#1: A Wimbledon-type (elimination) tennis tournament has 256 first-round entrants. How many total matches will there be in the tournament? You have three seconds to answer.
#2: An acquaintance tells you that he has two children. You ask him if at least one of the children is a boy and he says yes. What is the probability that he has two sons? (This is a mathematical, not a scientific, question, so assume that equal numbers of boys and girls are born and live to adulthood.)
#3: You have a scale that gives an accurate reading to the ounce. There are seven piles, each containing seven weights. The seven weights in any given pile are equal to each other. The first pile has weights labeled “one pound,” the second pile has weights labeled “two pounds,” and so on. One of the piles contains weights that are each either one ounce too light or one ounce too heavy. You may weigh whichever weights you like. How many weighings are required to determine which weights are mislabeled and whether they are an ounce heavy or an ounce light?
#4: I have heard it said that the earth is smoother than a billiard ball (i.e., if you expanded the billiard ball to the size of the earth, the highs and lows of its surface would be greater than the mountains and valleys of the earth). I have no idea whether this is true, but for this problem let us assume that the earth itself is perfectly smooth. Let us also assume that the circumference of the earth is exactly 25,000 miles (i.e., 132 million feet).
A piece of string 132 million feet long is stretched taut around the earth. You add a foot of string to the string, so the string is now 132,000,001 feet long. If you prop up the string equidistantly around the earth, how tall will the props have to be? Not only will you find this answer surprising, you will also find another surprise given with the answer.
#5: How many ways are there to change a dollar?
#6: You are on the TV game show, Let’s Make a Deal. There are three closed doors: behind one, a car has been randomly placed; behind the others are goats. You win whatever is behind the door that you end up choosing, and you want to win the car. The host, Monte Hall, plays fair; he does nothing to trick you or to help you.
Monte says, “You choose a door. Then I’ll open a different door, one that I know has a goat behind it. Then you may stick with your original choice or may switch to the other closed door.”
Should you: (A) stick with your original choice? (B) Switch? (C) It makes no difference?
#7: This is a relatively easy one to boost your spirits. In a two-mile race, going twice around a one-mile oval track, your car averages thirty miles an hour for the first lap. How many miles per hour must you average on the second lap to average sixty miles an hour for the entire race?
#8: How many randomly selected people would you have to invite to a dinner party to be certain that
At least three all know each other, or1
At least three all do not know each other.2
#9: An anthropologist and spouse go to a party at which there are ten people, including the anthropologist and spouse. Everyone at the party, except possibly the anthropologist, shakes hands with a different number of people (i.e., one person shakes hands with no one, one with one person, one with two people, etc.).
Who is the anthropologist married to (i.e., how many people did the anthropologist’s spouse shake hands with)?
#10: A hat contains three cards: one white on both sides; one white on one side and red on the other; and one red on both sides. You pick a card and place it on the table with white showing. You do not look at the other side.
I size up the situation and offer to give you 3–2 odds ($1.50 to $1.00) that the other side is white. Should you take the bet?
#11: Three men pay ten dollars each for a thirty-dollar hotel room. The concierge later realizes that he should have charged only twenty-five dollars for the room and returns five dollars to the men, who then tip him two dollars. So, each man paid nine dollars (the original ten minus one returned), and the concierge got two dollars. Where did the other dollar go?
#12: This may be the oldest of all brainteasers, and you may well know it in one of its many forms. But it is a beauty and worth including for those who are reading it here for the first time.
There are two doors. One is the “door of life.” The other is the “door of death.” In front of each door is a man. One of the men—you do not know which—always tells the truth. The other man always lies.
Your life hangs on the one question you are permitted to ask one of the men. After asking the question, you choose a door. If you choose the door of life, you live (and, what the hell, marry the princess, win billions of dollars, and get to meet Michael Jordan). If you choose the door of death, you are a memory.
What question do you ask?
#13: This is one of Martin Gardner’s favorites of the easier brainteasers.3
Miss Green, Miss Black, and Miss Blue are out for a stroll. One is wearing a green dress, one a black dress, and one a blue dress.
