azz8azz@L''# '@ j' ''#]'G@'@'''#'@@o''# M'?@A'P0''#'H@'5@'''#s'B@'%@5''# '@''#'H@' @?' ''''#'?'' '''#'< J'Gz''''#('G@'@'''#h'F@' @?'''#z'B@'-@'# '@ j' ''#]'G@'@'''#'@@o''# M'?@A'P0''#'H@'5@'''#s'B@'%@5''# '@''#'H@' @?' ''''#'?'' '''#'<`'Dz''''#('G@'@'''#h'F@' @?'''#z'B@'-@'+Another important contribution to the evolution of function came from the works of Fourier (1768-1830), who was concerned with the problem of heat flow in material bodies. Fourier considered temperature as a function of two variables, namely time and space. At some point, he conjectured that it would be possible to obtain a development of any function in a trigonometric series in a suitable interval. Fourier, however, never gave a mathematical proof of his assertion. The problem was later taken up by Dirichlet (1805-1859) who formulated the sufficient conditions so that a function may be represented by a Fourier series. To do so, Dirichlet needed to separate the concept of function from its analytical representation. He did this in 1837, casting the definition of function in terms of an arbitrary correspondence between variables representing numerical sets. A function, then, became a correspondence between two variables so that to any value of the independent variable, there is associated one and only one value of the dependent variable. Dirichlet also gave this well-known example of a function that is discontinuous in all the points of the domain [0,1]: 0 if x is a rational number 1 otherwise. With the development of set theory, initiated by Cantor (1845-1918), the notion of function continued to evolve. In the 20th century, function was extended to include all arbitrary correspondences satisfying the uniqueness condition between sets, numerical or non-numerical. The evolution of function has continued. From the notion of correspondence, mathematicians moved to the notion of relation. A close relative to the notion of function constitutes a primitive concept in category theory. In the theory of computation, for example, as in l-calculus, a function is not viewed as a relation but as a computational rule. In its beginnings, the notion of function was used to designate correspondences between geometrical entities. Through its association with the study of analytical expressions, function secured a fundamental place in the mainstream of mathematical thinking. To indicate the historical role of this association, Youschkevitch (1976/77) commented, It was the analytical method of introducing functions that revolutionized mathematics and, because of its extraordinary efficiency, secured a central place for the notion of function in all exact sciences. (p. 39) This association between analytical expressions and geometrical objects, in fact, revealed itself to be so highly fruitful that it still informs current mathematical practice. The Notion of Function and Physical Phenomena The notion of function did not appear in mathematics by chance. It arose, as Bento Caraa (1951) so masterfully has shown, as the necessary mathematical tool for the quantitative study of natural phenomena, begun by Galileo (1564-1642) and Kepler (1571-1630). Its development was based in the expressive possibilities enabled by the modern algebraic notation created by Vite (1540-1603) and, especially, by the analytic geometry introduced by Descartes and Fermat (1601-1665). In opposition to the verbalistic stance of the medieval scholastic thinking, Galileo underscored that mathematics was the most appropriate language for the study of nature. According to Galileo, to study a given phenomenon, it was necessary to measure quantities, identify regularities, and obtain relationships representing mathematical descriptions as simply as possible. The study of the movement of falling bodies, of the motion of planets, and more generally, of curvilinear motion, led to the consideration of direct and inverse proportionalities, as well as of polynomial and trigonometric functions. Mathematics and physics were at this point closely interconnected. Newton, regarded as one of the greatest mathematicians of all times, was also a prominent physicist. Many other mathematicians, such as Daniel Bernoulli, Euler, Lagrange, and Fourier were also very interested in physical problems. One of the most extraordinary discoveries of Newton provides an example on the nature of variation. Newton found that the law of variation of the motion of a material body of mass m, which in modern notation may be given by the function s(t), the space varying as a function of time, does not have a direct relationship with the force f acting upon that body. A simple relationship does not exist for the law of velocity given by v(t), in which v is the derivative of s, ds/dt. Such a relationship does exist, however, for the acceleration of the body, given by a(t), the second derivative of s, that is, dv/dt, and is expressed by a very simple law: f = ma. Functions are excellent tools to study problems of variation. A given quantity may vary in time, may vary in space, may vary with other quantities, and may even vary simultaneously in several dimensions. Such variation may be faster or slower, or may even disappear at some point. It may follow simple or complex patterns and obey very diverse restrictions. In its origin, the notion of function was associated with the notion of natural law. The idea of regularity was one of its more important constituent elements. In fact, we may consider three essential elements in the formation of the primitive 17th and 18th century concept of function: (a) the algebraic notation, carrying important aspects such as simplicity and rigor, allowing the manipulation of analytical expressions and condensing in itself a large quantity of information; (b) the geometrical representation, yielding a fundamental intuitive basis, of which a remarkable example is the association of the notions of tangent to one curve and derivative of a function; and (c) the connection with the concrete problems