[EM_20160204PBIE] “Hillarious” Mathematics of Politics?

Problem

Six coins are tossed to decide a result for either “C” or “S”.  Assuming that the coins are fair, and that the results of the tosses are independent, calculate the probability that all the tosses are in favour of “C”.

Introduction

     Hot in recent news is the story of the purported six coin tosses that were needed to determine certain county delegates in the race between Hillary Clinton and Bernie Sandersin the state of Iowa.  All six coin tosses were in favour of Hillary Clinton, and the result is so improbable that some people said it was “hillarious”.
     How is the probability calculated?  This is an example of mathematics in real life events that has the potential to affect the United States of America, and the whole world (including Singapore).

Solution

Discussion

     In order to make the calculation, we make two assumptions: 

          (1) that the coins were fair, and

          (2) that the coin toss results are independent.
     So, what does it mean that the coins are “fair”?  It means that the probability of getting a “heads” is the same as the probability of getting a “tails”, which means ½ for each.
     Coin toss results can be “heads” or “tails”.  These are examples of events.  An event is something that can happen or not happen, and we associate a probability with it.  The probability is a number that indicates how likely the event happens.  It is between  0  and  1  inclusive.  Zero probability means a practically impossible event.  A probability of  1  means a practically certain event.  [The reason for me using the word “practically” is technical, which I shall not discuss.]  If the events do not affect one another (i.e. in our case, the coin tosses are not affected by the other coin tosses) then the events are said to be independent.  If the events are independent, then we can simply multiply the individual probabilites together, as above.  If the events are dependent, the calculation would be more complicated.

     So are the coin toss results valid?  I do not know.  All I can say is: improbable does not mean impossible.  Mathematics cannot tell whether the above assumptions (1) and (2) are correct.  But at least I “lay all the cards on the table”, so that astute students of probability know the basis of these calculations.  It is up to you to decide, but at least you would have made a mathematically-informed decision.  There could be other twists to the story, which is beyond the scope of this article.  This is one of the reasons why you need to learn mathematics carefully and think critically, whether or not you would become a mathematician,engineer, teacher or have a mathematics-intensive career.

Heuristics Used

H08. Make suppositions

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)

* other syllabuses that involve probability

* anybody in the whole wide world!

H2Maths VJC/2007/P2/Q9 The Power of Rephrasing

Introduction

     Today we discuss a question taken from one of the top junior colleges in Singapore’s 2007 preliminary examinations (final school internal examinations before the actual A-levels).  The seemingly difficult problems dissolve quickly using the right heuristics (rules-of-thumb / guidelines / problem-solving tactics).  Once the correct approach is determined, the calculation is simply a matter pressing buttons on the graphing calculator (e.g. TI-84 Plus or Casio fx-9860G).  This question can be tackled using heuristics.

     Try to guess what heuristic(s) will be useful in solving the above problem.  In a typical examination question, they will not tell you what topic or concepts are being tested in that question.  Try to determine what concepts are involved.

     Let’s tackle the question part by part.  As usual, we use the 5 step problem-solving process with metacognition (self-prompting, self-monitoring).

Part (i)

 

Part (i) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     That seems to be something to do with probability …

Do you understand the question?  Can you put it in your own words?

When Ai Wan (爱玩 “loves to play”?) rolls the die for the eight time, he gets exactly three ‘6’ to win the prize.  What are the chances of this event happening?

Part (i) Stage 2:  Planning the approach

Is there a heuristic you can use?

     How about “9. Restate the problem in another way” …

So how do you rephrase the question?

     “win on 8^th roll” means “two ‘sixes’ on the first 7 rolls” and “a ‘six’ on the 8^th roll”.

How will you proceed?

Break down the problem into parts (“11. Solve part of the problem”).

     (1) find probability of “two ‘sixes’ on the first 7 rolls”

     (2) find probability of “a ‘six’ on the 8^th roll” (this is easy.  Obviously  ¹/₆)

Then multiply them together, since these are independent.

How do you write out probability of “two ‘sixes’ on the first 7 rolls” in symbols?

P(X = 2)

Think backwards:  What is X?  How do you define it?

X  is a random variable.  It is the number of ‘sixes’ in the first 7 rolls.

What distribution does it follow?  How do you know?

It follows a Binomial Distribution: Because there are only two outcomes: either you get a ‘six’, or you don’t.  There is a fixed number of trials and these are independent, giving a constant probability. 

So what’s the number of trials (n)?

There are 7 rolls, so seven.  Each roll is a trial.

What is the probability of “success” (p)?

