You Thik You ve Got Problems? Marc Brodie Associate Professor of Mathematics, WJU Itroductio. My life, like that of ay other s, has its share of problems. I cosider myself fortuate, however, to have more problems tha most. I the past several years I ve dealt with war, gamblig, ad drugs to ame but a few. I writig this article I ited to dump some of my problems o you. The problems that I refer to are mathematical, but accessible to a wider audiece (so keep readig). Several problems will be surveyed, with a overview of their origis ad a brief discussio of the processes ivolved i their solutios. Computatios will be kept to a miimum ad there will be plety of visuals to aid you i uderstadig the major ideas. Some of these problems have led to publicatios, others to presetatios at cofereces; some are still works i progress. Hopefully, you will fid these problems iterestig i their ow right. At least, this article will shed a little light o what kid of problems some mathematics professors experiece outside the classroom. My iterest i problem solvig (as opposed to the hard-core theoretical research ivolved i a Ph.D.) bega soo after I bega workig with mathematics majors. I mathematics it is very difficult, if ot impossible, to egage udergraduate studets i the traditioal research that a faculty member might be ivolved i. A typical udergraduate will ot have the backgroud to uderstad the questios asked, let aloe get excited about tryig to aswer them. Fortuately, there is o shortage of iterestig problems that arise from everyday experiece problems that studets ca easily relate to ad be itrigued by. These problems are elemetary i the sese that they are easy to state, ad require o mathematics more advaced tha a studet would ordiarily see i the course of the major (ad ofte quite a bit less). O the other had, they are good problems. Their solutios are ot kow, they are ope-eded, ca be difficult to solve ad they ofte lead to other aveues of ivestigatio. Teachig at liberal arts colleges, such problems have become cetral to my work. They have give me joy ad struggles, led to publicatios ad udergraduate research projects, ad allowed me to share my love of mathematics with studets ad colleagues alike. War. My vetures ito problem solvig bega i earest about 14 years ago, whe playig the card game War with oe of my sos. Adam, who was 4 years old at the time, was somewhat impatiet while waitig for the battles (whe the two players cards have the same face value), ad would make declaratios such as I m ot goig to bed util after aother battle. I, o the other had, would be thikig more alog the lies of is t this game ever goig to ed? Such iocet thoughts ad commets ca be the seeds of serious work to someoe i mathematics. Cosider that battles i the game of War are a geeralizatio of the well-kow Hat-Check Problem (see [3], for example): Suppose te getleme check their hats upo arrival at a restaurat. At the ed of the meal, the hats are retured to them i a radom order. What is the probability that oe of the me get the correct hat? If we restate the Hat-Check Problem i terms of cards, we get a simplified versio of War:
Suppose you have te cards i your had, the ace through te of spades. Your oppoet holds the ace through te of hearts. You each shuffle your cards the lay them dow oe at a time. What is the probability of o matches (battles)? 16481 The solutio is 0.367879464286, so that there is a greater tha 1 i 3 chace that o 44800 oe gets the correct hat. Furthermore, as the umber of getleme checkig hats icreases 1 without boud, the solutios approach 0.367879441171. (As a aside, the fact that these e two umbers are so close to each other idicates that for all practical purposes the aswers are the same whether te hats or te thousad hats are checked!) Studyig battles i the game of War is much more complicated for two reasos. First, there are four suits istead of two, allowig for multiple possible matches (imagie te couples checkig hats ad cosiderig whether or ot each perso is retured a hat that belogs to them or their parter). Secod, i a typical deal of the cards oe player might hold all the kigs, for example, so that o match of kigs is possible. It turs out that i the case of War, the limitig probability 1 is 3 / 2 0. 