“Isn’t it odd,” says Miss Blue, “that our dresses match our last names, but not one of us is wearing a dress that matches her own name?”
“So what?” replies the lady in black.
What was the color of each lady’s dress?
#14: This brainteaser is at least 150 years old.
Your boss offers you a choice of two bonuses: (1) fifty dollars after six months and a semiannual increase of five dollars, or (2) one hundred dollars after a year and an annual increase of twenty dollars. Which bonus will prove more lucrative?
#15: This is a great one discovered by Martin Gardner—one that shows how God forgot to wire our brains so that we know probability by intuition.
You are playing roulette. You pick any triplet you like (i.e., black-black-black, red-red-black, etc.). I then pick a different triplet. The game ends when someone wins, but we will play as many times as you like.
I will give you three dollars each time you win; you give me two dollars each time I win. There are even odds, and I am giving you 3–2. It is a great bet for you, right?
You choose black-black-black, and I choose red-black-black. You will win if the first three spins are black. Should you take the bet?
#16: Here is an old chestnut that should be easy, but nearly everyone gets it wrong, even after they have heard the answer ten times:
If a hundred chickens eat a hundred bushels in a hundred days, how many bushels will ten chickens eat in ten days?
#17: This famous brainteaser, “Dudeney’s Cigar Puzzle,” rivals the tennis brainteaser for elegance and requires absolutely no mathematical knowledge.
You and a friend are sitting at a normal-sized, square (topped) table. You each have more cigars than are needed for this problem. The cigars, which are identical, are normal cigars (i.e., normal size, flat at one end and pointed at the other). You place a cigar on the table. Your friend then places a cigar anywhere on the table he likes, as long as it does not touch your cigar. You then place a cigar anywhere on the table you wish, as long as it does not touch any other cigar. And so it goes. The winner is the last person to successfully place a cigar.
Who wins, and why?
Hint: The “why” is crucial. There are acceptable arguments for either person’s winning.
#18: You have two barrels. Barrel 1 is filled with 10 gallons of red paint. Barrel 2 is filled with 10 gallons of blue paint. You take a gallon of red paint from barrel 1 and pour it into barrel 2, mixing the red and blue paint thoroughly. You then take a gallon of the mixture in barrel 2 and pour it into barrel 1.
Which is now greater: the red paint in barrel 1 or the blue paint in barrel 2?
#19: Two poles, each fifty-feet tall, are constructed somewhere on a football field. A sixty-foot rope is strung from the top of one to the top of the other. The lowest point of the rope is twenty feet from the ground. How far apart are the poles?
#20: This is a great one from Martin Gardner (who heard it from University of Oregon emeritus professor of psychology Ray Hyman, who, in turn, read it in a monograph by Gestalt psychologist Karl Duncker).
One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. The narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit.
The monk ascended the path toward the temple at varying rates of speed, stopping many times along the way to rest and to eat the dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation, he began his journey back along the same path, starting at sunrise and again walking at variable speeds with many pauses along the way. His average speed descending was, of course, greater than his average climbing speed.
Is there any way that there could be a spot along the path that the monk will occupy on both trips at precisely the same time of day. If not, prove that there cannot be such a spot.
#21: This is a tough one. You have two pieces of string (different lengths and different thickness). Each burns at varying rates, and the two pieces burn at different varying rates. (For example, when you light the end of the first piece, it might burn a foot in the first minute, half a foot in the second minute, two feet in the third, whatever. When you light the end of the second piece of string, it might burn at entirely different rates.)
All that the two strings have in common is that they each burn completely in an hour. How can you use these strings to measure forty-five minutes?
#22: You and two other people, Tom and Harry, who are both very quick but not quite as quick as you, sit facing each other. Each of you has a red or a white hat on. Each person can see the hats of the other two people, but not his own. You see that Tom and Harry have red hats. Can you deduce the color of your hat?
#23: You come to a fork in the road. One path leads east, the other west. One of the paths leads to heaven, the other to hell, but you do not know which is which.
At the fork are two men; one always tells the truth, and one always lies. You do not know which man is which.
You can ask one question to learn which path is the path to heaven. What do you ask?
