# Tag Archives: probability

## Probability, Polynomials, and Sicherman Dice

Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.

BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES

Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:

$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$

But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.

The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.

The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.

FROM POLYNOMIALS TO PROBABILITY

Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by

The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.

Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.

TAKING POLYNOMIALS FROM ONE DIE TO MANY

Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this.

By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker.

He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.

NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before.

Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following.

Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.

While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this.

In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.

With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition.

ROLLING DIFFERENT DICE SIMULTANEOUSLY

What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die.

The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.

A BEAUTIFUL EXTENSION–SICHERMAN DICE

My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get

$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.

Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives

Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?

Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what?

Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.

What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business.

SOLUTION

Here are my insights over time:

1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values.

2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor.

3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die.

That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.

One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice.

If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways.

The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.

Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is

corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.

CONCLUSION:

There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.

Enjoy.

## Birthdays, CAS, Probability, and Student Creativity

Many readers are familiar with the very counter-intuitive Birthday Problem:

It is always fun to be in a group when two people suddenly discover that they share a birthday.  Should we be surprised when this happens?  Asked a different way, how large a group of randomly selected people is required to have at least a 50% probability of having a birthday match within the group?

I posed this question to both of my sections of AP Statistics in the first week of school this year.  In a quick poll, one section had a birthday match–two students who had taken classes together for a few years without even realizing what they had in common.  Was I lucky, or was this a commonplace occurrence?

Intrigue over this question motivated our early study of probability.  The remainder of this post follows what I believe is the traditional approach to the problem, supplemented by the computational power of a computer algebra system (CAS)–the TI Nspire CX CAS–available on each of my students’ laptops.

Initial Attempt:

Their first try at a solution was direct.  The difficulty was the number of ways a common birthday could occur.  After establishing that we wanted any common birthday to count as a match and not just an a priori specific birthday, we tried to find the number of ways birthday matches could happen for different sized groups.  Starting small, they reasoned that

• If there were 2 people in a room, there was only 1 possible birthday connection.
• If there were 3 people (A, B, and C), there were 4 possible birthday connections–three pairs (A-B, A-C, and B-C) and one triple (A-B-C).
• For four people (A, B, C, and D), they realized they had to look for pair, triple, and quad connections.  The latter two were easiest:  one quad (A-B-C-D) and four triples (A-B-C, A-B-D, A-C-D, and B-C-D).  For the pairs, we considered the problem as four points and looked for all the ways we could create segments.  That gave (A-B, A-C, A-D, B-C, B-D, and C-D).  These could also occur as double pairs in three ways (A-B & C-D, A-C & B-D, and A-D & B-C).  All together, this made 1+4+6+3=14 ways.

This required lots of support from me and was becoming VERY COMPLICATED VERY QUICKLY.  Two people had 1 connection, 3 people had 4 connections, and 4 people had 14 connections.  Tracking all of the possible connections as the group size expanded–and especially not losing track of any possibilities–was making this approach difficult.  This created a perfect opportunity to use complement probabilities.

While there were MANY ways to have a shared birthday, for every sized group, there is one and only one way to not have any shared birthdays–they all had to be different.  And computing a probability for a single possibility was a much simpler task.

We imagined an empty room with random people entering one at a time.  The first person entering could have any birthday without matching anyone, so $P \left( \text{no match with 1 person} \right) = \frac{365}{365}$ .  When the second person entered, there were 364 unchosen birthdays remaining, giving $P \left( \text{no match with 2 people} \right) = \frac{365}{365} \cdot \frac{364}{365}$, and $P \left( \text{no match with 3 people} \right) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$.  And the complements to each of these are the probabilities we sought:

$P \left( \text{birthday match with 1 person} \right) = 1- \frac{365}{365} = 0$
$P \left( \text{birthday match with 2 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \approx 0.002740$
$P \left( \text{birthday match with 3 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 0.008204$.

The probabilities were small, but with persistent data entry from a few classmates, they found that the 50% threshold was reached with 23 people.

The hard work was finished, but some wanted to find an easier way to compute the solution.  A few students noticed that the numerator looked like the start of a factorial and revised the equation:

$\begin{matrix} \displaystyle P \left( \text{birthday match with n people} \right ) & = & 1- \frac{365}{365} \cdot \frac{364}{365} \dots \frac{(366-n)}{365} \\ \\ & = & 1- \frac{365 \cdot 364 \dots (366-n)}{365^n} \\ \\ & = & 1- \frac{365\cdot 364 \dots (366-n)\cdot (366-n-1)!}{365^n \cdot (366-n-1)!} \\ \\ & = & 1- \frac{365!}{365^n \cdot (365-n)!} \end{matrix}$

It was much simpler to plug in values to this simplified equation, confirming the earlier result.

