- Shopping Bag ( 0 items )
What are your chances of dying on your next flight, being called for jury duty, or winning the lottery? We all encounter probability problems in our everyday lives. In this collection of twenty-one puzzles, Paul Nahin challenges us to think creatively about the laws of probability as they apply in playful, sometimes deceptive, ways to a fascinating array of speculative situations. Games of Russian roulette, problems involving the accumulation of insects on flypaper, and strategies for determining the odds of the ...
What are your chances of dying on your next flight, being called for jury duty, or winning the lottery? We all encounter probability problems in our everyday lives. In this collection of twenty-one puzzles, Paul Nahin challenges us to think creatively about the laws of probability as they apply in playful, sometimes deceptive, ways to a fascinating array of speculative situations. Games of Russian roulette, problems involving the accumulation of insects on flypaper, and strategies for determining the odds of the underdog winning the World Series all reveal intriguing dimensions to the workings of probability. Over the years, Nahin, a veteran writer and teacher of the subject, has collected these and other favorite puzzles designed to instruct and entertain math enthusiasts of all backgrounds.
If idiots A and B alternately take aim at each other with a six-shot revolver containing one bullet, what is the probability idiot A will win? What are the chances it will snow on your birthday in any given year? How can researchers use coin flipping and the laws of probability to obtain honest answers to embarrassing survey questions? The solutions are presented here in detail, and many contain a profound element of surprise. And some puzzles are beautiful illustrations of basic mathematical concepts: "The Blind Spider and the Fly," for example, is a clever variation of a "random walk" problem, and "Duelling Idiots" and "The Underdog and the World Series" are straightforward introductions to binomial distributions.
Written in an informal way and containing a plethora of interesting historical material, Duelling Idiots is ideal for those who are fascinated by mathematics and the role it plays in everyday life and in our imaginations.
This is a book for people who really like probability problems. There are, I think, a lot of people who fall into that category. Indeed, the editors of Parade, a magazine insert in millions of Sunday newspapers across America, thought a probabilistic question intriguing enough to put it on the cover of their issue of August 10, 1997. For the real connoisseur of probability, however, it was actually a pretty tame problem: "Your dog has a litter of four. Is it most likely that two are males and two are females?"
That question was posed in the "Ask Marilyn" column by the famously intelligent Marilyn vos Savant, who answered, "Nope! The most likely split is three males and one female, or three females and one male." That is correct, too, for the case of female/male births being equally likely. Two males and two females has a probability of 3/8, while the second case has probability of 1/2. Vos Savant, who has carved a successful writing career partly out of posing old math questions (with answers that have been known for centuries) to readers who find them new, doesn't give her fans the math behind the answer, but for this book, the doggy problem is just too elementary to be included as a legitimate "probability puzzler."
Please don't misunderstand me. While I occasionally think Vos Savant is just a bit too unrevealing of her debt to ancient mathematical lore (I suspect that many of her readers think she is the originator of the problems in her column), I do think she does provide a useful service by publishing such problems. Who could deny that it is a refreshing change to see math of any sort in a newspaper column, as compared to the more typical, seemingly endless rehashing of the supposed details of celebrity lives, or other similar sophomoric speculations? Vos Savant's column is written in the spirit of Laplace's famous dictum, "The theory of probabilities is at bottom nothing but common sense reduced to calculus," but, while clever, Laplace did overstate his argument just a bit. Once beyond the doggy type of question, probability theory can quickly become nonintuitive in the extreme, even for experienced analysts. That is, perhaps surprisingly, one of its most seductive and charming features.
It isn't hard to understand vos Savant's reluctance to put real math in her column, of course, as the quote from Agatha Christie at the beginning of this book accurately reflects how most math-innocents view technical analyses of any sort. As W. Somerset Maugham wrote in the first paragraph of his short story "Mr. Harrington's Washing," "Man has always found it easier to sacrifice his life than to learn the multiplication table." But if you have read this far, then you certainly don't fall into that category. The doggy problem is so simple (see Part II of this introduction) that one could literally write down all sixteen possibilities (sixteen, because there are four consecutive births, each with two possible outcomes, and 2 x 2 x 2 x 2 = 16) and then just count how many times each of the different situations occurs. In the jargon of mathematics, "we have several different events defined on a finite sample space, with each sample point the result of a Bernoulli sequence of four trials, with the probability of a success being one-half."
