Counting with Polynomials

Here is a little counting problem I just came across that makes me glad to live in the age of computer algebra systems. You already know that there are 538 electoral votes distributed among the fifty states, and probably also know that it takes 270 of them to get elected. If you didn't know, you get them in clumps. Winning the state of Texas gives you 34 votes, winning Utah only nets 5. The number of votes for each state is here in case you want to try to solve the problem yourself.

The problem involves a mathematical idea we call a "critical coalition". The idea is to find enough states that could be grouped together (hence the coalition part) to add up to 270 or more, but only JUST enough. The critical part means your coalition is balanced so that if any state drops out, you do NOT have 270 or more votes. That's it, just make a list of everyway you could combine states to make a critical coalition; no fuss about political realities... just pure counting math. If you can handle that problem by ANY method, you can probably stop reading now because I have little to teach you. I'm bringing it up because I saw a really great solution Isabel Lugo who writes the blog "God Plays Dice", which is very worth reading.(and if you are doing the problem, I will give her numeric solution at the bottom so that you can check your work, or hers). If you are young and not used to a lot of variables, pick up a pencil, make lots of notes, read slowly, reread, struggle with it...it really is powerful stuff.

Isabel's solution was to multiply (1+x^55)*(1+x^34)*(1+x^31)*(1+x^27)*(1+x^21)^2*(1+x^20)*(1+x^17)*(1+x^15)^3*(1+x^13)*(1+x^12)*(1+x^11)^4*(1+x^10)^4*(1+x^9)^3*(1+x^8)^2*(1+x^7)^4*(1+x^6)^3*(1+x^5)^5*(1+x^4)^5*((1+x^3)^8-1). Ok, this is only part of the solution, but when you multiply that big polynomial, what it will give you is the number of coallitions you can put together of EVERY size that includes at least one of the states with three electorial votes. This is a polynomial of degree 538, and it has encoded in the coefficients of the terms some of the numbers you need to answer the questions.

To try to make that understandable to my high school readers, here is a smaller, but similar problem. What are the critical coalitions of pools of 2, 2, 3, and 4? Those total to 11, and you need 6 to get a victory. I think you can figure this one out by hand, so let's do that first. The set 4+3 is a critical coalition, they add up to 6 or more, and if you take either of them away, it doesn't. Likewise 4+2 forms two coalitions, one with either 2. Another is found by 3+2+2, and I think that is all of them. So we have a total of four coalitions. Now doing this one with a polynomial is longer than doing it by hand, but it shows you how to get the answer by polynomials (with a computer's help of course).

To represent the solutions with a polynomial, we use (1+x^4) for the pool with four votes, (1+x^3) for the pool with three votes, and (1+x^2) for each of the pools with two votes. For a minute I will ignore the -1 at the end of one Isabel's solution, but fear not, we will bring it back in a few lines. When we multiply (1+x^4)(1+x^3)(1+x^2)^2 (the last is squred because there were two of them) we get x^11 + 2 x^9 + x^8 + 2 x^7 + 2x^6 + 2x^5 + 2x^4 + x^3 + 2x^2 + 1. So what?, you ask. If you look at the coefficients of each term (the numbers in front) you get the number of coalitions (not critical, just any old coalition) that can be formed that have as many votes as the power of x. For example, the 2 x^9 tells us that there are two coalitions that have 9 votes. We can see that would be 4+3+ 2 and 4+3+(the other )2. And what about 2 x^7? Well, we can have 4+3 for one coalition, and 3+2+2 for another. See how that works? Check the others to be sure I didn't mess up (and if I did, write a nice correction comment to help me out).

Now that doesn't answer the critical coalition question, but it is a great calculating tool. And if you are still trying to figure out how many eight letter words can be made using the letters in Mississippi that I asked in my recent blog, Yes, Barbie, Math is Hard, then the polynomial method described above should help.