Not everyone saw the “complete the factorial” manipulation, but one noticed in the first solution the linear pattern in the numerators of the probability fractions.  While it was easy enough to write a formula for the fractions, he didn’t know an easy way to multiply all the fractions together.  He had experience with Sigma Notation for sums, so I introduced him to Pi Notation–it works exactly the same as Sigma Notation, except Pi multiplies the individual terms instead of adding them.  On the TI-Nspire, the Pi Notation command is available in the template menu or under the calculus menu.

Conclusion:

I really like two things about this problem:  the extremely counterintuitive result (just 23 people gives a 50% chance of a birthday match) and discovering the multiple ways you could determine the solution.  Between student pattern recognition and my support in formalizing computation suggestions, students learned that translating different recognized patterns into mathematics symbols, supported by technology, can provide different equally valid ways to solve a problem.

Now I can answer the question I posed about the likelihood of me finding a birthday match among my two statistics classes.  The two sections have 15 and 21 students, respectively.  The probability of having at least one match is the complement of not having any matches.  Using the Pi Notation version of the solution gives

I wasn’t guaranteed a match, but the 58.4% probability gave me a decent chance of having a nice punch line to start the class.  It worked pretty well this time!

Extension:

My students are currently working on their first project, determining a way to simulate groups of people entering a room with randomly determined birthdays to see if the 23 person theoretical threshold bears out with experimental results.

## Monty Hall Continued

In my recent post describing a Monty Hall activity in my AP Statistics class, I shared an amazingly crystal-clear explanation of how one of my new students conceived of the solution:

If your strategy is staying, what’s your chance of winning?  You’d have to miraculously pick the money on the first shot, which is a 1/3 chance.  But if your strategy is switching, you’d have to pick a goat on the first shot.  Then that’s a 2/3 chance of winning.

Then I got a good follow-up question from @SteveWyborney on Twitter:

Returning to my student’s conclusion about the 3-door version of the problem, she said,

The fact that there are TWO goats actually can help you, which is counterintuitive on first glance.

Extending her insight and expanding the problem to any number of doors, including Steve’s proposed 1,000,000 doors, the more goats one adds to the problem statement, the more likely it becomes to win the treasure with a switching doors strategy.  This is very counterintuitive, I think.

For Steve’s formulation, only 1 initial guess from the 1,000,000 possible doors would have selected the treasure–the additional goats seem to diminish one’s hopes of ever finding the prize.  Each of the other 999,999 initial doors would have chosen a goat.  So if 999,998 goat-doors then are opened until all that remains is the original door and one other, the contestant would win by not switching doors iff the prize was initially randomly selected, giving P(win by staying) = 1/1000000.  The probability of winning with the switching strategy is the complement, 999999/1000000.

IN RETROSPECT:

My student’s solution statement reminds me on one hand how critically important it is for teachers to always listen to and celebrate their students’ clever new insights and questions, many possessing depth beyond what students realize.

The solution reminds me of a several variations on “Everything is obvious in retrospect.”  I once read an even better version but can’t track down the exact wording.  A crude paraphrasing is

The more profound a discovery or insight, the more obvious it appears after.

I’d love a lead from anyone with the original wording.

REALLY COOL FOOTNOTE:

Adding to the mystique of this problem, I read in the Wikipedia description that even the great problem poser and solver Paul Erdős didn’t believe the solution until he saw a computer simulation result detailing the solution.

## Probability and Monty Hall

I’m teaching AP Statistics for the first time this year, and my first week just ended.  I’ve taught statistics as portions of other secondary math courses and as a semester-long community college class, but never under the “AP” moniker.  The first week was a blast.

To connect even the very beginning of the course to previous knowledge of all of my students, I decided to start the year with a probability unit.  For an early class activity, I played the classic Monte Hall game with the classes.  Some readers will recall the rules, but here they are just in case you don’t know them.

1. A contestant faces three closed doors.  Behind one is a new car. There is a goat behind each of the other two.
2. The contestant chooses one of the doors and announces her choice.
3. The game show host then opens one of the other two doors to reveal a goat.
4. Now the contestant has a choice to make.  Should she
1. Always stay with the door she initially chose, or
2. Always change to the remaining unopened door, or
3. Flip a coin to choose which door because the problem essentially has become a 50-50 chance of pure luck.