#24: This is a nifty one from Marilyn vos Savant.
It is taking you too long to get to your workplace. You figure that you will leave later and arrive earlier. How do you accomplish this?
#25: You have a girlfriend who lives downtown and a girlfriend who lives uptown. The uptown and downtown trains alternate, first one and then the other. So, you figure, you will leave to chance which girlfriend you will visit each night, taking whichever train comes in first at whatever time you happen to get to the subway station. This way you will visit each girlfriend the same number of times.
But it does not work out that way. About five times out of every six, the uptown train comes. How come?
Answer #1: If you were permitted more time, you could add the matches played in the first round, the second round, etc.: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1. This is the way in which the intelligent nonmathematician would go about solving the problem. But solving the problem this way does not capture the elegance of the mathematical insight.
What is this mathematical solution? Consider this: There is one and only one loser per match and one and only one match per loser. Every player except the tournament winner (who does not lose) loses once and only once. Therefore, the number of matches equals the number of players who lose. There are 255 players who lose, so there are 255 matches in the tournament. (For the dubious, 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255.)
If you did not have this insight, do not worry. Neither do many mathematicians.
Answer #2: You know that this is not the obvious answer, 1 in 2 (i.e., fifty-fifty). This would be the answer if the acquaintance stated that the first child (or the second child) was a boy. If the acquaintance had said that the first was a boy, he would have left only two possibilities:
Boy-Boy
Boy-Girl
However, when the acquaintance answers that at least one is a boy (without telling you whether it is the first or second child), he slips in some new information. The new information is that there are three possibilities:
Boy-Boy
Boy-Girl
Girl-Boy
In two of these cases, the other child is a girl. In only one is the other child a boy. Therefore, the probability that the other child is a boy is 2 in 3, not 1 in 2.
If you still do not believe this, take a random-numbers table and consider only the first two digits of each number. Consider odd digits as boys and even digits as girls. Circle all the numbers in which one or both digits are odd (i.e., “at least one is a boy”). Finally, for all of these, compare the number of numbers in which one digit is even (i.e., a boy and a girl) with the number of numbers in which both digits are odd (i.e., two boys). You will find approximately twice as many of the latter.
This brainteaser is sometimes stated: “A couple has two children. One is a boy. What is the probability that the other is a boy?” Unfortunately, this form, which is undeniably more natural and felicitous, leaves an ambiguity concerning the method of randomization and the possibility that 1 in 2 can be considered a correct answer. The form given here avoids this problem.
Answer #3: One. Put on the scale: one one-pound weight, two two-pound weights, three three-pound weights, four four-pound weights, five five-pound weights, six six-pound weights, and seven seven-pound weights. If the scale reads one ounce over any exact pound weight, the one-pound weights are heavy. If it reads x pounds and two ounces, the two-pound weights are heavy, and so on. If the scale reads x minus one pound and fifteen ounces, the one-pound weights are light. If it reads x minus one pound and fourteen ounces, the two-pound weights are light, and so on.
Answer #4: Given the enormous length of the string, you probably assumed that the props would be tiny, say, one-millionth of an inch. However, the props must, in fact, be an impressive 1.91 … inches tall.
Here’s why. The circumference of a circle equals pi (3.14 …) times the diameter of the circle. This is true for every circle everywhere.4 If you add 12 inches (a foot) to the circumference of a circle, you add 12/3.14 … (i.e., twelve inches divided by pi inches, which equals 3.82.) Put another way, a circle with a circumference of 12 inches has a diameter of 12/pi inches; whether the circle has a circumference of 12 inches or 12 inches are added to the circumference of a larger circle does not matter. If you do the math for the earth example, you will find that the props must be 1.91 inches tall (1.91 is half of 3.82); the props are half of 3.82 because they are on both sides of the earth.
Now for the second surprise. Do the same problem with a sphere with a 9-inch circumference (which we will call a “baseball”), instead of with the earth (i.e., add a foot to the 9-inch taut string), so that the string is now 21 inches long. How tall must the props be?
You might think that this time the props will be much larger (because the foot of added string is much greater in relation to the baseball than the foot of added string in relation to the earth. You will be surprised to find that that the props must be the same, 1.91 … inches tall.