22313. Further details are available i Avoidig Your Spouse at a Party Leads to e War, published i Mathematics Magazie, a joural of the Mathematical Associatio of America [1]. The secod issue, that of a ever-edig game, was particularly suited to udergraduate level research, ad was give to two of my best studets. Specifically, they studied the questio of whether or ot a game ca go ito a ifiite loop, so that either player would ever ed up holdig all the cards. The questios were sufficietly ope-eded to allow the research to ivolve the sort of ivestigatio where examples were studied, cojectures were made, ad the theorems were proved. A fair amout of programmig i Mathematica was doe, so that the computer could simulate thousads of games of cards, determiig whether a particular deal of the cards would result i a game that eds up i a loop or ot. Subsequetly, specific games that eded i loops were chose ad the computer prited out the etirety of the games. We were thus able to study the structure of the loops ad to see patters that otherwise might ever have bee foud. The result: uder certai covetios of play, it turs out that ot oly are loops possible, but their structure ca be described quite elegatly, as my studets did i two idividual hoors theses. Gamblig. A itroductory course i probability ad statistics, such as MAT105, is ot the first place oe might expect to ru ito a coudrum of the sort that leads to a article, but it happes. Various state lotteries ca be used to illustrate the basic cocepts of (sophisticated) coutig ad probability. They are real world examples, ad the studets get some satisfactio from seeig that the aswers we arrive at i class match those prited o the play slips. To a mathematicia, there is o differece i the computatios from oe lotto game to aother (oly the umbers chage; ot the cocept), so I see o eed to work out the aswers i advace I kow I ll get them right i class. Such was the case whe I asked the studets to compute the probability of matchig 5 of 6 umbers i the Missouri Lotto. The exact aswer is
6 38 5 1 57 Pr( match 5 of 6) = =. 44 1764763 6 57 The studets foud that aswer, ad by takig the reciprocal of, foud that 1764763 1 Pr( match 5 of 6). They correctly claimed that the odds of matchig 5 of 6 umbers 30960.75 were, to the earest iteger, 30961:1. This is the umber that should have bee o the play slip. They the tured over their play slips oly to discover that 30961:1 was owhere to be foud. (So much for the studets sese of satisfactio.) After a brief momet of cofusio, a more careful look at the play slips revealed that i the Missouri Lotto, a player must buy two tickets at $.50 each, ad the odds are give per $1 play: Sice two games were played, the studets suggested that Pr( match 5 of 57 114 2 1 6) = 2 =. 1764763 1764763 30960.75 15480.38 The play slip stated the odds were 15480:1, ad the studets were satisfied. At this poit, class was over ad everyoe wet home happy. Everyoe but me, that is. I geeral, it is ot true that Pr(A or B) = Pr(A) + Pr(B). If it were, the purchasig eough tickets (say 30961 of them) would guaratee a wi. I order to fid the correct probability for the Missouri Lotto problem, it is ecessary to cosider the fact that both tickets could match 5 of the 6 umbers. The correct aswer is give by
Pr( A B) = 2 57 1764763 6 38 5 1 44 6 2 = 201179733 3114388446169 1 15480.62, a aswer which we arrived at ext class, after a discussio of the probability rules. Of course the practical differece betwee these umbers is egligible if you expect to wait 15480 weeks before wiig such a prize, what s aother week? O the other had, i the iterest of hoest advertisig, the odds should be stated o the play slip as 15481:1. This discrepacy seemed worthy of further ivestigatio. The result: A Tale of Two Tickets, published as a Classroom Capsule i The College Mathematics Joural (aother publicatio of the MAA). [2] I that paper, it was show that the discrepacy ca ever be larger tha that foud i the Missouri Lotto. Formally, the followig result was proved. Theorem: Suppose we have a game i which a player choosig k of umbers purchases two tickets, selectig the umbers at radom. Let A be the evet first ticket matches j of umbers, ad B be the evet secod ticket matches j of umbers. Let the approximate probability P ( A B) P( A) + P( B) be expressed as s:1, ad the correct probability P( A B) = P( A) + P( B) P( A B) be expressed as t:1, where s ad t are rouded to the earest iteger. The t s = 0 or t s = 1. It ca also be show that o matter how may tickets might hypothetically be required, the approximate odds ad the correct odds will ever differ by more tha 1. Drugs. No life is so complicated that addig a dog to the mix ca t brig ew problems. Ad so it was with Rhapsody, the first, ad most eurotic, of the five dogs I ve lived with. Amog Rhapsody s quirks i her later years was a debilitatig fear of thuderstorms. Durig such a storm she could typically be foud i the bathtub diggig for all she was worth, lookig for a place to hide. Occasioally, she would crawl ito a kitche cabiet ad cower behid the pots ad pas 1 : 1 I am capable of focusig a camera, but ot whe stumblig out of bed without my glasses to deal with a pathetically patig dog i a cabiet at two i the morig.