Historically, many people (including many very highly educated, degree flaunting PhDs) intuit the solution to be “pure luck”.  After all, don’t you have just two doors to choose from at the end?

In one class this week, I tried a few simulations before I posed the question about strategy.  In the other, I posed the question of strategy before any simulations.  In the end, very few students intuitively believed that staying was a good strategy, with the remainder more or less equally split between the “switch” and “pure luck” options.  I suspect the greater number of “switch” believers (and dearth of stays) may have been because of earlier exposure to the problem.

I ran my class simulation this way:

• Students split into pairs (one class had a single group of 3).
• One student was the host and secretly recorded a door number.
• The class decided in advance to always follow the “shift strategy”.  [Ultimately, following either stay or switch is irrelevant, but having all groups follow the same strategy gives you the same data in the end.]
• The contestant then chose a door, the host announced an open door, and the contestant switched doors.
• The host then declared a win or loss bast on his initial door choice in step two.
• Each group repeated this 10 times and reported their final number of wins to the entire class.
• This accomplished a reasonably large number of trials from the entire class in a very short time via division of labor.  Because they chose the shift strategy, my two classes ultimately reported 58% and 68% winning percentages.

Curiously, the class that had the 58% percentage had one group with just 1 win out of 10 and another winning only 4 of 10. It also had a group that reported winning 10 of 10.  Strange, but even with the low, unexpected probabilities, the long-run behavior from all groups still led to a plurality winning percentage for switching.

Here’s a verbatim explanation from one of my students written after class for why switching is the winning strategy.  It’s perhaps the cleanest reason I’ve ever heard.

The faster, logical explanation would be: if your strategy is staying, what’s your chance of winning?  You’d have to miraculously pick the money on the first shot, which is a 1/3 chance.  But if your strategy is switching, you’d have to pick a goat on the first shot.  Then that’s a 2/3 chance of winning.  In a sense, the fact that there are TWO goats actually can help you, which is counterintuitive on first glance.

Engaging students hands-on in the experiment made for a phenomenal pair of classes and discussions. While many left still a bit disturbed that the answer wasn’t 50-50, this was a spectacular introduction to simulations, conditional probability, and cool conversations about the inevitability of streaks in chance events.

For those who are interested, here’s another good YouTube demonstration & explanation.

## Multiplication Practice Plus Creativity

I hope this post is particularly helpful for parents and teachers of elementary school children.  Through my Twitter network last week I found via @Maths_Master‘s Great Maths Teaching Ideas ‘blog a 2010 post summarizing Dan Finkel’s Damult dice game. Recognizing that “practicing times tables can be unmotivated and boring for kids,” Damult is an attempt to make learning elementary multiplication facts more entertaining. I offer some game variations and strategies following a description of the game.

Here’s Dan’s game:

image via Wikipedia

Each player takes turns rolling 3 dice. First to break 200 (or 500, etc.) wins. On your turn, you get to choose two dice to add together, then you multiply the sum by the final die. That’s your score for that turn.

Simple; no bells, no whistles. For example, I roll a 3, a 4, and a 6 on my turn. I could either do (3+4) times 6 for 42 points, OR (3+6) times 4 for 36 points, OR (4+6) times 3 for 30 points. I’ll take the 42 points.

I spent some time playing this with kids the other day and I saw that (1) it was genuinely fun, and (2) it gives you almost all the multiplication practice you could ask for. In fact, it gives even more, because the choice of which dice to add and which to multiply reveals some interesting structure of numbers. Seriously, get a kid hooked on this game, and it’s the equivalent of dozens or hundreds of times table practice sheets.

It’s a fun activity idea by itself.  Damult combines a bit of luck and memory, and rewards the ability to recall multiplication facts.  As an added bonus, it requires players to be able to manipulate objects in their heads–how many different ways can the three given dice be manipulated in summation stage to create unique products? How can a player ensure that she has found the biggest product for her score?  Try the game!

CONNECTIONS AND EXTENSIONS:

This is a great opportunity for parents to engage with their  children as they learn multiplication facts.  Parents and teachers could play along, or the learner might be the only player, talking out loud so that the teacher or parent can “hear the thinking.”