The reason for this is because this problem concerns only the relationship between the addition to the radius and the addition to the taut string. Nowhere does the size of the original sphere, be it earth or baseball, enter into the problem; that is, the lengths of the original diameter and the original circumference are irrelevant in each case. Of course, the 3.82 … inches added to the diameter and the foot that was added to the circumferences are much greater in proportion to a baseball than to the earth, but this has nothing to do with the question.
Answer #5: There are 292 ways to change a dollar—293 if you count a silver dollar as change.
When my nephew, Matt Mayers, was in high school, he gave the answer of 293. The teacher gave no credit for this answer, arguing that a silver dollar is not “change for a dollar.” Even if this claim were not dubious in the extreme, the teacher’s considering Matt’s answer to be no better than no answer is strong evidence that this teacher is in the wrong field.
A computer method for finding the solution to this problem is given in Creative Computing, July–August 1977. A noncomputer method is given in G. Polya’s How to Solve It (Princeton, NJ: Princeton University Press, 1957).
Answer #6: Switch. The car must be in one of only three places:
Whichever door you select, Monte must open a (different) door with a goat behind it. (The rules Monte stated require this.) If the car is behind the door you first select, changing your selection to the other will turn a winning choice into a losing one. If the car is behind the other door, changing your selection will turn a losing choice into a winning one. Thus, you are twice as likely to win if you do change your selection as you are if you do not because there were two chances out of three that the car is behind one of the other two doors, and Monte has ruled out one of those two doors.
If you do not switch, your chances remain the same 1 in 3 they would be if Monte did not open a door with a goat behind it. They remain 1 in 3 because you ignore new information that improves the odds to 1 in 2.
Say you choose door 1 and do not switch. The car can be behind door 1, door 2, or door 3, so your chances of winning the car are the same 1 in 3 they would be if Monte did not open a door. You already knew that two doors had goats behind them, and Monte simply opened one of them. So the chances of winning do not change if you do not switch. Putting this another way, you are ignoring the information that reduces the odds from 1 in 3 to 1 in 2. (This would not be the case if Monte could open a door with the car behind it; but the problem stipulates that he is going to open one of the two doors with a goat behind it.)
Now, note that a new contestant, entering the game after Monte opens the door, chooses from only two doors, and therefore has a 1-in-2 chance of winning; knowing that the car is not behind the door that Monte opens, this new player will face odds of 1 in 2, not 1 in 3.) You are in the same situation if you switch.
Your chances of winning, then, do change if you switch. You become the equivalent of the “new player” just mentioned.
Say you chose door 1. The door Monte opened could be
A2, leaving A3 (goat) closed
A3, leaving A2 (goat) closed
B3, leaving B2 (car) closed
C2, leaving C3 (car) closed
Thus, when you switch you raise your chances of winning from 1 in 3 to 1 in 2 (i.e., 2 in 4).
The problem is sometimes explained in these simple terms: When you first chose a door, there was a one-third chance that door had the prize and a two-thirds chance that one of the other two doors had the prize. When Monte opened one of the other two doors, the remaining door—the one you can choose to switch to—has a two-thirds chance of being correct. While this is correct and simple, it usually fails to persuade because it is not intuitively clear.
Say that there are fifty doors, with a car behind one of them. This becomes more clear if we make two sets: (1) “Set 1,” containing only one door, the one you randomly guessed to hide the car, and (2) “Set 2,” containing the other forty-nine doors. Obviously, the odds are overwhelming that the car is behind one of the forty-nine doors in Set 2.
Now, Monte opens forty-eight of the forty-nine doors of Set 2; by the rules of the problem, none of these may have a car. This leaves two closed doors: the one you originally guessed and the remaining one of the forty-nine in Set 2.
The likelihood is still overwhelming (49/50) that the car is behind a door in Set 2. Because there is only one door left in Set 2, the likelihood is overwhelming that the car is behind this door and not behind the door you selected. Thus, you should change your selection. (You knew all along that there were at least forty-eight doors without cars in Set 2 and that Monte would have to open them; the 49/50 likelihood of having the car resided all along in the door in Set 2 that Monte would have to leave closed).