One out of six.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

This is just probability for a single value of  X, so it’s a p.d.f.  It can be found from the graphing calculator.  [ Or use the formula ⁿC_x p^x q^n–x = ⁷C₂ (¹/₆)²(⁵/₆)⁵ ]

Part (i) Stage 3:  Executing the plan

Part (i) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  It’s quite small, which tallies with my intuition, as getting three sixes in 8 rolls is highly unlikely.  If I get an answer like 0.4 something, I might smell a rat.  If I get a probability that is less than 0 or more than 1, then I know for sure it’s definitely wrong.

Part (ii)

Remark

     This problem is related to the Geometric Distribution, which is not in the syllabus.  Nevertheless it is solvable using the knowledge contained in the syllabus, so it is within the student’s reach.

Part (ii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     probably probability again … maybe Binomial or something related

Do you understand the question?  Can you put it in your own words?

Ai Ying (爱赢 “loves to win”?) already has got 2 ‘6’ and 2 ‘1’.  That means she needs just one more ‘6’.  She needs to throw exactly four more times (no more and no less).  What are the chances of this happening?

Part (ii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way” … but the problem still seems complicated because of the initial conditions (two “sixes” and two “ones” ) …

Is there a way to simplify the problem or another way to think about it?  Is there a feature about the situation … ?

     Ah!  Since all the trials are independent, what happens in the past does not affect the future.  This is the memoryless or forgetfulness property of the (Bernoulli) trials.  That means I don’t need to worry about the two “sixes” and two “ones”.  I can forget the past!  It is as though I can just start from a clean slate!

So what does it mean to say “Ai Ying needs to throw exactly four more times”?

     It means after throwing three more times, she does not get any six, and then she gets a six on the next throw.

How do you write this out mathematically?

P(S’, S’, S’, S)

What do you mean by S?  What is S’ ?

S  is the event of getting a ‘6’ in a roll of the die.  S’  means not getting a ‘6’.

Now, how do you proceed?

The chain of events can be broken down.  (“Split the problem into smaller parts”)

P(S) = ¹/₆.  P(S’) = 1 – ¹/₆ = ⁵/₆.  Since the four events are independent, I can just multiply their probabilities  all together!  This is easy!

Part (ii) Stage 3:  Executing the plan

 

Part (ii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  Again the answer is quite small, which tallies with my intuition.

Part (iii)

Part (iii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     Probability, Binomial Distribution and … er … rephrasing?

Do you understand the question?  Can you put it in your own words?

Ai Du (爱赢 “loves to win”?) cannot win in eight rolls, so he needs to roll some more.  What are the chances of this happening?

Part (iii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way”

So how do you restate the question?

 “requires more than eight rolls of the die to win a prize” means “after 8 rolls, he does not get three ‘sixes’”.

Which means?

Which means “after 8 rolls, he gets only 2 or less ‘sixes’”.

How do you write the required probability mathematically?

P(D < 2)

Think backwards:  What do you mean by D?

D  is a random variable that counts the number of ‘sixes’ within 8 rolls.

What distribution does it follow?  Why?

Binomial Distribution, again.  Same reasons as in part (i).

So what’s the number of trials (n)?  What is the probability of “success” (p)?

n = 8,  p = ¹/₆.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

Because it is a “<” probability, it’s a c.d.f, which can be found from the graphing calculator, or checking tables.  [ Calculation by hand is possible, but tedious.  Also, there is a recurrence formula that can help a bit, but this is out of syllabus. ]

Part (iii) Stage 3:  Executing the plan

 

Part (iii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  The answer is high, which tallies with my intuition.  Since it is hard to win, my gut feel is that you would very likely need more rolls of the die to win.

Stage 5:  Reflection

What did you learn from solving this problem?

     I learned that the memoryless (forgetfulness) property of Bernoulli trials (i.e. the type of trials involved in the Binomial distribution) is useful to allow me to disregard past events and simplify the problem.

     I learned that whenever I encounter a maths problem that seems difficult, I should not despair.  I should try the following heuristics:-

     * rephrasing the problem (heuristic 9)

     * breaking the problem into smaller parts (heuristic 11)

     * writing in mathematical notation (heuristic 13)

     * thinking backwards (heuristic 5)

     * simplify the problem (heuristic 10)

Rephrasing is particularly useful, as it helps to tackle all three parts of this exam question.  Even though part (ii) was on the fringes of the syllabus (just barely in the syllabus), we could still solve it by using heuristics, metacognition and using what we know already.

Conclusion

     Heuristics and metacognition will guide you when you seem to be in uncharted territory.  Use what you know (basic probability) to tackle what you do not know.  Do not be afraid.  Have faith.  May the Heuristics be with you!