I love that the game completely randomizes the multiplication tables.  This significantly enhances recall as memory is not tied to particular patterns or positions on fact pages. Players must adapt to each random roll.

In any variation, there obviously should be a discussion among all players about what products were found to confirm the results. Make the game more formative or more competitive, depending on the experience level of the players.  In more competitive variations with experienced learners, if a product was miscalculated and claimed, you might decide that no score should be recorded for that round.

If you’re guiding someone on this it is critical that you DO NOT give answers.  Students need to explore, hypothesize, discover errors, learn how to communicate their conclusions in clear and concise language, and to learn how to defend their findings while also learning how to admit flaws in their reasoning when faced with contradicting data.  Experimenting and discovery is always deeper, richer,and more long-lasting than just being told.  Remember the Chinese Proverb: “I hear and I forget. I see and I remember. I do and I understand.” Always seek understanding.

The first comments on Dan’s post noted that while one player was summing and multiplying, the other player(s) were largely disengaged. Also, the game could drag on as unconfident players tried to make sure  they had explored every possibility.  To address that and several other possibilities, I offer the following Damult variations.  Some more complex variations are toward the end. Read on!

Finally, if you’ve read my ‘blog much, you know that I’m a huge fan of leveraging technology for math learning, but this is one of those situations where I think you should 100% unplug. To learn multiplication facts is to learn some of the basic grammar and vocabulary that makes the language of mathematics work.  You simply can’t communicate mathematically with an underlying awareness of how the structure of the language works.

GAME VARIATIONS:

Variation 1: Adding a timer to the game could cure the slow-down issue. Depending on the age of the child and his/her familiarity with multiplication, the timer can be longer or shorter.  If the skill levels of the players are unequal, make the timer unequal.  (I love the adage, “Fair is seldom equal, and equal is seldom fair.”)

Variation 2: Why must only one player be active? The players could take turns rolling the dice while both record scores based on what they find.  If a particular combination was not noticed by one player, that player doesn’t get to consider it for his/her score.

Variation 3 – As an aside, notice that Dan implicitly claims there are only 3 possible sums from a 3-dice roll. Will that always be the case?  Can you convince someone why your solution is correct?

(For 3 dice the maximum number of possible sums is 3. When and why would there be fewer products?)

Variation 4 – How many multiplication facts are possible using only 3 dice?

This would be a great number sense exploration.  Some may try it by gathering lots of data, others may have more sophisticated reasoning.  I suggest that you or your students hypothesize an answer first along with some reason why you think your hypothesis is correct.  Different answers are OK, and you can always revise your hypotheses if you get evidence leaning in another direction. No matter what, have fun exploring and learning.

(Middle School extension: Damult creates products of axb where a can be any integer 1 – 6 and b can be any integer 2 – 12.  That gives 66 different products if you count different arrangements (3×4 and 4×3) as different products. Can you or your student see why? How many outcomes are possible if you look only at the product result and not at the factors which created it?)

Variation 5 – After discovering or just using the answer to the last variation, you could use a table of multiplication facts and see how quickly different facts and be “discovered” from rolls of the dice.  After rolling 3 dice, mark off all multiplication facts you can using the sum-then-multiply combination rules posed at the beginning.  This might be a fun way for early learners to familiarize themselves with multiplication patterns.

NOTE: If you play variations 4 or 5 as a game, you’ll likely want (or need) to stop before all possibilities are found.  Some (eg, 6×12 and 1×2) will be pretty uncommon from dice rolls.

Variation 6 – You could make a Bingo-like or a 4 or 5-in-a-row game.  The first person to mark off a certain number of facts or the first to get a certain number in a row would be a winner.

Variation 6 – If you try the last few variations, you’ll see that some products occur much more frequently from the dice rolls than others.  This could be used to introduce probability. Which products are more likely and why?

As an example, I suspect 3×7 could happen six times more often 1×2.  Can you convince yourself why 3×7 is so much more likely?  Can you see why 3×7 is exactly six times more likely than 1×2?

Variation 7 – Why restrict yourself to 3 dice? When just starting out, using more than 3 dice would definitely be a frustration factor, but once you’ve got a good grip on the game, consider rolling 4 dice and allow players to multiply the sum of any 2 or 3 of the dice by the sum of the remaining dice.

By my computation, using 4 dice means there are up to 7 possible combinations in a given roll.  Can you prove that? Being able to consistently find them all is likely to be a very difficult challenge, but it is a phenomenal and early opportunity to stretch a young person’s mind into considering multiple outcomes and reliable ways to guarantee that you’ve considered all possibilities.