It is, of course, possible that you happened to choose the door with the car behind it and that Monte’s forty-ninth door did not have a car behind it. But forty-nine out of fifty times (on average) this will not be the case.
The three-door problem is entirely analogous to this. However, because there are so few doors, it does not occur to us to think of the three-door problem in this way. It is the intuition that, whenever there are two doors, each must have a fifty-fifty chance that is responsible for so many people incorrectly assuming that the two of the three doors remaining in the three-door version each have a fifty-fifty chance of having the car.
The fifty-door analogy is helpful because it enables intellect to counter this incorrect intuition by making obvious the fact that Monte’s opening the forty-eight doors gives us new information: it tells us that the door remaining in his set after he has opened forty-eight doors still carries the original 49/50 probability that the set of forty-nine had—much more obviously—before Monte opened the forty-eight doors.
In short, Monte does give us new information: one of the doors (the last remaining door of Monte’s group) comes from a set with a higher probability of containing the winning door than does the door we selected at first. In the fifty-door version, the door we choose has a 1-in-50 chance of winning, while the remaining one of Monte’s doors has a 49-in-50 chance. In the three-door version, the door we choose has a 1-in-3 chance, while the remaining of Monte’s two doors has a 2-in-3 chance.
The three-door problem is entirely analogous to the fifty-door problem, but, because there are only three doors, the reality is much harder to see in the three-door version.
If you are still not persuaded, go to www.stat.sc.edu/west/javahtml/LetsMakeaDeal.html. There you can play the game many times per minute. Play for five minutes and the probability is overwhelming that you will win twice as often when you switch.
Answer #7: It cannot be done. To average sixty miles an hour for the two miles, you would have to complete the two miles in two minutes. You have already used two minutes in doing the first mile (1 mile at 30 mph = 2 min.). Unless you complete the second lap in no time at all, you cannot complete the two miles in the two minutes required for a sixty-mile-per-hour average for the entire race.
Answer #8: Obviously there must be at least three people at the party. But three are not sufficient, because, for example, one may know two, and three may know neither. If this is the case, then no three people know each other, and no three people do not know each other.
Four people are not sufficient because one may know only two, and three may know only four. If this is the case, then no three people know each other, and no three people do not know each other.
Five people are not sufficient because one may know only two and five; two may know only three and one; three may know only four and two; four may know only five and three; and five may know only one and four. If this is the case, then no three people know each other, and no three people do not know each other.
However, what if there are six people at dinner? Person one must either know at least three of the remaining five people or not know at least three of the five remaining people.
Now, here is an interesting fact: let’s say we want to find out how many people must be present to guarantee that five people all know each other or five people do not all know each other. To guarantee that four people know each other or four people not know each other, there must be eighteen people at the party. To guarantee that four people know each other or five people do not know each other, there must be twenty-five people at the party. (The proof of this takes eleven years of desktop computer time.)
So how many to guarantee that six people know each other or do not know each other? No one knows; the answer is between forty-two and fifty-five persons.
All this is known as Ramsey Theory (for the seventy-year-old work of British mathematician Frank Ramsey; the leading current mathematician in the area—and many others—is Ronald Graham of AT&T Lab).
Answer #9: Let’s call the person who shook hands with no one “A,” the person who shook hands with one person “B,” and so on. (See the chart below.)
| A shook with | 0 people |
| B | 1 |
| C | 2 |
| D | 3 |
| E | 4 |
| F | 5 |
| G | 6 |
| H | 7 |
| I | 8 |
The central insight is this: If the person who shakes hands with eight people (I) is not the spouse of the person who shakes hands with no one (A), then there will be at least three people with whom the person who shakes hands with eight people (I) does not shake hands: himself/herself, or his/her spouse, and the person who shakes hands with no one (A). But this leaves only seven people, and makes shaking hands with eight people impossible. Therefore, the person who shakes hands with eight people (I) must be the person who is the spouse of the person who shakes hands with none (A).