of the physical world, associated with the idea of regularity, providing the fundamental motivation and interest for the study of families of functions. After some time, mathematicians began considering functions to which no analytical expression corresponded, functions that did not have a simple geometrical representation, or even functions that did not have any relation to concrete physical situations. Thus, the concept of function began evolving on its own, moving away from its origins. That evolution had its dramatic moments. We may remember the moments of horror created by the monster-functions, such as continuous functions with no derivative at any point, that challenged the mathematicians prevailing intuitions and seemed to have no other purpose than to steal from them their confidence in their own reasoning. The evolution of the concept of function in this early phase was marked by two aspects: a concern with coherence and another with generality. The discussion, however, did not proceed at just an abstract level; it went along the major mathematical problems of the time. The Notion of Function and New Developments in Mathematics Today, mathematics is no longer as exclusively tied to the physical sciences as in the past. It has seen a significant increase in its domains of application, becoming an instrument for the study of the phenomena and situations of biological sciences, human and social sciences, business, communications, engineering, and technology. Mathematics constitutes an essential means of description, explanation, prediction, and control. All the applications of mathematics presuppose the notion of model. A mathematical model is a representation through relations and structures that intends to describe the elements found fundamental in a given situation while deliberately omitting secondary elements. A mathematical model can take several forms, but usually it is constituted by variables, relations among those variables, and their respective rates of change. @@@ @ ` BB80B B bK b ;Ja(`Aa'ppE(B n&("B (+( 'JPJ'The History of the Concept of Function and Some Educational Implications Joo Pedro PonteJGMMxGMMGMMJxx2dxx2dxx2dJaB 1/2. Clearly, the princess should choose to wait in Room A. The initial branching makes selecting Room A more likely. At the first branching, regardless which path is selected, it is still possible to choose a door into Room A. However, if the middle path is selected, it would not be possible to find Room B. These facts make the selection of Room A more likely. As in the game show problem where you cannot disconnect yourself from your first step of choosing an initial door, here you cannot disconnect yourself from the initial three branches and look at new problem of a simple random selection of one door from the six. Such a disconnection could be visualized as: figure 3 As to whether the young man actually found the room with the princess, well one wonders. Perhaps he found a car instead. Ideas for the Classroom For explaining the difference between dependent events and independent events and for experiencing the idea of conditional probabilities in the classroom, draw branching tree diagrams for probability experiments with multiple steps. For some of these experiments, make the second step dependent upon the outcome of the first step like: Roll a fair die. If the outcome is more than 3, you lose. If the outcome is 3 or less, then flip a fair coin (which can be simulated by selecting the middle digit of a number produced by a random number generator on a calculator) and if you get Heads (or say an even digit), then you win. Otherwise, you lose. Then work with other experiments, where the steps are independent of each other like: Flip a fair coin a first, second, and then a third time. If you get Heads on the third flip, you win. First consider probabilities like P(lose) for each game. Then consider conditional probabilities by conditioning on the first step like P(lose|odd on die) and P(lose|Heads on first flip). Observe what happens to the P(lose) for the different experiments when conditioning is added. Such classroom projects are fun and will help students avoid future traps of using independence at the wrong time. 1 Ask Marilyn by Marilyn Vos Savant; PARADE, 750 Third Ave., New York, New York 10017; page 22 of Sept. 9, 1990; page 28 of Dec. 2, 1990; page 12 of Feb. 17, 1991; pages 29 and 29 of July 7, 1991. @ @xGMMxxGMM%xGMMxxGMMnxGMMnxGMMCnxGMMnxMnxMnxMnxMnxMnxMnxMSnxMnxMhnxMnMnxMxxMnxMxxMZnxMxxMvnxMZxMZxMM)ZxMMZxMMZxMM;xx2dxx2dfhxx2dhxx2dŝxx2dxx2d/xx2dxx2dwxx2d`xx2d?xx2dxx2dʝxx2dxx2dxx2dxx2dxx2d#xx2dxx2dxx2d xx2dExx2d>xx2dxx2d)xx2d0xx2dxx2dRxx2d9xx2dXxx2djxx2dxx2d͝xx2d,vep #xprWEwdxprWE j Behind Closed Doors Gerald Thompson and Michele Benedict Many students and teachers have followed with great interest the continuing saga of the game show problem controversy presented in four different issues of PARADE magazine in the Ask Marilyn column written by Marilyn Vos Savant1. Marilyn is in fact correct for her solution to that problem which was first posed to her by one Craig F. Whitaker of Columbia, Maryland on page 22 of the Sept. 9, 1990 issue of PARADE in the following way: Suppose youre on a game show, and youre given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door -say, No.1- and the host, who knows whats behind the doors, opens another door -say, No.3- which has a goat. He then says to you, Do you want to pick door No.2? Is it to your advantage to switch your choice? Marilyn claims that in the game show problem, you should switch because switching doors will result in a 2/3 probability of winning whereas most people tend to believe that the probability is 1/2. In fact, readers have claimed the answer is 1/2 with tremendous amounts of emotional energy expended in the process. One reader even went so far as to tell Marilyn You are the goat! The probability of winning given that you switch is in fact 2/3. When those readers who are students or