Variation 8 – Why go for maximum products and being the first to get to 200 or 500 points?  Why not try for a low score (like golf), seeking minimum products  and being the last to exceed 100 or 200?

Variation 9 – Stealthy Calculus:  OK, my analysis on this one goes way deeper than is necessary to play the game, but sometimes knowing more than is necessary can give insights and can help you lead others toward developing “math sense”–a truly invaluable skill.

LOW LEVEL – After you’ve played this a few times, ask the player(s) if there is some strategy that could be used to guarantee the biggest (or smallest) possible product for any roll.  This could be a great mathematical experiment for which the solutions are not at all intuitive, I think.  Some might figure it out quickly and others might need to gather lots of data, comparing products from lots of rolls before distilling the relationship.

If you’re guiding someone on this it is critical that you DO NOT give the answer.  Students need to explore, hypothesize, learn how to communicate their conclusions in clear and concise language, and to learn how to defend their findings while also learning how to admit flaws in their reasoning when faced with contradicting data.  If you don’t know the answer, stop reading now and figure it out for yourself. I provide an answer in the next paragraphs, but experimenting and discovery is always deeper, richer,and more long-lasting than just being told.  Remember the Chinese Proverb: “I hear and I forget. I see and I remember. I do and I understand.” Always seek understanding.

MUCH HIGHER LEVEL – As a calculus teacher, the very first fact that struck me was Damult’s implied goal: Getting the largest possible product from any roll of three dice.  That’s an optimization problem, and I knew from calculus that the greatest possible product of two numbers whose sum was constant happens when the two numbers are as close as possible to being equal.  Likewise, the smallest possible product happens when the two factors are as far apart as possible.  (If you recall some calculus of derivatives, I encourage you to prove these for yourself.  If anyone asks, I could write a future post with the proof.)

In Dan’s initial example above in which 3, 4, and 6 were rolled, I stopped reading after the first sentence of paragraph 2 (pausing to think and draw your own conclusions is a great habit of the mind) for a few moments as I thought, “I know 3+4 and 6 are as close to equivalent as I can get, so 7*6=42 is the greatest possible product.”  I didn’t even look at the other possibilities, I knew they were less. This fact was established (unnecessarily for me) in the end of the paragraph.

Without calculus, I propose students try making tables of their data.  They’ll have up to three unique products (Variation 3) and will need to explore the data before hopefully discovering the relationship. If a young person doesn’t discover the relationship, Don’t tell him/her! it is far better to leave a question as unanswered to think on and answer another day than to have a relationship given unearned.  Value comes from effort and discovery. Don’t cheat young learners out of that experience or lesson.

Conclusion: Don’t just play a game. Be creative! Strategize! Encourage young ones not just to play, but to play well. Children are quite creative in free play as they continually make new and adapt old “rules”.  Why should intellectual play be any different?  I’d love to see what variations others discover or have to offer.

## Binomial Probability and CAS

I posted previously about a year ago an idea for using CAS in a statistics course with probability.  I’ve finally had an opportunity to use it with students in my senior one-semester statistics course over the last few weeks, so I thought I’d share some refinements.  To demonstrate the mathematics, I’ll use the following problem situation.

Assume in a given country that women represent 40% of the total work force.  A company in that country has 10 employees, only 2 of which are women.
1) What is the probability that by pure chance a 10-employee company in that country might employ exactly 2 women?
2) What is the probability that by pure chance a 10-employee company in that country might employ 2 or fewer women?

Over a decade ago, I used binomial probability situations like this as an application of polynomial expansions, tapping Pascal’s Triangle and combinatorics to find the number of ways a group of exactly 2 women can appear in a total group size of 10.  Historically, I encouraged students to approach this problem by defining m=men and w=women and expand $(m+w)^{10}$ where the exponent was the number of employees, or more generally, the number of trials.  Because question 1 asks about the probability of exactly 2 women, I was interested in the specific term in the binomial expansion that contained $w^2$.  Whether you use Pascal’s Triangle or combinations, that term is $45w^2m^8$.  Substituting in given percentages of women and men in the workforce, $P(w)=0.4$ and $P(m)=0.6$, answers the first question.  I used a TI-nSpire to determine that there is a 12.1% chance of this.

That was 10-20 years ago and I hadn’t taught a statistics course in a very long time.  I suspect most statistics classes using TI-nSpires (CAS or non-CAS) today use the binompdf command to get this probability.