But what about this problem? We can count all the coalitions, and even tell if they have a majority (2+2 is still a coalition, even 2 or 3 by itself is a considered a coalition, but they don't have a majority), but we can't figure out if it is critical or not. And that is where the -1 in the last term of Isabel's polynomial comes in; it allows us to eliminate all the coalitions that don't contain some solution. In our mini-problem, if we change the last term to (1+x^4)(1+x^3) ((1+x^2)^2 -1) we get a listing of all the coalitions that contain a 2. x^11 + 2 x^9 + x^8 + 2 x^7 + 2x^6 + 2x^5 + 2x^4 + x^3 + 2x^2 + 1 - ((1+x^4)(1+x^3)) . That leaves us x^11 + 2 x^9 + x^8 + 1 x^7 + 2x^6 + 2x^5 + 1x^4 + 2x^2. Now each of these coalitions have at least one member who had two votes, so we want to only count the ones that have a total of 6 or 7 since those are the only ones which would be critical. The two is the smallest member, so his absence would only impact the success of coalitions with 7 or less votes. Looking at the terms with powers of 6 and 7 we see that there are 3 such critical coalitions, (3 + 2 + 2; 4+2 and 4+ (other)2). Ok.. so what about coalitions that DON"T have a two... We just take the x^2 term completely out of the mix, and multiply again with -1 on the x^3 term.....(1+x^4)((1+x^3)-1) gives us x^7 + x^3. Since only the x^7 is a winning coalition, and it must be critical because if we took away the 3, the total is less than 6. And that is all of the winning coalittions, if we leave out all that have a 2 and a 3, we are left with 4, which will not win. Add the three that include a 2, and the one that does not, and you get all four coalitions. When Isable did that with her BIG equation, she came up with ...wait for it...51,199,463,116,367 ... that is how many different critical coalitions of states could be combined to get the 270 electoral votes needed to be elected President.

Ok, so now a practice problem for you.. suppose we had voting pools of 2, 3, 3, 5, and 7 for a total of 20.. that means 11 for a majority. Can you figure out how many critical coalitions there are.. if you want to submit a comment, we can walk you towards the answer with some hints and (if needed) corrections. Good luck.

A Little Arithmetic Test

OK, quick, evaluate the two arithmetic expressions below

3 + 5*7............. 2+5 * 8

Ok, if you remember high school algebra, the answers are 38 and 42. If you don't remember high school algebra, you may have gotten 56 on both. The formal order of operations that we try to get across to high school students is that these questions should be done by first completing the multiplication or division operations, and then do the addition or subtractions. (for the one on the left, you should multiply 5 * 7 first, then add the 3)

The amazing thing, according to research by cognitive psychologists David Landy and Robert Goldstone from Indiana University, is that you were more likely to miss the second than the first, and even if you are an old hand at simple arithmetic, the second may have taken a micro-scosh longer... because of the way they are spaced. Notice that in the first I have grouped the multiplication operation more tightly and in the second the addition operation is grouped tightly. The researchers write,"When physical spacing was inconsistent with
order of precedence rules, six times as many errors were made relative to when the spacing was
consistent."

The grouping by proximity seems to be a natural process affecting how we read and write expressions. In the authors own words:
"Specifically, we propose that formal notations are diagrammatic as well as sentential, and that the property conventionally described as syntactic structure is cognitively mediated, in part, by spatial information. Elements of expressions are “bound” together through perceptual grouping— often induced by simple spatial proximity."

According to the research, it even effects the way you write the expressions. They tested people by giving them random mixed expressions using words with addition and multiplication in both orders, such as "Four plus seven times three". and asking people to write them down. It seems that people wrote them down with the multiplication operation more closely grouped than the addition.

The suggestion of the researchers is that this grouping tendency may be observed from parents and teachers and become part of our reading and writing style even if we do NOT remember the order of operations. However, there was a greater accuracy of the results of people who tended to group, "Consistently spaced submissions were also more likely than either inconsistent or
unspaced expressions to be correct; 53% of consistent equations were correct, compared with
only 37% of inconsistently spaced and 50% of unspaced equations." To me this would seem to support a written response that indicates a cognitive bias based on knowing the order of operations more profoundly.

Yes, Barbie, Math IS Hard!

A short while ago in in “Average, Percent, and other Misunderstood Math Terms”, I wrote about the fact that even simple math was particularly hard in application. I got three replies that suggested I had committed some unpardonable sin, and two even reminded me of the hassle when Mattel released the talking Teen-talk Barbie, who expressed a similar concern in her tiny little recorded voice by adding “Math is Hard” to her other stereotypical exaggerations of modern teens, such as “I like to shop.” I figured if a merchandising mishap 17 years ago by a toy company is the best they can do to challenge my statement, I MAY actually be right.