Durig these episodes, Rhapsody was icosolable. As the frequecy of such evets icreased, somethig had to be doe. Rhapsody s vet recommeded a daily 20mg dose of Clomicalm, or as I liked to call it, puppy Paxil. All that was left to do was buy the pills, I aively thought. But the I wet o lie, searched for Clomicalm ad foud the followig ad [4]: Clomicalm For Dogs 20mg 30 Tablets Best Price: $22.23 Buy It! Clomicalm For Dogs 40mg 30 Tablets Best Price: $24.56 Buy It! Havig some familiarity with the cocept of uit pricig, I chose to purchase the 40mg pills. But ow I had to break the pills, which gave me problems. Specifically, a) What is the expected umber of days util we select a half-pill from the bottle? b) What is the probability that o day 59 there is a whole pill left? c) What is the expected umber of days util the last pill is broke? Oe practical way to solve these problems is to come home from the vet, dump the pills o the couter ad break them all. This solutio is mathematically uiterestig. Istead, we assume that each day Rhapsody takes half a pill ad that o ay give day a selectio is made at radom from the pill bottle. If the selectio is a half-pill, it is give to Rhapsody; if the selectio is a full pill, the pill is cut i half. Oe half is give to Rhapsody, ad the other is put back i the bottle. There are two reasoable models that ca be applied. 1. Assume that the probability of selectig ay particular whole pill is twice that of selectig ay particular half pill. 2. Assume that each pill ad half pill is equally likely to be selected. Oly the first model will be discussed here. Furthermore, it will be assumed that istead of a specific umber of pills (30), there is a arbitrary umber,. That way the limitig behavior ca be discussed, as i The Hat Check Problem ad War. To aid with uderstadig of the problem ad its solutios, we reformulate it i the traditioal settig of marbles ad urs. Imagie a ur cotaiig two marbles of each of differet colors. The two marbles of a give color represet the two halves of a give pill. Marbles are selected oe at a time at radom without replacemet. The three questios ca ow be rephrased thus: a) How log ca oe expect to wait to remove a secod marble of some color? b) What is the probability that whe oly two marbles are left, they are the same color? c) How log ca oe expect to wait util the ur o loger cotais two marbles of the same color? Let X represet the draw o which the first repeated color occurs. To address questio a), cosider the orderigs of the 2 marbles i which the first repeated color is at positio k.