teachers try computer or classroom simulations of the problem they discover, using the Law of Large Numbers, that the statistical probability converges to 2/3. One amazed reader said Youll have to help rewrite the chapters on probability. Actually, this problem does not require a new chapter on probability but it does require, as an excellent starting place, an old chapter on probability, which can be found in any good introductory probability and statistics textbook. In such, refer to the sections on conditional probability, independent events, and dependent events. Conditional probability is denoted with symbols like P(A|B) representing the probability of A given B, and is defined to be the probability that both A and B happen divided by the probability that B happens, denoted P(AB)/P(B). Events A and B are independent if and only if P(A@B) = P(A)P(B) so for the independent events A and B, the probability that both A and B happen is simply the product of the separate probabilities. If A and B are independent, P(A|B) = P(A@B)/P(B) = P(A)P(B)/P(B) = P(A). Hence, intuitively speaking, events are independent from the perspective of conditional probability, if conditioning on one of the events does not affect the probability of the other event. Treating independent events as if they are dependent and visa versa are some of the common intuitive errors students make. For example, in three flips of a fair coin, given that the coin comes up showing Heads on the first two flips, students may feel that surely the third flip will most likely be Tails. After all, it seems only fair. Actually, the coin has no memory of being flipped those prior times and hence the coin will not cooperate with that sort of mistaken intuition. In symbols we see that P(Heads on third flip|first two flips were Heads) = 1/2 = P(Heads). The other case, treating dependent events as if they are independent, leads to errors like the ones in the game show problem. It is time to pay attention to and understand the subtleties of the game show problem. An old Chasidic saying is ...the hand, held before the eye, can hide the tallest mountain... which gives us a metaphor for the error. Knowing that the contestant will eventually be staring at exactly two doors, and knowing that behind one of those doors is a car and behind the other is a goat, Marilyns readers are blinded by another probability question. It is true that if a person randomly selects one door from two available doors, then the probability of selecting a particular door is 1/2. Many of the readers feel that what happened on the first door selection is independent of what will happen next. However, the rules of the game are cleverly concealing an initial branching step that changes the problem subtly but significantly away from the simple random selection of one from two doors and into a problem in which choosing the car on the second step, the switching step, depends on what you picked in the first place. Thus the game show problem has a sleight of pen twist to it. First, let us try to explain why the probability of winning, given that you switch, is 2/3. Instead of focusing on doors, which gets a bit confusing, focus on randomly selecting a prize from the following list: George the goat , Fred the goat, and a car. You could pick George, Fred, or the car with equally likely probabilities, specifically 1/3 each. After you pick a prize, the game show host must reveal, and remove from the list, one of the goats AND (now this is that fine detail that is so easy to overlook) THE HOST CANT TOUCH YOUR INITIAL PRIZE. If you were so lucky as to select a goat on the first stage of the game (which has a 2/3 probability of happening since you could select either George or Fred) you force the host to reveal and remove the only remaining goat from the list, leaving the car behind the untouched door. If you were so unlucky as to pick a car on the first stage (the probability of which is 1/3), the host has a choice of revealing and removing either George or Fred leaving the untouched door with the other goat. If you decide on the I will switch strategy, the problem of winning simply involves choosing and thereby reserving a goat on the first stage of the game which has the 2/3 probability of happening. This forces the host to remove the only remaining goat from the list allowing you to then switch to the door guaranteed to have a car. That is, in the switching strategy, to win you first have to lose (i.e. choose a goat) and to lose you first have to win (i.e. choose a car). Written below is the tree diagram for the problem showing that the conditional probability P(win|switch) = 2/3. Given you will switch pick George pick Fred pick the Car (p=1/3) (p=1/3) (p=1/3) Host throws Host throws Host throws out Fred and out George and out George or Fred you switch to you switch to end you switch to the the car and Win the car and Win. other goat and Lose. Interestingly enough, it is easy to get the answer 1/2 as the solution to a new but related and more probabilistic problem that allows for a switching probability. In this version, the probability of winning will vary from 2/3 to 1/3 depending on your likelihood of switching with 1/2 being the answer given that the contestant randomly switches or not with a 50% probability. Suppose the novice contestant has a fixed probability of switching prizes that is known to be p. That is, given a choice of switching, P(contestant will switch) = p and P(contestant wont switch) = 1-p. What is the probability that the contestant wins the car? In this problem, the contestant can win in one of two ways. The contestant can switch and win or the contestant can decide not to switch and win. P(wins car) = P(switches and wins) + P(doesnt switch and wins). Now for the and probabilities, we use the conditional equation after having solved for P(A@B), namely P(A@B) = P(A|B)P(B). So P(wins) = P(wins|switches)*P(switches)+P(wins|doesnt switch)*P(doesnt switch) We have just seen that