But to be fair, I wanted to explain why I thought math was hard, and while I was thinking about it, a couple of VERY good mathematicians brought up a minor variation of a somewhat easy and common high-school text book problem.. Only this problem is VERY hard. Its a high school problem that IS in a high school textbook, although it probably shouldn’t be, and it had the wrong answer in the back of the book .. we know it is wrong, …too big.. but we don’t have the right answer yet… keep in mind that teen talk Barbie is older than most of the students in my classes who would have been expected to solve the problem. You can still buy her on the Amazon link in the last sentence, and she still says “Math is hard”,… hang in there Barbie.

So here is how an easy problem crosses the border from easy to incredibly hard in a few steps. We begin with a linear permutation. To make the problem even easier than the one in the textbook, lets deal with only five beads.

Question one, easiest… Five different colored beads are to be placed in a row. How many ways can it be done…Easy, five factorial or 120 ways. You have five choices for the first color, then four choices for the next .. etc.. so the answer is 5(4)(3)(2)(1) = 120…

Question two, a minor variation… How many ways can the five colors be laid in a circle? Now we have a clinker (think Christmas Story and the furnace). Red, orange, yellow, green, blue is the same as orange, yellow, green, blue, red. It is also the same if we write them starting with yellow, green… etc.. so in the case of a circle, we can only get 5 factorial divided by 5 or 24 different circles.

Question three, a spatial flip.. Now instead of putting them in a circle on the ground, we put them on a hoop of wire, for a necklace or bracelet or something. Ok, still a circle so 24,… but wait.. we can reverse the necklace so that now blue, green, yellow, orange, red is the same as red, orange, yellow, green, blue.. and similarly for any list of colors, so we have to divide by 24 by two and now we are down to twelve unique necklaces that can be made… think goodness it can’t be any harder than that….(gotcha’)

Question four (really hard and if you go up to 20 beads, as the text book did, a total mess) What if there are not five colors, let’s be really easy, what if there are three red and two blue? Now the problem requires us to look at every possible order of the beads and decide how many ways it can be reflected in space or rotated around the wire… You could have 3 reds together and 2 blue…or you could have two reds, a blue, a red, and another blue… We could also have red, blue, red, blue, red… is that it? What if we had a blue, two reds, a blue, and another red… NOPE, that one is already counted.. can there only be three ways to order these five beads on a necklace. Math got hard because all the way down to the last problem we had some formula approach we could use if we understood the problem… and then… it all fell away..

If you need a challenge, the original problem had fourteen beads of one color and six of the other.. try your luck.

A very similar problem uses the letters in the spelling of Mississippi.. Writing words is a linear permutation, so we don’t have to worry about flipping the necklace or circular similarities, but that doesn’t make it easy. Students are taught, using the same rule we used for the beads in a row, that with six distinct letters there are 6 factorial orders (that’s 720) . With 11 letters there are 11 factorial (a lot more than 720, in fact almost forty million) . But then if we repeat letters, we show them that they have to divide by how many letters appear more than once..

If you write out all the 11 factorial arrangements of the eleven letters, all of them have two p’s that are indistinguishable. If we switched the two p’s no one would know; so we have to divide 11 factorial by 2. But then, there are four letters s in the word, so you have to divide by the 24 ways they could be switched around, and then there are four letters i, so we have to divide by 24 yet again. Finally we have reduced the unique orderings of Mississippi down to only 34,650 (IF I did that right) . You can list them out to check, but that may take a while… If you want to try an easier one that you can see how, try finding all the possible orders of TOOTH; there should be only thirty…

Now we make a subtle change in the problem and make it horrible, what if we don’t use all the letters? What if we only use eight of the letters in Mississippi at a time? One word has all four of the letters s, another might only have three, and some would only have one s (with eight letters, there has to be at least one of the letters s in the word, can you see why?) Try your luck figuring out how many three letter “words” you can make with tooth.