Now that is a mouthful. And I am not going to define any of those terms; if you know what that last sentence means, then you just passed the test qualifying you to get the most out of this book. If you don't know what that sentence said, then this book may be just a bit too much for you, at least for now. But that doesn't mean you shouldn't buy it. Do buy it and use it as a study supplement as you take an elementary course in probability.
As I declared before, the doggy problem is really just a routine drill problem, the sort of question that textbook authors put a dozen or so of at the end of each section of their books. Such problems are important to do as learning exercises, and every beginning student should do a number of them when first learning any new math topic. There are lots of such problems in all of the mostly excellent probability textbooks available today; so many, in fact, that Vos Savant will never run out of recyclable drill problems with which to dazzle her readers.
Once beyond the drill problem stage, however, most probability students are eager to try their new and powerful skills on more challenging, more interesting problems. That is the sort of problem you'll find in this book. And where did these problems come from, you may wonder. During the past twenty-five years, I have taught (and continue to teach) probability theory to undergraduate electrical engineering students at the University of New Hampshire. (My debt to hundreds of students who have patiently listened to me talk and scribble on the blackboard in EE647 is a very large one, indeed.) At the end of each term there are always pleas to provide some sort of extra credit work with which to bolster grades, and I have responded by offering what I call "Challenge Problems." These are optional problems (students have to accept the challenge before seeing the problem, and after seeing it, they can't change their minds) to be done as "take-homes" during the week before the final exam (independent work only), with unlimited time, and no partial credit. If a student gets the problem right, then I add five points to his or her final exam score. But if he or she gets it wrong, then I subtract five points.
Over the last twenty-five years, I have created perhaps a hundred or so such questions, and the ones I think are the best are included here. None, to my knowledge, has ever appeared in print before, at least not in the way posed here. The level of these problems is elementary, but that simply means that they all can be done with no mathematics beyond freshman calculus (and at least one of them can be solved with just arithmetic). Each problem has a detailed solution and extended discussion (often including computer illustrations using the powerful scientific application software called MATLAB) in the second half of the book. The problems are much like the famous Birthday Problem or the Buffon Needle Problem, neither of which is included here because they have become so easy to find in textbooks. The problems here are actually no more conceptually difficult than are those two classics, however, and I hope you have fun trying your luck on them.
Binary Numbers, the Doggy Problem, the Gulf War, and Shooting at Targets
The doggy problem can be solved by simply counting. If we let 1 denote a female birth and 0 denote a male birth, then the sixteen possibilities for a litter of four dogs are represented by the sixteen four-digit binary numbers from 0 to 15:
0 = 0000 8 = 1000 1 = 0001 9 = 1001 2 = 0010 10 = 1010 3 = 0011 11 = 1011 4 = 0100 12 = 1100 5 = 0101 13 = 1101 6 = 0110 14 = 1110 7 = 0111 15 = 1111.
As a mathematician would put it, "Each number is a sample point in the sample space of the experiment of having a litter of four puppies." If a male and female are equally likely to be born, then each sample point is equally likely, with a probability of 1/16. There are six sample points with two 1's and two 0's (the numbers 3, 5, 6, 9, 10, and 12), and there are eight sample points with either three 1's and one 0 or one 1 and three 0's (the numbers 7, 11, 13, and 14, and the numbers 1, 2, 4, and 8, respectively). Thus, the probability of two males and two females is 6/16 = 3/8, and the probability of either three males and one female or one male and three females is 8/16 = 1/2.
It was surprising to me that as elementary as this counting technique is, it at first appeared as if Marilyn vos Savant had been unaware of it. I say this because in her Parade column of June 14, 1998, she printed a letter asking her a question (the details of which are not important here); the correspondent ended by informing vos Savant that he was also asking "my other two heroes-Stephen Hawking and Kurt Vonnegut-[the same question]. I figure that my chances are 7 out of 8 that I will get at least one response from the three of you." To that, vos Savant replied, "I have a question for you: How in the world did you figure those chances?!"
The answer to Marilyn's question is an even simpler binary counting exercise than is the doggy problem. Just let 1 denote a response and 0 denote no response, and we see immediately that there are eight possibilities; the binary numbers 000 (no responses at all) to 111 (three responses). Seven of those eight binary numbers have at least one 1, so if the correspondent was assuming each of the eight sample points is equally likely (the probability of each response is 1/2 and that the responses are independent of each other), we then have the stated chances by inspection. If the probability of a response is not 1/2, then the chances are different but no more difficult to calculate. Suppose p is the probability of a response from a hero; thus, 1 - p is the probability of no response. Then the probability of no responses at all is [(1 - p).sup.3] and thus the probability of at least one response is 1 - [(1 - p).sup.3]. For example, if p - 1/6 then the chances for at least one response are 91 out of 216.