P(wins|switches) = 2/3. Since winning the car without switching would simply involve selecting the car on the first stage, P(wins|doesnt switch) = 1/3. Hence P(wins car) = 2/3p + 1/3(1-p). We claim the solution of this version of the problem should make everybody happy. When the probability of switching is 1, the probability of winning becomes 2/3 (Marilyns solution since she said the contestant should definitely switch). Adding a random switching probability of 0.5, P(wins) = 2/6 + 1/6 = 1/2 (the Marilyns readers favorite solution). This pleases the intuition of the reader who is more focused on the random selection of one of the two doors at the end anyway. Finally, for the worst possible game strategy, if the probability of switching is 0, the probability of winning becomes 1/3. The game show problem is very similar to the famous Lady or the Tiger? problem. In the Lady or the Tiger? problem the reader will once again be faced with a problem with decisions to make that seem to have 50-50 probabilities associated with them. Once again, because of an initial selection step, the dependent events may seem to be independent. Instead of should you switch or not the question becomes should the princess wait in Room A or not? One version of the problem is given with the following diagram and explanation. figure 1 Imagine that the diagram shows paths through the garden of an exotic castle. Each path leads to a door which opens into one of two rooms in the castle. The rooms are labeled in the diagram as Room A and Room B. As legend has it, there was a princess in the castle who fell madly in love with a young man. Unfortunately, the king, father of said princess, despised the young man. In fact, the king decided to try to do away with the young man. The king told the princess that she should wait in one of the rooms shown above. In the other room, the king would place tigers. The young man would then be forced to pick a path through the garden and open one of the doors into one of the two rooms. If the young man found the room with the princess, he would get to keep the princess. If, however, he found the room with the tigers, then the tigers would get to keep him. The princess is now supposed to choose which room she wants to use for herself and which room is to be used by the tigers. Which room should she choose for herself to maximize the probability of being found? Once again we see a problem for which there seems to be a 50% probability for randomly selecting a prize. There are, after all, three paths to doors into Room A and three paths to doors into Room B and at any decision step we assume the random selection of a branch of one of those paths. It is however once again the initial branching process that changes to probability of finding each door so that they are not equally doors. Numbering the doors from top to bottom, and placing the values of the conditional probabilities of selecting each branch we see that: figure 2 P(Room A) = P(door 2) + P(door 3) + P(door 4) = P(I and b) + P(II) + P(III and c) = P(b|I)*P(I) + P(II) + P(c|III)*P(III) = (1/2)*(1/3) + (1/3) + (1/3)*(1/3) = 11/18 which is > 1/2. Clearly, the princess should choose to wait in Room A. The initial branching makes selecting Room A more likely. At the first branching, regardless which path is selected, it is still possible to choose a door into Room A. However, if the middle path is selected, it would not be possible to find Room B. These facts make the selection of Room A more likely. As in the game show problem where you cannot disconnect yourself from your first step of choosing an initial door, here you cannot disconnect yourself from the initial three branches and look at new problem of a simple random selection of one door from the six. Such a disconnection could be visualized as: figure 3 As to whether the young man actually found the room with the princess, well one wonders. Perhaps he found a car instead. Ideas for the Classroom For explaining the difference between dependent events and independent events and for experiencing the idea of conditional probabilities in the classroom, draw branching tree diagrams for probability experiments with multiple steps. For some of these experiments, make the second step dependent upon the outcome of the first step like: Roll a fair die. If the outcome is more than 3, you lose. If the outcome is 3 or less, then flip a fair coin (which can be simulated by selecting the middle digit of a number produced by a random number generator on a calculator) and if you get Heads (or say an even digit), then you win. Otherwise, you lose. Then work with other experiments, where the steps are independent of each other like: Flip a fair coin a first, second, and then a third time. If you get Heads on the third flip, you win. First consider probabilities like P(lose) for each game. Then consider conditional probabilities by conditioning on the first step like P(lose|odd on die) and P(lose|Heads on first flip). Observe what happens to the P(lose) for the different experiments when conditioning is added. Such classroom projects are fun and will help students avoid future traps of using independence at the wrong time. 1 Ask Marilyn by Marilyn Vos Savant; PARADE, 750 Third Ave., New York, New York 10017; page 22 of Sept. 9, 1990; page 28 of Dec. 2, 1990; page 12 of Feb. 17, 1991; pages 29 and 29 of July 7, 1991. ـ xGMMxxGMM%xGMMxxGMMnxGMMnxGMMCnxGMMnxMnxMnxMnxMnxMnxMnxMSnxMnxMhnxMnMnxMxxMnxMxxMZnxMxxMvnxMZxMZxMM)ZxMMZxMMZxMM;xx2dxx2dfhxx2dhxx2dŝxx2dxx2d/xx2dxx2dwxx2d`xx2d?xx2dxx2dʝxx2dxx2dxx2dxx2dxx2d#xx2dxx2dxx2d xx2dExx2d>xx2dxx2d)xx2d0xx2dxx2dRxx2d9xx2dXxx2djxx2dxx2d͝xx2d,vep #xprWEwdxprWE s !Given you will switch pick George pick Fred pick the Car (p=1/3) (p=1/3) (p=1/3) Host throws Host throws Host throws out Fred and out George and out George or Fred you switch to you switch to end you switch to the the car and Win the car and Win. other goat and Lose. Interestingly enough, it is easy to get the answer 