I think that is the big difference between math and other subjects, that razor fine line where easy problems become totally intractable. Maybe math is easy, and I (and Barbie) are just not too bright). By the way, how many eight letter "words" could YOU find using the letters in MISSISSIPPI?

The B-2 Theorem

In my B2 Pre-calc class today, we re-discovered a theorem about Pascal's triangle that I had not known. It began, appropriately enough, with a question Jacob C. asked about dealing cards from a standard deck; "How many 13 card hands can be dealt that contain exactly two suits. " As we were working through the problem, I began by attacking the somewhat easier problem, "how many hands can be dealt with only hearts and diamonds, but at least one of each." We began writing out the possibilities of 12 hearts, 1 diamond plus 11 hearts two diamonds ..etc To make life easy, for this short while let’s let (n,r) mean “n choose r” , the combinations of n things taken r at a time. So we needed to find (13,1)(13,12) for the first part, 12 hearts and 1 diamond. Then we needed to add on (13,2)(13,11) + (13,3)(13,10)…. And all the way down to (13,1)(13,12). One of the clever ones quickly realized that each of these pairs were just the same number due to the symmetry of Pascal's triangle, and so we were really looking for (13,1)² + (13,2)²... etc. While some of the kids were adding these on their calculators, I wrote out several lines of the arithmetic triangle and began to write the sums of squares on the right....

As I wrote the totals of each row, 1, 2, 6, 20, 70.. it struck me that they were all the center number of an even numbered row, (2n,n). I remembered them from working with Catalan’s Numbers (another cool pattern that shows up in Pascal’s triangle). About the time the first students were coming up with an answer, I asked them to check (26, 13) and compare it to the answer they got for the actual squares of the thirteenth row…

Close, but not right, was the reply.... huh??? … , oh yeah, we had avoided the case of (13,0) and (13,13) because we wanted to ignore the case where all were hears or all were diamonds, so the answer to our mini-problem was (26,13) - 2; and the only thing needed to solve the original problem was to multiply by 6, to account for all the ways we could pick two suits to be in the hand out of the four possible suits.

When I showed them the result, and we checked a couple of more cases to be more sure, I admitted that I had never seen this theorem. One kid suggests it should be a test question… I countered with, “and extra credit for the person who comes up with the best name for it. Several played to my ego, “Ballew’s theorem, of course!” but then they thought they might deserve partial credit, and hence the name, B-2 theorem, at the top.

Unfortunately, we were not the first to stumble across this little gem. I haven’t had time to chase it down fully, but it may actually date back to the Chinese around the 12th century. So fame and fortune will have to wait, but when you walk in the footsteps of greatness, you’re taking pretty big steps; so congratulations class, I’m proud of you, and it will always be the B-2 theorem when I teach it. Dennis was going to send me a class picture we took on his phone, so if it turns out, I will add that later,

While I was searching for the history of the sum of the binomial coefficients, I came across another place where the triangle is related to squares. One of those theorems we teach when we get to sequence and series in high school is the sum of the integers, 1 + 2 + 3 + … + n, and the sum of the squares of the integers, 1² + 2² + 3² + ….. + n². Usually we present the formula for this last without proof since it occurs before they are introduced to inductive proofs. As I was researching I came across this neat little relation to the arithmetic triangles. To find the sum of the squares of the first ten integers, just go down to 10 at (10,1) and turn right and follow the diagonal down two numbers to (12,3) and add this to the number on the diagonal above it (11,3), the sum of 220 + 165 = 385 which is the same as 1² + 2² + … + 10²

In general you can find

And I think they can accept that as evidence, at least until we get to inductive proofs.

l'Hospital's Rule

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In 1694, Johann Bernoulli wrote a letter to Guillaume-Francois-Antoine de l'Hospital that included the theorem now known as l'Hospital's Rule (The alterante spelling "l'Hôpital" is often used in France) . There is little doubt for most math historians that a) Bernoulli first discovered it, and b) that L'Hospital first published it. l'Hospital is sometimes discredited because he published someone else's theorem, and paid for the privilege. My calulus teacher in college would rant about l'Hospital being an "inept" mathematician and "buying his fame", and gave me the impression that he had published it as if it were his own creation. The truth is, in the 1696 differential calculus book in which he published the theorem, L'Hospital thanks the Bernoulli brothers for their assistance and their discoveries. And in addition, he was far from inept as a mathematician. The MacTutor History of Math site comments that, "L'Hôpital was a very competent mathematician and solved thebrachystochrone problem."