Once you have had this insight, the rest follows easily, if not quickly. If the person who shakes hands with seven people (H) is not the spouse of the person who shakes hands with one person (B), then there are at least four people with whom the person who shakes hands with seven people (H) does not shake hands: himself/herself, his/her spouse, A (who shakes hands with no one), and B (the person who shakes only one person’s hand; the one person must be the person who shakes the hands of all eight possible people (I). But this leaves only six people and makes shaking hands with seven people impossible. Therefore, the person who shakes hands with seven people (H) must be the spouse of the person who shakes hands with one (B). H cannot be the spouse of A because A is the spouse of I.
You continue until you discover that the person who shook hands with four people (E) must be the spouse of the anthropologist (who also shook the hands of four people).
Answer #10: Intuition and common sense would have you reason that (A) the card on the table must be either the card that is white on both sides or the card that is white on one side and red on the other, and (B) there is one chance in two that the hidden side is white (if it is the card that is white on both sides) and one chance in two that the other side is red (if it is the card that is white on one side and red on the other). Thus, (C) as the odds are even that the other side is white, you should take the 3–2 odds.
You will not be shocked to learn that intuition and common sense are wrong. (After all, it would not be much of a brainteaser if they were not.)
Probability problems are notoriously counterintuitive because intuition fails to include some of the possibilities and compares events that are not equally probable. And probability calculation must always be based on equally probable events.
There are six possible situations in which the card on the table shows white:
In four of the six cases, the underside is white. Therefore, if you get 3–2 odds, you will lose because you will lose twice for every once you win. A third of the time you will win $1.50 for a $1.00 bet, and two-thirds of the time you will lose $1.00 on a $1.00 bet.
Answer #11: There is no coherent answer to this question because the question is incoherent. Say that the concierge discovered that the room should have cost twenty dollars and gave the bellhop ten singles to return to the men. Say the men took three dollars each and left the bellhop the remaining dollar as a tip.
It would not and should not occur to us to say: Thus, each man spent seven dollars, and the bellhop got one dollar, for a total of twenty-two dollars. Where did the other eight dollars go?
It is only because the numbers in the original problem are so close that we (incorrectly) expect these numbers to add up to the money originally paid by the men.
Answer #12: Ask either man, “If I were to ask you if your door is the door of death, would you say yes?”
If you are asking the truthful man and his door is the door of life, he will truthfully say that he would answer no.
If you are asking the truthful man and his door is the door of death, he will truthfully say that he would answer yes.
If you are asking the liar and his door is the door of life, he will lie and say that he would answer no. (That is, the liar would, in fact, lie and say, “Yes, this is the door of death.” Because he would, in fact, say “yes,” he answers your question about what he would say with a lying “no.”)
If you are asking the liar and his door is the door of death, he will lie and say that he would answer yes. (That is, the liar would, in fact, lie and say, “No, this is the door of life.” Because he would, in fact, say “no,” he answers your question about what he would say with a lying “yes.”)
Thus, a “yes” answer from either man means that his door is the door of death, and a “no” answer from either man means that his door is the door of life. So, if the man says “yes,” choose the other man’s door; if the man says “no,” choose his door.
You never do find out whether the man who answered is the truth teller or a liar, but with the princess, the money, and the chance to meet Michael Jordan, who cares?
Answer #13: There are only two possibilities:
Miss Green wears blue, Miss Blue wears black, and Miss Black wears green.
Or
Miss Green wears black, Miss Blue wears green, and Miss Black wears blue.
Miss Blue cannot be wearing black because the woman wearing black says “So what” to Miss Blue. So, Miss Green wears black, Miss Blue wears green, and Miss Black wears blue.
Answer #14: Compare the offers over six-month periods:
| Offer A | Offer B | |
| End of 6 months | 50 | |
| End of 12 months | 55 | 100 |
| Total 1 Year | 105 | 100 |
| End of 18 months | 60 | |
| End of 24 months | 65 | 120 |
| Total 2 years | 230 | 220 |
| End of 30 months | 70 | |
| End of 36 months | 75 | 140 |
| Total 3 years | 375 | 360 |
Things continue in this way, so it is clear that A is the better choice.