The probability that the first time we remove a secod marble of a particular color is o the k th k 1 2 ( k 1)( k 1)!(2 k)! k 1 try is P( X k) = =. For Rhapsody s pills, = 30, (2)! the probabilities are show i the histogram below. I a probability histogram, the height of a bar gives the probability of a particular outcome. I this example, the left-most bar represets fidig a half-pill o day two. That evet has a probability of about 0.017 (a 1.7% chace of occurrig). probability k 0.08 0.06 0.04 0.02 0.00 5 10 15 20 25 30 day of first half pill Note that it is impossible to select a secod marble of a give color o the first try, as well as o ay try after the 31 st (Oe caot select more tha 30 differet colored marbles, so a match must occur o or before the 31 st draw.) The expected umber of marbles removed before drawig the secod of some color is give by 2 2!! E( X ) =. Whe = 30, the expected value is approximately 9.74866. The iterpretatio (2)! of this umber is that, while it will vary pill bottle to pill bottle, over may years the average umber of days I will have to wait after startig a ew bottle, util that lucky day whe I do t have to break a pill i the morig, is 9.74866. Below is a graph of E(X) vs. the umber of pills (i blue) alog with the umbers 2 (i red). days 200 150 100 50 20 40 60 80 100 umber of pills
It appears that the umber of days oe expects to wait to fid a half-pill grows very slowly. 2 2!! However, lim =, which meas that if we could buy bottles cotaiig more ad more (2)! pills, there would be o boud to how log we would have to wait for that half-pill. O the other had, it is true that the expected umber of days to wait grows very slowly compared to the umber of days the pills will last. The followig graph shows the ratio of E(X) to 2. ratio 0.4 0.3 0.2 0.1 0 20 40 60 80 100 umber of pills E( X ) I fact, lim = 0, so that the proportio of the time a pill bottle will last util we get a halfpill gets arbitrarily 2 small. Now cosider questio b), what is the probability that whe oly two marbles are left, they are the same color? I this case, orderigs of the marbles such as i which the last two marbles are the same color (represetig two halves of the same pill) are 1 relevat. The probability of such a orderig is =. For = 30, the chace of havig a 2 1 whole pill left at 59 is 1/59 or approximately 0.0169492. More geerally, oe ca ask for the probability that the last pill is broke o day k. Let Y represet the draw o which the last pill is broke. The 2 k 1 2 ( k 1)!(2 k)! k P( Y k) = =. The probabilities for = 30 pills are show i the (2)! probability histogram below. Sice it is impossible to break the last pill before day 30, or o day 60, the probabilities for these values of k are zero.
probability 0.08 0.06 0.04 0.02 0.00 35 40 45 50 55 60 day of last pill break The expected umber of days util the last pill is broke is the give by the expressio 2 k 1 k 2 ( k 1)!(2 k)! 2 k E( Y ) =. Whe = 30, the value is E(Y) = 51.2513, which as k = (2)! oe might expect is rather late i the 60 day process. The behavior for geeral is show i the graph o the left below (i blue) alog with the umbers 2 (i red). O the right is the ratio E(Y)/2. days 200 150 100 ratio 0.90 0.85 0.80 50 0.75 20 40 60 80 100 umber of pills 20 40 60 80 100 umber of pills The expressio for E(Y) does ot simplify icely as did the expressio for E(X), but all umerical E( Y ) evidece at this poit idicates that lim = 1. It appears that for bottles with a large umber 2 of pills, the proportio of the time we must wait util all pills are broke approaches 100% as a limit. This work has bee preseted to the Sectioal Meetig of the Mathematical Associatio of America, but may questios remai uaswered. I fact, i the process of my puttig together the graphs, I oticed that the two histograms may be exact mirror images of each other. Somethig ew to ivestigate! Crosswords. My curret obsessio is crossword puzzles. I have spet so much time lookig at the grids that I recetly bega to woder just how may differet grids are possible. A stadard New York Times crossword puzzle is a 15 15 array of black ad white squares, so that there are 225 total squares. Sice each square ca be oe of two colors, the total umber of possibilities is give by