The slight differences in the input parameters determine whether you get the probability of the single event or the probabilities for all of the events in the entire sample space.  The challenge for the latter is remembering that the order of the probabilities starts at 0 occurrences of the event whose probability is defined by the second parameter.  Counting over carefully from the correct end of the sequence gives the desired probability.

With my exploration of CAS in the classroom over the past decade, I saw this problem very differently when I posted last year.  The binompdf command works well, but you need to remember what the outputs mean.  The earlier algebra does this, but it is clearly more cumbersome.  Together, all of this screams (IMO) for a CAS.  A CAS could enable me to see the number of ways each event in the sample space could occur.  The TI-nSpire CAS‘s output using an expand command follows.

The cool part is that all 11 terms in this expansion appear simultaneously.  It would be nice if I could see all of the terms at once, but a little scrolling leads to the highlighted term which could then be evaluated using a substitute command.

The insight from my previous post was that when expanding binomials, any coefficients of the individual terms “received” the same exponents as the individual variables in the expansion.  With that in mind, I repeated the expansion.

The resulting polynomial now shows all the possible combinations of men and women, but now each coefficient is the probability of its corresponding event.  In other words, in a single command this approach defines the entire probability distribution!  The highlighted portion above shows the answer to question 1 in a single step.

Last week one of my students reminded me that TI-nSpire CAS variables need not be restricted to a single character.  Some didn’t like the extra typing, but others really liked the fully descriptive output.

To answer question 2, TI-nSpire users could add up the individual binompdf outputs -OR- use a binomcdf command.

This gets the answer quickly, but suffers somewhat from the lack of descriptives noted earlier.  Some of my students this year preferred to copy the binomial expansion terms from the CAS expand command results above, delete the variable terms, and sum the results.  Then one suggested a cool way around the somewhat cumbersome algebra would be to substitute 1s for both variables.

CONCLUSION:  I’ve loved the way my students have developed a very organic understanding of binomial probabilities over this last unit.  They are using technology as a scaffold to support cumbersome, repetitive computations and have enhanced in a few directions my initial presentations of optional ways to incorporate CAS.  This is technology serving its appropriate role as a supporter of student learning.

OTHER CAS:  I focused on the TI-nSpire CAS for the examples above because that is the technology is my students have.  Obviously any CAS system would do.  For a free, Web-based CAS system, I always investigate what Wolfram Alpha has to offer.  Surprisingly, it didn’t deal well with the expanded variable names in $(0.4women+0.6men)^{10}$.  Perhaps I could have used a syntax variation, but what to do wasn’t intuitive, so I simplified the variables here to get

Huge Pro:  The entire probability distribution with its descriptors is shown.
Very minor Con:  Variables aren’t as fully readable as with the fully expanded variables on the nSpire CAS.

## Running into Math

Here’s a real-world math problem I just found.

For the last two years, the AJC Peachtree Road Race in Atlanta, the “World’s Largest 10K” (it happens every July 4th), has been using a lottery system to determine which non-invited runners get race numbers.

To accommodate those who would like to participate in the AJC Peachtree Road Race with their family and friends, the lottery registration system allows groups of up to 10 people to enter the lottery as a “Group”.  During the selection process, if a “Group” is selected everyone in the group will receive an entry.  If a “Group” is not selected through the lottery, no one in the group will receive entry into the event.  Those entering the lottery as a group have an equal chance of getting into the event as those entering as individuals (source, emphasis added).

Assume a full group of 10 runners enters as a group.  If any 1 runner in the group is selected in the lottery, every runner in the group gets a race number even if no one else in the group is chosen.  On the surface, this seems like it ought to give a runner a better chance of getting a lottery number if entering as a group.  But … the organizers claim that individuals seeking race numbers have an equal probability of getting into the race whether entering solo or in a group.  So how do they do it?

I didn’t find this problem at the right point in my class’ curriculum sequence this year (I get that I raise lots of rightfully debatable curriculum & teaching issues here), but maybe it will work for one of you.  Even so, I’m trying to create a 10-15 minute gap in an upcoming class to give this problem as a “cool (or real) math moment” that I have from time to time in my courses.  If I can get some student results, I’ll post them here.  I’ll provide links/posts from here to any pages or tweets that tackle this.  Enjoy.

## Probability musings

From Futility Closet:

A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?