L'Hospital never called the rule by his own name, and in fact, it appears that noone else did for several hundred years. Jeff Miller's web page on the first use of mathematical terms gives the first citation for the use as "de l'Hospital's theorem on indeterminate forms is found in approximately 1904 in the E. R. Hedrick translation of volume I of A Course in Mathematical Analysis by Edouard Goursat. The translation carries the date 1904, although a footnote references a work dated 1905 "

The rule is a method for finding the limiting behavior of a rational function whose numerator and denominator tend to zero at a point.(The rule is somewhat expanded today from its original form and can be used if both functions diverge to infinity also) In a traditional Calculus course, a student might use the rule to find, for example, the limit of the function as the value of x approaches 2. This meets the conditions since both the numerator and denominator approach zero as the value of x approaches two. The rule states that in such cases, one can take the derivative of the numerator and denominator independently and then find the limit of this ratio. Since the derivative of the numerator is 2x, and the derivative of the denominator is 1, the ratio of the derivatives is the experession 2x/1 or just 2x; and the limit as x approaches two for this function is just 2(2) = 4. In the area NEAR x=2, the value of the original ratio is very near 4.

The Stolz-Cesaro Theorem

Until recently I had never heard of this discrete analogue of l'Hospital's rule, and I thank the folks at Topological Musings blog for the lesson. The adjusted rule can apply to sequences (l'Hospital's rule is for continuous functions) under certain circumstances, and allow us to calculate the limit of the ratio of two divergent (they both go to infinity) seqences. If we think of the function above as a sequence in which the numerator ( x²-4) diverges to infinity as x grows larger and larger, and likewise the denominator (x-2) also grows without bound as x goes toward infinity, then the Stolz-Cesaro theorem says that .

So for out example, we need to find ((n+1)² -4) - (n²-4) for the numerator, and ((n+1)-2)-(n-2) for the denominator. The numerator simplifies to (2n+1) and the denominator to (2n+2).. since these both still meet the conditions of the theorem, we can apply it once more to get 4/2 = 2... Using l'Hosptials rule for the same function as x-> infinity, we get (2x/x) and applying it once more we get two by l'Hospital as well.... (nice if they both get the same limit)..

The theorem is named after mathematicians Otto Stolz and Ernesto Cesàro.

Average, Percent,

Average, Percent,

and

Other Misunderstood Math Terms

“Misunderstood? Surely, you jest!” you reply…. But hear me out.. Okay, I agree that almost every seventh grader knows how to compute averages, but the question is, do they understand them… Consider the following:
A school is asked to report the average student/teacher ratio. Now we could do that two ways; the school could count how many students there are, then count how many teachers there are, and in the end, divide the number of students by the number of teachers to get the average… and that is how it is normally done. But you could also have each student count the number of students in each class they go to (assuming one teacher per class) and then average the class sizes reported by all the students. That takes a little more work, but it should give us the average number of students per teacher also… except, in most cases the two numbers will be different.

Let’s illustrate with a simple (no,, I mean REALLY simple) example. Suppose there is a school where there are only two teachers and two classes. The total enrollment is 6 kids in one class and 4 in the other. From the school point of view, there are 10 students and two teachers, so the average is 5 students per teacher. But when we survey the students, the class sizes reported by the ten students is {6,6,6,6,6,6,4,4,4,4} for an average of 5.2 students per teacher. Go ahead, make up your own numbers, but the only way to get it to agree is if EVERY class has the same number of students….and now the big question…. Which is the CORRECT average??? (“pssst… say ‘both’ ”)… But if they are not the same number, then they can’t both be the average of the same thing… Now you start thinking…. Tick… tick… tick… So what is is that each is the average of?