1/2 as the solution to a new but related and more probabilistic problem that allows for a switching probability. In this version, the probability of winning will vary from 2/3 to 1/3 depending on your likelihood of switching with 1/2 being the answer given that the contestant randomly switches or not with a 50% probability. Suppose the novice contestant has a fixed probability of switching prizes that is known to be p. That is, given a choice of switching, P(contestant will switch) = p and P(contestant wont switch) = 1-p. What is the probability that the contestant wins the car? In this problem, the contestant can win in one of two ways. The contestant can switch and win or the contestant can decide not to switch and win. P(wins car) = P(switches and wins) + P(doesnt switch and wins). Now for the and probabilities, we use the conditional equation after having solved for P(AXB), namely P(AXB) = P(A|B)P(B). So P(wins) = P(wins|switches)*P(switches)+P(wins|doesnt switch)*P(doesnt switch) We have just seen that P(wins|switches) = 2/3. Since winning the car without switching would simply involve selecting the car on the first stage, P(wins|doesnt switch) = 1/3. Hence P(wins car) = 2/3p + 1/3(1-p). We claim the solution of this version of the problem should make everybody happy. When the probability of switching is 1, the probability of winning becomes 2/3 (Marilyns solution since she said the contestant should definitely switch). Adding a random switching probability of 0.5, P(wins) = 2/6 + 1/6 = 1/2 (the Marilyns readers favorite solution). This pleases the intuition of the reader who is more focused on the random selection of one of the two doors at the end anyway. Finally, for the worst possible game strategy, if the probability of switching is 0, the probability of winning becomes 1/3. The game show problem is very similar to the famous Lady or the Tiger? problem. In the Lady or the Tiger? problem the reader will once again be faced with a problem with decisions to make that seem to have 50-50 probabilities associated with them. Once again, because of an initial selection step, the dependent events may seem to be independent. Instead of should you switch or not the question becomes should the princess wait in Room A or not? One version of the problem is given with the following diagram and explanation. figure 1 Imagine that the diagram shows paths through the garden of an exotic castle. Each path leads to a door which opens into one of two rooms in the castle. The rooms are labeled in the diagram as Room A and Room B. As legend has it, there was a princess in the castle who fell madly in love with a young man. Unfortunately, the king, father of said princess, despised the young man. In fact, the king decided to try to do away with the young man. The king told the princess that she should wait in one of the rooms shown above. In the other room, the king would place tigers. The young man would then be forced to pick a path through the garden and open one of the doors into one of the two rooms. If the young man found the room with the princess, he would get to keep the princess. If, however, he found the room with the tigers, then the tigers would get to keep him. The princess is now supposed to choose which room she wants to use for herself and which room is to be used by the tigers. Which room should she choose for herself to maximize the probability of being found? Once again we see a problem for which there seems to be a 50% probability for randomly selecting a prize. There are, after all, three paths to doors into Room A and three paths to doors into Room B and at any decision step we assume the random selection of a branch of one of those paths. It is however once again the initial branching process that changes to probability of finding each door so that they are not equally doors. Numbering the doors from top to bottom, and placing the values of the conditional probabilities of selecting each branch we see that: figure 2 P(Room A) = P(door 2) + P(door 3) + P(door 4) = P(I and b) + P(II) + P(III and c) = P(b|I)*P(I) + P(II) + P(c|III)*P(III) = (1/2)*(1/3) + (1/3) + (1/3)*(1/3) = 11/18 which is > 1/2. Clearly, the princess should choose to wait in Room A. The initial branching makes selecting Room A more likely. At the first branching, regardless which path is selected, it is still possible to choose a door into Room A. However, if the middle path is selected, it would not be possible to find Room B. These facts make the selection of Room A more likely. As in the game show problem where you cannot disconnect yourself from your first step of choosing an initial door, here you cannot disconnect yourself from the initial three branches and look at new problem of a simple random selection of one door from the six. Such a disconnection could be visualized as: figure 3 As to whether the young man actually found the room with the princess, well one wonders. Perhaps he found a car instead. Ideas for the Classroom For explaining the difference between dependent events and independent events and for experiencing the idea of conditional probabilities in the classroom, draw branching tree diagrams for probability experiments with multiple steps. For some of these experiments, make the second step dependent upon the outcome of the first step like: Roll a fair die. If the outcome is more than 3, you lose. If the outcome is 3 or less, then flip a fair coin (which can be simulated by selecting the middle digit of a number produced by a random number generator on a calculator) and if you get Heads (or say an even digit), then you win. Otherwise, you lose. Then work with other experiments, where the steps are independent of each other like: Flip a fair coin a first, second, and then a third time. If you get Heads on the third flip, you win. First consider probabilities like P(lose) for each game. Then consider conditional probabilities by conditioning on the first step like P(lose|odd on die) and P(lose|Heads on first