2 225 = 53919893334301279589334030174039261347274288845081144962207220498432. However, ot all arragemets of black ad white squares are acceptable as crosswords. Most solvers would be shocked to ope up the paper ad fid oly two white squares, for example. Thus we must decide exactly which of the myriad possible arrays are acceptable. I other words, we must defie exactly what a crossword puzzle is, mathematically. A quick glace at some puzzles will help. These puzzles all have certai features i commo, motivatig the followig as the mathematical defiitio of a crossword puzzle: A array of black ad white squares The white squares are coected Each row (ad colum) has at least oe ru of white squares Every ru of white squares has legth at least three Rotatioal Symmetry
As far as I ca tell at this stage i my work, determiig the umber of crossword puzzles is goig to be extremely difficult, ad a log-term project. As is ofte the case i mathematical edeavors, a modest start was ecessary, ad I bega with the determiatio of the umber of possible rows. I keepig with our defiitio above, each row must cotai at least oe ru of white squares, ad every ru of white squares must have legth at least three. A few such rows are show. Oe approach to coutig the umber of rows is recurrece fid a rule that will give the umber of rows of legth 15, for example, if the umber of rows of each shorter legth is kow. It turs out that such a procedure is successful i this problem. To get started, let a : umber of acceptable rows of legth b : umber of acceptable rows of legth startig with a black square w : umber of acceptable rows of legth startig with a white square ad ote that a = b + w. If a row starts with a black square, the that square ca be removed, resultig i a acceptable row of legth 1; coversely to ay acceptable row of legth 1, a black square ca be added. Thus b a, so that = 1 a = a 1 + w (1) Similar reasoig leads to w a + w 1 (2) = 4 1 + Combiig equatios (1) ad (2) gives the desired recurrece relatio: a a a + a 1. (3) = 2 1 2 4 +
If the umber of rows of legth 1, 2, ad 4 are kow, the umber of rows of legth ca be computed. It is a simple matter to list all rows of legth 3, 4, 5 ad 6, so the umber of rows of ay desired legth ca be foud. Startig with = 3, the sequece 1, 3, 6, 10, 16, 26, 43, 71, 116, 188, 304, 492, 797, gives the umber of possible rows i a crossword. Upo fidig a sequece such as that above, it is advisable to check the O-Lie Ecyclopedia of Iteger Sequeces [5] to determie whether the sequece has bee foud by someoe else workig o the same problem, or perhaps as the solutio to a differet problem. I this case, the sequece was ukow, so I submitted it. This sequece is ow officially sequece A130578, further evidece that this is a good problem. A recurrece relatio is ice, but if oe wats to kow how may rows of legth 100 are possible, the umber of rows of every legth up to 100 would have to be computed. It would be preferable to have a explicit formula. There are stadard techiques for solvig recurrece relatios such as (3), which i this case yield the impressive as the formula for the umber of possible rows of legth. Thus the umber of rows of legth 100 is (i slightly simplified form) or 463686346096539499587. The atural ext step is to determie how may two-row arragemets, such as the oe show below, are possible. I this case the recurrece relatio is a bit more complicated: τ 2 τ τ + τ + τ + ϕ, = 1 2 3 4 2 2 2 2 2 where ϕ = a 2a 1 + a 2 a 3 a 4 2a 3 ad a is the umber of possible rows as give above. It is a fairly simple matter to implemet this sequece o a computer to fid the sequece (startig with = 3), 1, 9, 36, 98, 246, 646, 1777, 4883, 13120, 34642, 90976, 239160, 629427
OEIS: A133226. O the other had, a explicit formula is at this poit ukow. Mathematica ra for 24 hours tryig to fid oe with o luck. It seems certai that a recurrece relatio for three-row arragemets ca be foud, but due to the complexity of the two-row case, it may ot be the best path to pursue. I have programmed Mathematica to cout such arragemets by brute force, but this is extremely iefficiet, ad i fact the program s memory capacity is exceeded eve at the relatively small value = 12. If I wat to kow about the stadard 15 15 puzzle, I m goig to have to fid a ew approach. Thakfully, after all this time, I ve still got problems. Refereces: [1] Brodie, Marc. Avoidig Your Spouse at a Party Leads to War Mathematics Magazie Jue, 2002 [2] Brodie, Marc. A Tale of Two Tickets The College Mathematics Joural May, 2004. [3] Brualdi, Richard. Itroductory Combiatorics, 3 rd editio, Pretice Hall, New Jersey, 1999. [4] http://store.yahoo.com/cgi-bi/clik?etirelypets+gvwdfg+clomicalm1.html [5] The O-Lie Ecyclopedia of Iteger Sequeces http://www.research.att.com/~jas/sequeces/