A solution appears on the page, but it doesn’t suggest how it was determined or if the solution is unique.  I’ll provide that below plus an extension.

Assume there were w white balls originally, then the number of ways to draw two white is $\displaystyle \frac{w*(w-1)}{2!}$, two black is $\displaystyle \frac{(16-w)*(15-w)}{2!}$, and one of each is $\displaystyle \frac{w*(16-w)}{1!1!}$.  As the likelihood of drawing one of each color is to be the same as drawing only one color,

$\displaystyle \frac{w*(w-1)}{2}+ \frac{(16-w)*(15-w)}{2}= w*(16-w)$

This is a quadratic equation with solutions $w=\{6,10\}$, verifying Futility Closet’s solution, and establishing those solutions as the only ones possible.

Extension:  For what total number(s) of white and black billiard balls is the likelihood  for two drawn balls to be one of color the same as them being one of each?

Let n be the total number of balls.  Repeating the procedure above, the number of ways to draw two white is $\displaystyle \frac{w*(w-1)}{2!}$, two black is $\displaystyle \frac{(n-w)*((n-1)-w)}{2!}$, and one of each is $\displaystyle \frac{w*(n-w)}{1!1!}$.  Because the likelihoods are still equal,

$\displaystyle \frac{w*(w-1)}{2}+ \frac{(n-w)*(n-1-w)}{2}= w*(n-w)$

I solved this with my CAS (below) to get $\displaystyle w=\frac{n}{2}\pm\frac{\sqrt{n}}{2}$.

I noticed two things from this solution.

1. Because w must be a whole number, the denominators require n to be an even number and to guarantee that, the square root means n must be 4 times any perfect square.
2. For any value of n, there are only two solutions for any of these situations, a point that could have been obvious before seeing the solution if you noticed the equation was quadratic, but something I suspect many (most?) students would initially miss.

The original problem gave $n=4*4=16$ with solutions $\displaystyle w=\frac{16}{2}\pm\frac{\sqrt{16}}{2}=\{10,6\}$ confirming the patterns predicted by the general solution.  For another possible solution, let $n=4*25=100$ with solutions $\displaystyle w=\frac{100}{2}\pm\frac{\sqrt{100}}{2}=\{55,45\}$.

A further extension of the general solution would be to show that as $n\rightarrow\infty$, the likelihoods become essentially equivalent for drawing either two different colors or both the same color.

Pedagogical side note:  Both equations in this problem are quadratic and could be solved “by hand”.  But, why?  The point of these questions focuses on probability and not symbolic manipulation, so I argue that use of a CAS is entirely appropriate and perhaps should be the ideal approach in this problem. Keep the focus appropriately on the problem.  If a teacher needs to assess a student’s ability to solve quadratic equations, that should be done in a different question, not here.

## CAS and Probability

A recent thread on the TI-Nspire Google Group asked about uses of CAS in probability.  There are so many possibilities–one uses CAS for binomial probabilities. For example, what’s the probability of getting exactly 3 heads in 5 tosses of a fair coin?  A CAS approach expands $(\frac{1}{2}h+\frac{1}{2}t)^5$.  The $\frac{1}{2}$ coefficients of h (heads) and t (tails) are the respective probabilities of each outcome and the exponent is the number of trials. Obviously, there’s lots to unpack here to prevent this from being a black box tool, but note the power of the output.  The three heads event is represented by the $\frac{5h^3t^2}{16}$ term, and the coefficient is the desired probability, $\frac{5}{16}$. Early in my career, I taught this by expanding $(h+t)^5$, picking the appropriate term, and substituting for each variable its probability.  The great power of this approach is that the meaning of each fractional term remains by the presence of the variables while you gain the answers simultaneously.  Also note that while the problem asked only for the probability of exactly 3 heads, the CAS output gives the result of every possibility in the entire sample space. Variations 1)  What is the probability of 3 heads in five tosses if the coin was bent in a way that $P(h)=0.4$?  Adjust the coefficients to get 0.2304. 2)  The technique is not restricted binomial probabilities.  If there are three possible outcomes (a, b, and c) where $P(a)=0.4$, $P(b)=0.35$, and $P(c)=0.25$, then what is the probability of exactly 2 as in 3 trials? Because only 2 outcomes are specified for the 3 trials, the third could be either b or c.  These two outcomes are highlighted above, giving a total probability of 0.288. While these values certainly could be computed without a CAS, the point here is to use technology for computations, freeing users to think.