flip). Observe what happens to the P(lose) for the different experiments when conditioning is added. Such classroom projects are fun and will help students avoid future traps of using independence at the wrong time. 1 Ask Marilyn by Marilyn Vos Savant; PARADE, 750 Third Ave., New York, New York 10017; page 22 of Sept. 9, 1990; page 28 of Dec. 2, 1990; page 12 of Feb. 17, 1991; pages 29 and 29 of July 7, 1991. nnxGMMnxGMMhnxGMMnGMMnxGMMxxGMMnxGMMxxGMMZnxGMMxxGMMvnxGMMZxGMMZxMM)ZxMMZxMMZxMM`xx2d?xx2dxx2dʝxx2dxx2dxx2dxx2dxx2d#xx2dxx2dxx2d xx2dExx2d>xx2dxx2d)xx2d0xx2dxx2dRxx2d9xx2dXxx2djxx2dxx2d͝xx2d0vep #xprWEwdxprWE $Behind Closed Doors Gerald Thompson and Michele Benedict Many students and teachers have followed with great interest the continuing saga of the game show problem controversy presented in four different issues of PARADE magazine in the Ask Marilyn column written by Marilyn Vos Savant1. Marilyn is in fact correct for her solution to that problem which was first posed to her by one Craig F. Whitaker of Columbia, Maryland on page 22 of the Sept. 9, 1990 issue of PARADE in the following way: Suppose youre on a game show, and youre given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door -say, No.1- and the host, who knows whats behind the doors, opens another door -say, No.3- which has a goat. He then says to you, Do you want to pick door No.2? Is it to your advantage to switch your choice? Marilyn claims that in the game show problem, you should switch because switching doors will result in a 2/3 probability of winning whereas most people tend to believe that the probability is 1/2. In fact, readers have claimed the answer is 1/2 with tremendous amounts of emotional energy expended in the process. One reader even went so far as to tell Marilyn You are the goat! The probability of winning given that you switch is in fact 2/3. When those readers who are students or teachers try computer or classroom simulations of the problem they discover, using the Law of Large Numbers, that the statistical probability converges to 2/3. One amazed reader said Youll have to help rewrite the chapters on probability. Actually, this problem does not require a new chapter on probability but it does require, as an excellent starting place, an old chapter on probability, which can be found in any good introductory probability and statistics textbook. In such, refer to the sections on conditional probability, independent events, and dependent events. Conditional probability is denoted with symbols like P(A|B) representing the probability of A given B, and is defined to be the probability that both A and B happen divided by the probability that B happens, denoted P(AB)/P(B). Events A and B are independent if and only if P(AxB) = P(A)P(B) so for the independent events A and B, the probability that both A and B happen is simply the product of the separate probabilities. If A and B are independent, P(A|B) = P(AXB)/P(B) = P(A)P(B)/P(B) = P(A). Hence, intuitively speaking, events are independent from the perspective of conditional probability, if conditioning on one of the events does not affect the probability of the other event. Treating independent events as if they are dependent and visa versa are some of the common intuitive errors students make. For example, in three flips of a fair coin, given that the coin comes up showing Heads on the first two flips, students may feel that surely the third flip will most likely be Tails. After all, it seems only fair. Actually, the coin has no memory of being flipped those prior times and hence the coin will not cooperate with that sort of mistaken intuition. In symbols we see that P(Heads on third flip|first two flips were Heads) = 1/2 = P(Heads). The other case, treating dependent events as if they are independent, leads to errors like the ones in the game show problem. It is time to pay attention to and understand the subtleties of the game show problem. An old Chasidic saying is ...the hand, held before the eye, can hide the tallest mountain... which gives us a metaphor for the error. Knowing that the contestant will eventually be staring at exactly two doors, and knowing that behind one of those doors is a car and behind the other is a goat, Marilyns readers are blinded by another probability question. It is true that if a person randomly selects one door from two available doors, then the probability of selecting a particular door is 1/2. Many of the readers feel that what happened on the first door selection is independent of what will happen next. However, the rules of the game are cleverly concealing an initial branching step that changes the problem subtly but significantly away from the simple random selection of one from two doors and into a problem in which choosing the car on the second step, the switching step, depends on what you picked in the first place. Thus the game show problem has a sleight of pen twist to it. First, let us try to explain why the probability of winning, given that you switch, is 2/3. Instead of focusing on doors, which gets a bit confusing, focus on randomly selecting a prize from the following list: George the goat , Fred the goat, and a car. You could pick George, Fred, or the car with equally likely probabilities, specifically 1/3 each. After you pick a prize, the game show host must reveal, and remove from the list, one of the goats AND (now this is that fine detail that is so easy to overlook) THE HOST CANT TOUCH YOUR INITIAL PRIZE. If you were so lucky as to select a goat on the first stage of the game (which has a 2/3 probability of happening since you could select either George or Fred) you force the host to reveal and remove the only remaining goat from the list, leaving the car behind the untouched door. If you were so unlucky as to pick a car on the first stage (the probability of which is 1/3), the host has a choice of revealing and removing either George or Fred leaving the untouched door with the other goat. If you decide