Ok, let’s go to something even easier; fifth grade percentages. Here is the problem:
Three weeks ago 87% of the students were in favor of the new football coach. Then the team lost and now only 67% of the people support the football coach. Ok, So which is correct Newspaper headline:
…..Support for Coach drops 20%
……Support for Coach drops 23%
…….Support for Coach drops 30%

All of them are a valid percent.. but the question with percentages, like the one with averages, and like most of the questions in applied math, resolve down to the object of a preposition…..percent of WHAT?
See, Math really is Hard.

It Ain't Fair............ is it???

Had a chance to take ten of my students down to the Center for Mathematical Sciences in Cambridge last night to hear a lecture by Richard Weber of Queen’s College. He is the one in the right on the picture during his appearance on the British TV game show, Who Wants to be a Millionaire. He also happens to be the Churchill Professor of Mathematics for Operational Research. His primary work is in problems in communications and systems, and the mathematics of optimization, algorithms, probability and game theory.

On this particular night, he was talking about “The Disputed Garment Problem”, an ancient problem involving the concept of fair division. It seems a simple idea. We all think we know when the division is un-fair, and if we picked an honest, unbiased third person to make the division, it would seem that a fair result would follow, and yet, it seldom does. If you ask people to describe the rules that a fair division would entail, most people can come up with several rules, what a mathematician would call the axiom set, and probably agree with each other. But then when you make a decision consistent with those rules, the same people will scratch their heads and say, “Wait, that’s not fair.” What is even more scary is that often they can make five or six rules, and you find out that it is Impossible to meet all those rules.

So let’s look at a simple example. The Babylonian Talmud is a compilation of the ancient laws from an oral tradition set down during the first five centuries after Christ. They serve as the basis of Jewish religious, criminal and civil law. In one of the problems the Talmud offers:
Two people dispute the possession over a garment. One claims that he should get half, the other claims all of it. The solution of the Talmud is that the first gets ¼, and the other gets ¾ .

The Talmud is a terse document, (sort of like 19th century math books). It tells you what to do, but not why. In this case however, the reasoning seems somewhat clear. Since half the garment is not in dispute, it should go to the one who claims the whole. The other half, claimed by both, shall be shared equally, giving the 3:1 split.

Given that beginning, a more interesting problem involves sharing by three wives in what is called the marriage contract problem:
A man has three wives whose marriage contracts specify that in the case of his death they receive 100, 200 and 300 units respectively. But when he dies he leaves less than a sufficient amount to cover the 600 debt. How should the estate be divided.

The answers the Talmud gives for different amounts is surprising, and somewhat confusing. If the man leaves an estate of 100 units, the three wives will divide it equally (see below). If the estate has a value of 300 the wives divide it proportional to their claim, each getting ½ the amount they were promised. But what is happening in the case with 200 remaining? The question for the reader is, what will be the division of an estate with 400 remaining? (I will give the answer later, but don’t cheat, try to work it out. And additional clues will come if you are really stuck…. I was)

Estate..wife 1...wife 2....wife 3
100.....33 1/3...33 1/3....33 1/3
200.....50 ......75........75
300.....50 ......100......150


Ok, so which of those seem like a fair division to you, and which do you think most people would agree with. With your answer tucked away, here's a similar problem with a different twist. Three people want to build a runway for their private planes. For Pilot A, the cost of a runway for his plane would be 100 units. For Pilot B, with a bigger plane and therefore a longer runway needed, the cost would be 200 units. And Pilot C, needing an even longer runway, would have to pay 300 units. If they form a coalition, they can build one runway big enough for all three for the 300 units. How should the share the costs?

Should they pay 100 units each? Probably not, the first guy would see no advantage to the coalition. What about each paying half of what they would pay for their separate runways? If that sounds fair, then look again at the marriage contract problem above for an estate of 300; are these really the same problem? The numbers are all the same. The only difference is that in one case they are sharing a debt of 600 units and paying a combined 300, and in the earlier case they had a combined credit of 600 and sharing a combined payout of 300. Is fair in one case the same as fair in the other case?

