## Monday, October 2, 2017

### Lesson 3-3: Justifying Conclusions (Day 33)

This is what Theoni Pappas writes on page 275 of her Magic of Mathematics:

"Cut out these pieces, or rearrange them in your head. Some may need to be rotated, flipped, or translated. They form a capital letter of the English alphabet. Good luck!"

This is the only page of "The Block Letter Puzzle." Unfortunately, it's yet another puzzle that's completely dependent on the accompanying picture. And you can't see that picture. But don't worry -- tomorrow's puzzle will be solvable without having to see any picture, so be ready for that.

Before we leave page 275, though, notice what Pappas writes about the puzzle pieces -- "Some may need to be rotated, flipped [reflected -- dw], or translated." So here we go again with rotations, reflections, and translations -- and this was a quarter-century before the Common Core.

Last weekend marked the death of a game show legend -- Monty Hall. I've written about Hall several times on the blog, including last year on the second day of school. But I've yet to mention something mathematical that actually named for him -- the Monty Hall Problem.

The Monty Hall Problem works as follows -- suppose you are a player on a game show such as Let's Make a Deal, and the famous host offers you a choice of one of three doors. One door contains a car, and the other two contain Zonks. After you make a choice, Monty will always reveal one of the Zonks behind one of the other two doors, and then offers you the choice of keeping your door or switching to the remaining door.

We know that since there are three doors, you have a 1/3 probability of choosing the car. Now here's the paradox -- the probability that the car is behind your door never changes from 1/3, and so the probability of winning the car by switching is in fact 2/3.

Much has been written about the paradox, and so here's a link to one of my favorites, Numberphile:

In the video, it's explained that if there were 100 doors, with one car and 99 goats, you have only a 1% probability of choosing the car, and this never changes, so switching actually gives you a 99% probability of winning the car.

The following is how we avoid falling for the paradox. The only way a probability can increase is if there's some event that can cause it to decrease -- and the fact that such an event doesn't occur is what causes it to increase. In other words, suppose Monty had a nonzero probability of opening the door with the car in it. Then the fact that he doesn't reveal the car would raise the probability that your door has the car to 1/2.  But in this problem, Monty will reveal the car with probability 0, and reveal a goat with probability 1. Since the probability can never decrease, it can't increase either, and so the probability that your door has the car remains at either 1/3 (3 doors) or 1% (100 doors).

In reality, neither Monty Hall nor the current host, Wayne Brady, actually made this offer on Let's Make a Deal. The doors only appear during the Big Deal of the Day (as curtains are used earlier on the show), and there are no Zonks, nor an opportunity to switch doors, during the Big Deal. (I know about the Monty Hall version, even though it aired well before I was born, since the old episodes aired a few months ago on the BUZZR channel.)

But there is a game on the current show where the Monty Hall Problem actually occurs. This is the game "Three of a Kind." In this game, there are six cards dealt face down, with three of one kind and three of another kind. Usually, three would be matching face cards (let's say kings) and the other three would be number cards (let's say fours). You are to choose three cards, and if they match (either trip kings or trip fours) then you win the car, otherwise you win nothing.

But there is a possibility of switching. First Wayne Brady will reveal that two of your cards match (without loss of generality, let's say a pair of kings), and then offer you the opportunity to switch to a smaller prize -- let's say it's worth about a quarter of the car's value. If you refuse to switch, then he will reveal that two of the cards you didn't choose also match (pair of fours). Then Wayne will offer you the possibility to switch again to the smaller prize plus a few hundred dollars.

We're led to believe that the probability of winning the car at this point is 1/2. But let's calculate what the actual probability is. We know that the originally there are 20 combinations of three cards that you could choose from the original six (6 choose 3 is 20, found on Pascal's triangle). Of these, two combos will win you the car -- trip kings and trip fours. So the car probability is 2/20, or 10%.

Now the question is, can this probability increase? We know that it can increase only if there's some event that would cause it to decrease. Now no matter which three cards you choose, at least two of those will match (either pair of kings or pair of fours), which Wayne reveals with probability 1. And so this event can't change the probability. I recall an actual player on one episode cheering that his win probability was now 1/4 (since if two kings are showing, there is one king out of the four cards still face down). That player obviously fell for the paradox, since the actual probability is still 1/10.