on the I will switch strategy, the problem of winning simply involves choosing and thereby reserving a goat on the first stage of the game which has the 2/3 probability of happening. This forces the host to remove the only remaining goat from the list allowing you to then switch to the door guaranteed to have a car. That is, in the switching strategy, to win you first have to lose (i.e. choose a goat) and to lose you first have to win (i.e. choose a car). Written below is the tree diagram for the problem showing that the conditional probability P(win|switch) = 2/3. Interestingly enough, it is easy to get the answer 1/2 as the solution to a new but related and more probabilistic problem that allows for a switching probability. In this version, the probability of winning will vary from 2/3 to 1/3 depending on your likelihood of switching with 1/2 being the answer given that the contestant randomly switches or not with a 50% probability. Suppose the novice contestant has a fixed probability of switching prizes that is known to be p. That is, given a choice of switching, P(contestant will switch) = p and P(contestant wont switch) = 1-p. What is the probability that the contestant wins the car? In this problem, the contestant can win in one of two ways. The contestant can switch and win or the contestant can decide not to switch and win. P(wins car) = P(switches and wins) + P(doesnt switch and wins). Now for the and probabilities, we use the conditional equation after having solved for P(AXB), namely P(AXB) = P(A|B)P(B). So P(wins) = P(wins|switches)*P(switches)+P(wins|doesnt switch)*P(doesnt switch) We have just seen that P(wins|switches) = 2/3. Since winning the car without switching would simply involve selecting the car on the first stage, P(wins|doesnt switch) = 1/3. Hence P(wins car) = 2/3p + 1/3(1-p). We claim the solution of this version of the problem should make everybody happy. When the probability of switching is 1, the probability of winning becomes 2/3 (Marilyns solution since she said the contestant should definitely switch). Adding a random switching probability of 0.5, P(wins) = 2/6 + 1/6 = 1/2 (the Marilyns readers favorite solution). This pleases the intuition of the reader who is more focused on the random selection of one of the two doors at the end anyway. Finally, for the worst possible game strategy, if the probability of switching is 0, the probability of winning becomes 1/3. The game show problem is very similar to the famous Lady or the Tiger? problem. In the Lady or the Tiger? problem the reader will once again be faced with a problem with decisions to make that seem to have 50-50 probabilities associated with them. Once again, because of an initial selection step, the dependent events may seem to be independent. Instead of should you switch or not the question becomes should the princess wait in Room A or not? One version of the problem is given with the following diagram and explanation. figure 1 Imagine that the diagram shows paths through the garden of an exotic castle. Each path leads to a door which opens into one of two rooms in the castle. The rooms are labeled in the diagram as Room A and Room B. As legend has it, there was a princess in the castle who fell madly in love with a young man. Unfortunately, the king, father of said princess, despised the young man. In fact, the king decided to try to do away with the young man. The king told the princess that she should wait in one of the rooms shown above. In the other room, the king would place tigers. The young man would then be forced to pick a path through the garden and open one of the doors into one of the two rooms. If the young man found the room with the princess, he would get to keep the princess. If, however, he found the room with the tigers, then the tigers would get to keep him. The princess is now supposed to choose which room she wants to use for herself and which room is to be used by the tigers. Which room should she choose for herself to maximize the probability of being found? Once again we see a problem for which there seems to be a 50% probability for randomly selecting a prize. There are, after all, three paths to doors into Room A and three paths to doors into Room B and at any decision step we assume the random selection of a branch of one of those paths. It is however once again the initial branching process that changes to probability of finding each door so that they are not equally doors. Numbering the doors from top to bottom, and placing the values of the conditional probabilities of selecting each branch we see that: figure 2 P(Room A) = P(door 2) + P(door 3) + P(door 4) = P(I and b) + P(II) + P(III and c) = P(b|I)*P(I) + P(II) + P(c|III)*P(III) = (1/2)*(1/3) + (1/3) + (1/3)*(1/3) = 11/18 which is > 1/2. Clearly, the princess should choose to wait in Room A. The initial branching makes selecting Room A more likely. At the first branching, regardless which path is selected, it is still possible to choose a door into Room A. However, if the middle path is selected, it would not be possible to find Room B. These facts make the selection of Room A more likely. As in the game show problem where you cannot disconnect yourself from your first step of choosing an initial door, here you cannot disconnect yourself from the initial three branches and look at new problem of a simple random selection of one door from the six. Such a disconnection could be visualized as: figure 3 As to whether the young man actually found the room with the princess, well one wonders. Perhaps he found a car instead. Ideas for the Classroom For explaining the difference between dependent events and independent events and for experiencing the idea of conditional probabilities in the classroom, draw branching tree diagrams for probability experiments with multiple steps. For some of these experiments, make the second step dependent upon the outcome of the first step like: Roll a fair die. If the outcome is more than 3, you