The good professor provided another alternative to the airplane problem, using a method called the Shapley value, named for Lloyd Stowell Shapley, from UCLA. The method is founded on objection and counter-objection of each party being balanced. But an easier way to understand the value is to take every possible permuation (different orders of the pilots) and how much each would pay if they joined the coalition in each order. For instance if they joined in the order 3,2,1, the large plane pilot would have to pay for the whole runway, the other two would pay nothing. Here is a list of ALL the possible orders and the amount each pays:

Order.......Pilot 1...Pilot 2...Pilot 3
1,2,3.......100.......100.......100
1,3,2.......100 .......0 .......200
2,1,3 ....... 0.......200.......100
2,3,1 ....... 0.......200.......100
3,1,2 ....... 0.......0.......300
3,2,1....... 0 ......0.......300

Now if you add up all the possible amounts, the six ways add up to 1800 dollars. Since pilot one pays 200 of the 1800, his share of 300 should be 1/9 of the 300 units, or 33.33 units. Pilot 2 should bay 5/18 of the 300 or 83.33 units, and the last pilot shoud pay 11/18 of the 300, or 183.33 units.
Now that’s what I'd like for public service, good progressive taxation… make the rich guys carry the burden.

I expressed my dislike for the three wives solution on the night. I explained that if my bank went bankrupt and offered me 10 cents on the dollar, and paid another guy 50 cents on the dollar because he had more money, I would be screaming like a stuck pig. But if fair on one problem is fair on the other, then I’m REALLY not liking the rich guy getting 61 percent of his money back, and the little guy gets 33 % return.

Ok, so I’m going to make it a little more complicated in a minute, but I want to go back to the case of the three wives in the Talmud. If the rule is consistent, then any two wives should get the same amount they would get if there were only two of them and the estate was the sum of what they shared and the division was by the disputed garment solution. Ummm, let me try again. In the case for 300 in the estate, wife 1 and wife 2 get a total of 150 units, and their claim is for 100 and 200. If we use the logic of the disputed garment, then since wife 1 only claims 100 units, wife 2 should get the 50 extra right off, and they should share the remaining 100 equally giving a solution for the two of 50, 150… ok, that works, and if you check the other two possible groupings (wives 1 and 3 and wives 2 and 3) it also works out. So maybe this is the Talmudic approach. Can you find the solution for 400 units in the estate now?

For over 1500 years scholars studied the Talmud and couldn’t find a rule to explain the determination of values that was more specific than the general rule I just gave. Then Robert Aumann, a professor at the Hebrew University of Jerusalem, showed that the solutions were the ones you would get by using the Nucleolus for the coalitional game (I will spare the general reader the several pages of equations with subscripts and other assorted devices of mathematical torture that explain Nuclelus, but if you are, like me, one who shouts “Show me the Math”, you may find an extended description here. .)

Now, if you STILL haven’t figured out how much each person gets with the division of an estate of 400 units, here is a physical method that, remarkably, gives the solution for any amount less than the 600 units promised the three wives.


Imagine the image shows six interconnected tanks. The two tanks on the left hold fifty units of liquid each and represent the share of wife 1. The second set has two tanks of 100 units each, and represent the second wife's total promised 200 units, and the third set of two each hold 150 units, to represent the 300 units of wife three. Now we pour an amount of liquid equal to the value of the estate into the top three tanks, divided equally between them. Behold, what is left when the connection across the bottom is allowed to do its thing, is the Talmudic solution. The solution above shows the answer for the division of an estate of 100 units. Here is the one for 300 units to help you see the physical solution method.

Now admit it, even if your not a math/science person, even if you don’t think the solution is fair, that’s a pretty cute trick… ahh go on, admit it. Ok, if you STILL don't have the answer for 400, the division is 50, 125, 225....

Ok, here is a little postscript, one more thing that disturbs me about the Talmudic settlement. Suppose hubby dies, and they have 200 units in the estate. Everyone gets a settlement, and then they find more money hidden away under the mattress, say 200 units. How do we handle that. If we treat it like a single 200 dollar estate, then the wives each double what they had after the first settlement. Wife one gets another 50 units, wife two gets another 75 units, and wife three gets another 75 units. But now wife 1 has her entire promised amount back, while the other two have only a partial amount. But if we treat it as if we are adjusting for a 400 unit estate, now wife one gets nothing, even though the settlement is twice as large as origninally thought.
Do you see my point? A fair division added to a fair division does not seem fair. Is THAT fair?? hmmm.... Sometimes I like math better when there are only pure numbers... reality makes things complicated.