My original impression that vos Savant was unaware of all this turned out to be wrong, however, because in her September 6, 1998, column in Parade, she addressed the problem again by printing a letter from a reader asking just how the original correspondent calculated his odds. Marilyn's answer made it clear that she hadn't really meant to imply that she didn't know how to count in binary, as she correctly listed all eight possible combinations of reply possibilities from herself, Hawking, and Vonnegut. But then she stumbled again, with the following words about the original correspondent:
Then he incorrectly figured that these eight possibilities were equally likely. This is a lot like saying there are two possibilities regarding sunrise tomorrow: (1) The sun will rise in the morning; or (2) the sun will not rise in the morning. And, therefore, the chances are only fifty-fifty that the sun will rise!
Marilyn's first sentence is simply not a valid objection, and the rest of that passage is just irrelevant. The original correspondent certainly did not "figure incorrectly" by assuming each reply had a probability of 1/2. The probability of each reply could be anything from 0 to 1, and he was well within his rights to make the special (if perhaps overly optimistic) assumption of 1/2. Vos Savant's remarks concerning the sun are beside the point: The probabilities of the individual replies are due to individual human decisions, while the event of the sun rising tomorrow is the result of the physical laws of gravity and orbital mechanics. Probability has nothing to do with it. But, curiously, her particular imagery is reminiscent of Laplace's famously incorrect use of Bayes's theorem of conditional probability to compute the odds of the sun's rising tomorrow. Could this be the case of yet another classic puzzler that she once read about, and of which she has since forgotten the proper historical setting? (Laplace presented this calculation in his famous 1814 "Essai Philosophique des Probabilités.")
The counting method works well for problems that involve a small number of different possibilities, but in general we need a more powerful approach. The sophisticated math behind the doggy problem is simply the binomial theorem applied to a Bernoulli sequence of trials (which is a sequence with two characteristics: The trials are independent, and each trial has precisely two possible outcomes). If there are n births, with the probability of a female birth being p (and so the probability of a male birth is 1 - p), then the probability of k females and n - k males is given by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
where the binomial coefficient [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] denotes the number of different ways of selecting k things from n things. The factorial function is defined for positive integers as n! = n(n-1)(n-2) ... (2)(1). Notice, too, that since ([MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]) = 1, i.e., there is just one way to select all n things, then the binomial coefficient formula reduces to the special and important result that 0! = 1, not the zero that beginning students so often write.
For the doggy problem we have n = 4 and p = 1/2. So, the probability of two females (k = 2) and two males is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and the probability of either one female (k = 1) and three males, or three females (k = 3) and one male, is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
A far more interesting application of this simple math was reported in the Boston Globe on January 24, 1992 (p. 3), as part of the retrospective analyses then being conducted on the Gulf War. In particular, the Pentagon had gone on record with a claim that the Patriot antiaircraft missile system had "successfully engaged over 80 percent" of the Scud missiles Iraq had launched at Saudi Arabia. An MIT physicist, Theodore Postol, disputed that claim; it was a remarkable claim, too, as the Patriot was designed to counter relatively slow manned aircraft, not supersonic ballistic missiles.
Postol based his skepticism on what he saw after watching videotapes of fourteen Patriot-Scud engagements. There were thirteen misses and one probable hit. The Globe article ended with this quote from Professor Postol: "What are the odds I would see 13 misses and one hit if the Patriot was successfully shooting down 80 percent of the Scuds?"
No answer was given in the newspaper, but we can easily calculate it for ourselves using the doggy-problem math. Simply think of a hit as having the claimed probability of 0.8 (thus, a miss has a probability of 0.2), and if we assume independent engagements, we thus have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
as the probability that Professor Postol would see what he saw. The numbers work out to give a probability of less than [10.sup.-8], a value so small that most people would reject the Pentagon's claim of p = 0.8. Flipping a fair coin and getting twenty-six consecutive heads is more likely.
Excerpted from Duelling Idiots and other Probability Puzzlers by Paul J. Nahin
Copyright © 2000 by Princeton University Press
Excerpted by permission. All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.