Answer #15: No. Consider this: you win if you choose black-black-black and the first three spins are black-black-black. This will happen one-eighth of the time.* Any time after the first three spins, three blacks must be preceded by a red. But this means that the three spins before the last spin would have to have been red-black-black, the combination I chose to counter your black-black-black. Thus, I will win seven out of eight times (on average). The same strategy works whatever the combination you choose.
Some choices other than black-black-black improve your odds. But the best you (i.e., the person who goes first) can do is lower the odds to 2–1 against yourself.
*Because there are eight equally likely possibilities:
b-b-b
b-b-r
b-r-r
b-r-b
r-b-r
r-b-b
r-r-b
r-r-r
Answer #16: It is easy to conclude that, if a hundred chickens eat a hundred bushels in a hundred days, one chicken will eat one bushel in one day. This would imply that ten chickens will eat ten bushels in a day and a hundred chickens will eat a hundred bushels in ten days. Obviously something is wrong with this because we know that it takes a hundred chickens a hundred days to eat a hundred bushels.
The above reasoning is incorrect because if a hundred chickens eat a hundred bushels in a hundred days, then a hundred chickens eat one bushel in one day. Thus, ten chickens will eat one bushel in ten days. (And ten chickens will eat ten bushels in a hundred days. A hundred chickens will eat one bushel in one day, ten bushels in ten days, and a hundred bushels in a hundred days.)
Answer #17: This is a perfect example of the idea of symmetry. If you must place the cigar on its side, the person who goes second always wins (assuming that he plays correctly). The reason for this is that whenever the person who goes first places a cigar, the person who goes second places his cigar in precisely the “opposite” place on the table. Therefore, whenever the first player can place a cigar on the table, the second player can do so as well. (Note that it does not matter whether a cigar is permitted to be partly off the table; the same logic applies.)
However, if a player is permitted to stand a cigar on its flat end, then the first player always wins. The first player places his cigar, standing on the flat end, in the center of the table. This captures the one spot that has no symmetrical partner (it contains its own symmetry). By doing this, the first player effectively becomes, in the terms discussed above, the second player, and he always wins. (No, you cannot stand a cigar on its point. It is in the Bible. Look it up.)
Answer #18: The amount of red paint in barrel 1 equals the amount of blue paint in barrel 2. It seems that this should not be the case because you take undiluted red paint out of barrel 1 and put it in barrel 2 and then take the mixture out of barrel 2 and put it in barrel 1.
But look at it this way. The problem with that logic is that the 1 gallon from the red paint comes from 10 gallons. The diluted blue paint comes from 11 gallons, of which 10/11 is blue. There are 10 gallons in 10/11 of 11 gallons, the same amount as in the barrel of red paint at the start. Ten gallons of paint taken from 11 gallons of paint that is 10/11 blue leaves barrel 2 with the same amount of blue paint that barrel 1 has after its missing gallon is replaced with the 10 gallons of paint taken from barrel 2.
All numbers below refer to gallons
| Barrel 1 10 red |
Start | Barrel 2 10 blue |
| 1 gallon from barrel 1 to barrel 2 | ||
| 9 red | Mix paint in barrel 2; Mixture is 10/11 blue and 1/11 red |
10 blue + 1 red |
| 10/11 blue + 1/11 red | ||
| 1 gallon from barrel 2 to barrel 1 | ||
| 9 red + 1 × 10/11 blue + 1 × 1/11 red | ||
| 9 + 1/11 red + 10/11 blue red | 9 + 1/11 blue + 10/11 | |
This problem easier to understand if we consider an analogous problem that appeared in Marilyn vos Savant’s column, Ask Marilyn, in PARADE magazine. Say there is a boys’ college right next to a girls’ college. Each has a dorm that holds 1,000 people. One night, 500 girls raid the boys’ dorm. At this point, there are 1,500 students stuffed into the boys’ dorm. The fire marshal arrives and orders (any) 500 of the 1,500 to go to the girls’ dorm. Some girls stay and others leave. Which dorm now has the greater concentration of its original sex?