Likewise, two of the other cards you didn't choose will always match, and so Wayne will reveal these cards with probability 1. Therefore the car probability can never change from 1/10. Whenever I watched this game on the show, the player almost never won the car -- which makes sense, as there is a 90% chance of losing. But since the player is led to believe that they have a 1/2 probability of winning the car, the player inevitably chooses to go for the car, rather than a sure deal that's worth less than 1/2 the car's value.

If I were on the show, I'd always take the sure thing, knowing that I'm nine times as likely not to have the winning card than to have it, and since the small prize is worth more than 1/9 the car's value. Of course, I also know that Wayne will offer the extra couple hundred bucks with probability 1, and so I wait for him to offer it before taking the deal. In the end, "Three of a Kind" is equivalent to the original Monty Hall problem with ten doors -- one with the car and the others with nothing, with the possibility to switching to a smaller prize.

On the game show The Price Is Right, there was a game called "Barker's Markers." Players are offered three prizes, which they win by "marking" their prices from a list of four prices. Then Bob Barker, the host, would reveal that two of the markers are correct with probability 1, and they have the opportunity to switch the third marker. This is equivalent to the Monty Hall problem with four doors, with the three prizes behind one door and nothing behind the others, so you have a 1/4 chance of being correct and a 3/4 chance of winning by switching.

The twist was that Bob would offer the player \$500. The only way to keep the \$500 was to win the game without switching. If you switch you win the three prizes but not \$500, and if you stay and lose, you win nothing, not even \$500. And so players are already led to believe that they have a 1/2 chance of winning by staying, and then they inevitably choose to stay away since there's a \$500 incentive not to switch. Probability theory suggests that players should switch, since there's a 3/4 chance of winning the three prizes (which are each worth much more than \$500).

There's actually one final secret twist to this game. In the original Monty Hall Problem, if you are lucky enough to choose the car with the original selection (which happens 1/3 of the time), Monty will reveal a goat from one of the other doors at random. But on "Barker's Markers," if the initial three markers are all correct (probability 1/4), then the two markers revealed to be correct are not random -- instead, they are chosen so that the two remaining prices are as close as possible to each other (so that you'd be tempted to switch and lose).

So suppose the prices are \$1000, \$2111, \$3444, and \$4999. If you mark the three highest prices and you're correct, then the two revealed to be correct are \$3444 and \$4999, so that you'd be tempted to switch \$2111 to \$1000 and lose. So if we mark the highest three and the two revealed correct are \$2111 and \$3444, then I'd know that \$4999 is wrong (since had it been right, they would have revealed it instead of \$2111). Then switching wins with probability 1. On the other hand, if the two remaining prices are close enough to each other, then the win probability really could be 1/2. So if I were playing this game, I'd switch if the two remaining prices are far apart (win probability of 1) and stay if the two remaining prices are close (1/2 probability of win + extra \$500). (Suppose that in the original problem, you somehow knew that Monty always opened the higher of the two remaining doors if your choice is correct.)

On the other hand, Deal or No Deal is not equivalent to a Monty Hall Problem. In the game, when you are down to two cases and one has the million dollars, host Howie Mandel would offer you the opportunity to switch cases. The initial probability of choosing the million is 1/26. But in this case, there really is an event that decreases the probability -- if you choose to open another case early and it contains the million, then your probability of having the million in your case drops to 0. Therefore, the fact that you avoid this and make it to the final two cases really does improve your chances of winning to 1/2. In reality, no player ever switched cases -- probably because it feels so much worse to have the million and give it away than it is never to have had the million in the first place.

And so Perhaps one way to convert naysayers who think that the probability of winning Monty Hall is 1/2 is to compare it to the Howie Mandel problem and explain why the probability when you're down to two cases really is 1/2.

Rest in peace, Monty Hall. The BUZZR channel will soon play a tribute to the game show legend, most likely with a marathon of some of his episodes.

Lesson 3-3 of the U of Chicago text is called "Justifying Conclusions." This actually corresponds to Lesson 3-5 of the new Third Edition. (Meanwhile Lesson 3-4 in the new edition is called "Algebra Properties Used in Geometry." This is broken off from Lesson 1-7, "Postulates," in my old edition.)

