you who are not familiar with these formulas, we have

After going through these explorations, I had students again turn to the original problem from last class. Students found 792 ways to form groups of 7 from 12 people. They also found they could form groups of 3 from 5 people 10 different ways and concluded that the last group of 2 could only be formed in one way.

When I asked people what the final number was, students either calculated 803 or 7920. This was a situation where students had added, so there was still some confusion on this point. We discussed this again and students decided that 7920 was the correct result.

Some students said they started using the groups of 2 and 3 first, i.e. they calculated ₁₂C₂ and ₁₀C₃. They were concerned that they would get a different answer. We examined this and students started to see that the results all turned out the same and that the symmetries present in the calculations were coming into play.

I then had students work on a second grouping problem. For this problem everyone seemed to readily grasp what to do and the problem was quickly dispatched.

With a few minutes remaining, I previewed the ideas of counting and probability. Discrete probability will be the focus of the next couple of class.

Visit the class summary to read a student's perspective of the class and to view the lesson slides.

IPS - Day 13

Today we worked on an investigation problem. The question was whether or not a series of 60 digits was random or not? This gave students a break from calculating probabilities and a chance to work again with random number tables and generation.

The digits came from one of Danica McKellar's books (see below).



Students observed different characteristics of the number sequences. They were making statements about what they believe this showed. My comments back to them centered on asking how they could gather evidence (data) that would either support or contradict their claim. Some were able to immediately determine a course of action using random numbers. Others needed more prompting and guidance. All realized that they needed to compare the given data set to truly random digits to see if the characteristics they were observing were present in the truly random digits.

I had students examine the ratio of odd to even numbers, the number of repeated digits, palindromes of five numbers, sandwiches (a number squeezed between two numbers of the same value), and the distribution of digits present. Two groups felt the evidence showed the data was random, three thought the data was human generated, and two groups felt the results were inconclusive.

The most compelling evidence came from looking at the counts of digits present in the random numbers. The distributions looked much more variable in the truly random sets whereas the given data showed a much more uniform distribution.

One thought that came from a student was the issue that in an infinite long series of random numbers, the given values should appear. This is true. But at the heart of statistics is how frequently should you expect to see that sequence. Is it that one in a million chance occurrence or is it a one in five or one in ten chance?

We will continue to explore these ideas as the course progresses. This investigation was their first look at true statistical analysis and the students were engaged and interested in what their data was showing. It was a good close to the week.

Visit the class summary to read a student's perspective of the class.

Discrete Math - Day 13

Well, after yesterday's outstanding class, things were a little flat today. It sometimes takes a little more to get things going in a first period on a Friday morning.

We started out looking at how many ways five cards could be drawn from a deck. We compared this result with the possibilities when drawing four cards from yesterday. Students are amazed that there are over 2.5 million different hands that can be drawn. We used the full-house count we derived to show that probability of being dealt a full-house is very, very small. Although we haven't talked about discrete probability formally yet, this is our next topic, so I like to start introducing the connections between the counts we are performing and calculating probabilities.

We then looked at a grouping problem. This is where things really started to stall out. Students tackled the problem, but without much energy. The problem asks to split 12 people into one group of 2, one group of 3, and one group of 7. Many approached the idea that combinations were involved. The issue came in trying to figure out what to do with these values.

I used a messenger service and had one member of each group rotate through several other groups so they could get different perspectives on whether you should add or multiply values.

Students were still stuck but one student did look at a smaller problem. He looked at splitting 5 students into one group of 2 and one group of 3. He listed out the group of 2 options and saw that it was 10, which fit his use of ₅C₂. He also saw that there were 10 options for forming the group of 3. He couldn't figure out why they must be equal.

I had him share his work with the class and asked them why if there were 10 groups of 2 that we must have 10 groups of 3. Some students thought that it was because the 5 was an odd number. They had looked at ₆C₂ and ₆C₃ and these were different. I then posted up several combinations that I told them would be equal and asked why they must be equal.

     ₆C₄ = ₆C_{2     15}C₃ = ₁₅C_{12     7}C₁ = ₇C₆

I again asked students why these must be equal. Students observed that the groups being formed summed to the total but that's as far as they got.

At this point the class was ending. I asked students to think about what was happening in the formation of the groups, to connect the idea to the physical reality of what was happening as groups were formed.

Their homework was to try to figure out why these values must be equal and how this idea connected to the grouping problem that we were working on.

Hopefully some students will actually spend some time on the problem and come back with some ideas. This discussion will continue on Monday.

Visit the class summary for a student's perspective of the class and to view the lesson slides.