lose. If the outcome is 3 or less, then flip a fair coin (which can be simulated by selecting the middle digit of a number produced by a random number generator on a calculator) and if you get Heads (or say an even digit), then you win. Otherwise, you lose. Then work with other experiments, where the steps are independent of each other like: Flip a fair coin a first, second, and then a third time. If you get Heads on the third flip, you win. First consider probabilities like P(lose) for each game. Then consider conditional probabilities by conditioning on the first step like P(lose|odd on die) and P(lose|Heads on first flip). Observe what happens to the P(lose) for the different experiments when conditioning is added. Such classroom projects are fun and will help students avoid future traps of using independence at the wrong time. 1 Ask Marilyn by Marilyn Vos Savant; PARADE, 750 Third Ave., New York, New York 10017; page 22 of Sept. 9, 1990; page 28 of Dec. 2, 1990; page 12 of Feb. 17, 1991; pages 29 and 29 of July 7, 1991. ;C gcxGMMxxGMM%xGMMxxGMMnxGMMnxGMMCnxGMMnxMnxMnxMnxMnxMnxMnxM.nxMnxMhnxMnMnxMxxMxMnxMnMnxMxxMZnxMnMxMxxMvnxMZxMZxMM)ZxMMZxMMZxMMxMMnxMM;xx2dxx2dfhxx2dhxx2dŝxx2dxx2d/xx2dxx2dvxx2dxx2dxx2dxx2dxx2d"xx2dxx2dxx2d xx2dExx2d?xx2dxx2dxx2dxx2dxx2d0xx2dxx2dRxx2d9xx2dXxx2djxx2dxx2dΝxx2dxx2d+vep #xprWEwdxprWE + HH , 5m5m -h >> .QHd SPNT @"% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "Ie6 72 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "234 184 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "d SPNTd SPNT0H~d SPNT0Z~ơd SPNTdSPNT&~d SPNTP2_b 0 gx x6 end 1 setTxMode +Room Ad SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *$Room Bd SPNTQP|UXd SPNTQa|fXd SPNTQr|wXd SPNTQ|Xd SPNTQ|Xd SPNTQ|Xd SPNTd SPNT"t~d SPNT"td SPNT"Y-d SPNT"Y-Qd SPNT"Y-P d SPNT"-Q d SPNT"-Qd SPNT"-QdSPNT5md SPNT @ UUUU"% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "6g53 249 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "l283 365 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "d SPNTd SPNT ")d SPNT")Ad SPNT"\d SPNT"\Qd SPNT"\P d SPNT"D\Q d SPNT"D\Qd SPNT"D\Qd SPNTdSPNT& F \d SPNTP2_b 0 gx x6 end 1 setTxMode (LId SPNTdSPNT&V.qd SPNTP2_b 0 gx x6 end 1 setTxMode +IId SPNTdSPNT&4@JVd SPNTP2_b 0 gx x6 end 1 setTxMode (BFIIId SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode (ad SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode +bd SPNTdSPNT&-Cd SPNTP2_b 0 gx x6 end 1 setTxMode (;cd SPNTdSPNT&<Rd SPNTP2_b 0 gx x6 end 1 setTxMode +dd SPNTdSPNT&Ukd SPNTP2_b 0 gx x6 end 1 setTxMode (ced SPNTd SPNTd SPNT0id SPNT0Ed SPNTdSPNT&3Id SPNTP2_b 0 gx x6 end 1 setTxMode (ARoom Ad SPNTdSPNT&Wmd SPNTP2_b 0 gx x6 end 1 setTxMode *$Room Bd SPNTQ Xd SPNTQXd SPNTQ',Xd SPNTQ8=Xd SPNTQKPXd SPNTQ\aXd SPNTd SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode (1d SPNTdSPNT&$d SPNTP2_b 0 gx x6 end 1 setTxMode *2d SPNTdSPNT&5d SPNTP2_b 0 gx x6 end 1 setTxMode *3d SPNTdSPNT&0Fd SPNTP2_b 0 gx x6 end 1 setTxMode *4d SPNTdSPNT&DZd SPNTP2_b 0 gx x6 end 1 setTxMode *5d SPNTdSPNT&Tjd SPNTP2_b 0 gx x6 end 1 setTxMode *6d SPNTd SPNTdSPNT>d SPNT @ UUUU"% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "?g62 417 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "241 529 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "d SPNTd SPNT 0 d SPNT0͡d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode (Room Ad SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *$Room Bd SPNTQXd SPNTQXd SPNTQXd SPNTQXd SPNTQXd SPNTQXd SPNTd SPNT"d SPNT"d SPNT"d SPNT"d SPNT"d SPNT"d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode (1d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *2d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *3d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *4d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *5d SPNTdSPNT&d SPNTP2_b 0 gx x6 end 1 setTxMode *6dSPNTM;xx2d(xx2d;xx2d,;xx2d%xx2d;xx2dם#;xx2d%xx2d;xx2d.xx2d;xx2d<xx2d;xx2d՝xx2d;xx2d'xx2d;xx2dxx2d?Ixx2dϝ;xx2dxx2d;xx2dxx2d$xx2d;xx2dxx2d (;xx2d(;xx2d(hxx2d7,7̀%%nn@&# @$C$#@5@ @ [@ @@@ !@ @ @ % @ @ ' RX bR $@ o.@$@$@ @ @2C@ @ ɀ!$$p@ !@$@$p 79@@$B@$pBC u %u y$$p@)+!@$@$p 79@@$B@$pBC u %u y$$p@)+WDBN5DOORS (Word)version1A0Ll9Direct Drive 70:MSWDMESA JOURNAL:Winter 1992:DOORS.113:.) i !WDBN5DOORS (Word))uK_Direct Drive 70:MSWDMESA JOURNAL:Winter 1992:DOORS.113:.) r"i$WDBNdoors pt 2d)!-ADirect Drive 70:MSWDMESA JOURNAL:Winter 1992:DOORS.113:.m) %#+$,%-&.WDBNbdoors - working version+P-ADirect Drive 70:MSWDMESA JOURNAL:Winter 1992:DOORS.113:.-() /WDBNiARTICLE.PONTE te0ng?.Ponte:MSWD71M<) DWDBNcalendar jan 91Direct Drive 70:MSWD[y) ;%u y}'r̀%%nn@&# @$C$#@5@ @ [@ @@@ !@ @ @ % @ @ ' RX bR $@ o.@$@$@ @ @2C@ @ ɀ!$$p@ !@$@$p 79@@$B@$pBC u %u y$$p@)-4Bd&3**`Dd0\ӎ ,rMD`**3&dB4e02 ӎ ,rެS**3&r@ 0@ ' R\@ _4 'oR$@$@ @ @2C@ @ ɀ!$$p@ !@$@$p 79@@$B@$pBC u %u y$$p@)+u@%%nn@3# @$C$#@1@ @ [@ @@@ !@ @ @ / @ 0@ ' Q U 4@ s$@$@ @ @2C@ @ ɀ!$$p@ !@$@$p 79@@$B@$pBC u %u y$$p@)+'̀%%nn@3# @$C$#@5@ @ [@ @@@ !@ @ @ / @ 0@ ' RX bR $@ o.@$@$@ @ @2C@ @ ɀ!$$p@ !@$@$p 79@@$B@$pBC u %u y$$p@)+nd New Developments in Mathematics Today, mathematics is no longer as exclusively tied to the physical sciences as in the past. It has seen a significant increase in its domains of application, becoming an instrument for the study of the phenomena and situations of biological sciences, human and social sciences, business, communications, engineering, and technology. Mathematics constitutes an essential means of description, explanation, prediction, and control. All the applications of mathematics presuppose the notion of model. A mathematical model is a representation through relations and structures that intends to describe the elements found fundamental in a given situation while deliberately omitting secondary elements. A mathematical model can take several forms, but usually it is constituted by variables, relations among those variables, and their respective rates of change. 5ZdM?ZdM;;;;(;;;;;+;_;;;;;;;;;;;.;j;,;;i;};[;T;;;<;;;#;zz8azzazz6azz:q zzazz8azz