This is the last of the three lessons that I changed from two years ago. Therefore, this is what I wrote three years ago on this lesson:

Section 3-3 of the U of Chicago text is an introduction to proof. Because the Common Core Standards specifically mention statements that students are supposed to prove, that makes this section one of the most important sections of the book -- and that's why I cover this section before skipping to Chapter 4.

But what exactly is a proof? The following definition of proof comes from a professional mathematician:

A proof of a statement Phi consists of a finite sequence of statements, each of which is either an axiom, or follows from previous statements by logical inference such that Phi is the last statement in the sequence.

Notice that this is not that much different from the definition given in the U of Chicago text. Of course, a proof must be a finite sequence of statements -- proofs can't go on forever! The U of Chicago states that valid justifications in proofs include postulates (note that "axiom" is basically another word for "postulate") and theorems already proved (the "previous statements" mentioned above). But what about the third justification mentioned in the text -- definitions? Strictly speaking, a definition is also a special type of axiom, called a "definitional axiom." And of course, the last statement in the sequence is "Phi" -- that is, the statement that we're trying to prove!

Now Section 3-2 of the text mentions two theorems, the Linear Pair and Vertical Angle Theorems. But I left these out, since they didn't fit on my Frayer model page from last week. But those theorems certainly fit here in 3-3, for after all, the first example of a proof in the text is that of the Vertical Angle Theorem.

The proof of the Vertical Angle Theorem in the text is sort of a hybrid between a paragraph proof and a two-column proof. The conclusions and justifications aren't written in two-column form, but since each conclusion is followed by its justification, it might as well be a two-column proof. Subsequent proofs in this section are essentially paragraph proofs -- actual two-column proofs don't appear until Section 4-4.

One thing I like about this section is that it gives the reasons why anyone would want to write a proof. The first one is:

"What is obvious to one person may not be obvious to another person. Sometimes people disagree." [2017 update: like the Monty Hall Problem, for example.]

As I mentioned before, when asked what the least favorite part of geometry class is, a very common answer is, proofs. But this is what professional mathematicians do all day -- a common joke is that a mathematician is a machine for turning coffee into theorems. (This line is usually attributed to the 20th-century Hungarian mathematician Paul Erdős.) Recall that a theorem is a statement that has been proved -- so Paul is telling us that mathematicians are machines that prove things.

Indeed, some of the most famous math problems in the world are proofs. About 400 years ago, a French mathematician named Pierre de Fermat (actually he was a lawyer -- but then again, both lawyers and mathematicians are known for proving things) made a very innocent-looking statement:

"No three positive integers ab, and c can satisfy the equation an + bn = cn for any integer value of n greater than two."

But Fermat was unable to write a proof of this statement -- at least, not a proof that he could fit in the margin of the book he was reading. It was not until 20 years ago when a British mathematician named Andrew Wiles finally proved of Fermat's Last Theorem. His proof is extremely complicated -- no wonder it took over 350 years for anyone to prove it! [2017 update: Yes, we discussed FLT with Ogilvy two weeks ago!]

Even today there are statements that appear to be true, but no one has proved them yet. The Clay Mathematics Institute has offered a prize of one million dollars to the first person who can prove each of the seven Millennium Problems (so called because the prize was first offered at the start of this millennium). So far, only one of the problems has been proved, so six million dollars remain unclaimed. [2017 update: I actually have more to say about the solved Millennium Problem very soon!]

And we can go from problems that take years -- or even centuries -- to prove, to some which take a few hours to solve. Every year on the first Saturday in December, college students from around the country participate in the Putnam competition. There are twelve questions -- most or all of which are proofs -- and six hours in which to solve them. And if you can get even one of the twelve questions correct, then you will have one of the top scores in the country!

Now let's compare this to the attitude of many high school geometry students -- mathematicians may spend hours, years, even centuries to write a proof, yet the students can't spend a few minutes proving the Vertical Angle Theorem?

There's a wide range of beliefs on how much proof there should be in a geometry course -- from David Joyce, who believes that anything that can be proved should be proved as soon as possible, all the way to Michael Serra, who doesn't prove anything in his text until Chapter 14. On this blog, I'll take Joyce's approach, but only for proofs emphasized by the Common Core.