In a situation comparable to that in the paint problem, each has the same number of the “other sex” students and the same number of “original” students. Because each must end up with 1,000 students, for every girl who stays in the boys’ dorm (i.e., every girl who is not one of the 500 students who go to the girls’ dorm), there must be a boy who goes to the girls’ dorm (i.e., every boy who is one of the 500). Thus, each dorm will end up with the same number of “other sex” students.
Marilyn vos Savant also provides another version of this. You are blindfolded and given a deck of cards with (any) ten cards facing up and the other forty-two facing down. You must divide the deck into two stacks, each containing the same number of cards facing up. Can you do this?
Yes. Simply deal yourself ten cards and turn all of them over. If, say, none of the first ten cards of the original deck had been facing up, then the remaining forty-two card stack will have ten cards facing up, as will your ten-card stack (because the ten cards you dealt yourself would have no up-facing cards before you turn them over and, therefore, ten up-facing cards when you turn them over). If one of the first ten cards of the original deck had been facing up, there will now be nine cards facing up in the forty-two card stack and nine facing up in the ten-card stack.
Answer #19: There is no space between the poles; they are touching. If the lowest point of the rope is twenty feet off the ground, the rope must go down thirty feet on one pole and up thirty feet on the other. That takes care of the rope’s sixty feet, so the poles must be next to each other. (You are also correct if you answered that this situation cannot exist. You can reasonably argue that, even if the poles are touching, a slight bit of rope must be taken by the distance between the two attachment points, in which case there would not be quite enough rope to go down and up sixty feet—which it must if it is to reach a point only twenty feet from the ground.)
Answer #20: Forget all those equations, formulae, and diagrams you have just committed to paper. Instead, picture two monks, one starting up at the same time that other starts down. Obviously, they must meet at some point.
Answer #21: Light one piece of string at both ends. Light the other piece at one end. Whatever the rate of burning of the first string, the two flames will meet after half an hour. (You do not know where they will meet, but it does not matter.)
When the two flames of the first string meet, light the second end of the second piece of string. In fifteen minutes, the two flames of the second piece will meet. The burning rates of the two flames of the second piece of string do not matter; the two flames burning on the second string will meet in fifteen minutes.
Thirty minutes plus fifteen minutes is forty-five minutes.
Answer #22:
Note that whomever of the three is quickest can tell the color of his hat no matter what color of any of the hats.
Answer #23: Obviously, you cannot simply ask, “Which path leads to heaven?” You do not know whether the answer is coming from the truth teller or the liar.
But how about trying this. Let’s assume the path to heaven, the one you would like to take, is the west path. You ask either man, “If I ask the other man which path is the path to heaven, what will he say?”
If the man you happen to ask is the truth teller, meaning the other man is the liar, he will truthfully answer, “East.” (That is, he will truthfully tell you that the liar will lie and tell you “east.”)
If the man you happen to ask is the liar, he will lie and say, “East.” (That is, he will lie and tell you that the truth teller would tell you “east.”)
You still do not know which man is the truth teller and which is the liar, but you do not care. You do know what you wanted to know: the west path is the path to heaven. (Of course, this works whichever one is the path to heaven.)
Answer #24: You cannot. Think in terms of two people, rather than just you. The one who leaves later must catch up to the one who leaves earlier. At that point, they meet the same traffic and arrive at work at the same time. In other words, the best you can do by leaving later is to arrive at the same time (which, of course, is a good idea when you can in fact catch up to the earlier guy).
Answer #25: The downtown train comes at one minute after the hour and every six minutes after that. The downtown train comes at two minutes after the hour and every six minutes after that. So, unless you get to the subway within the minute after an uptown train has come, you will always get the uptown train.
Note: I used to include among my favorite brainteasers the following:
You have a box six feet by six feet by six feet. You want to fill it with balls in such a way that the least empty space is left. All the balls must be the same (finite) size and whole (i.e., no parts of balls). For example, you may fill the box with one ball with a six-foot diameter or four balls with three-foot diameters, and so on.
What size ball(s) would you choose?
It used to be thought that the correct answer is that it makes no difference; there is always the same amount of total empty space left. This answer was surprising to most people and fun—in other words, a fine brainteaser. Alas, some mathematician pointed out this is true only if one of a number of possible methods of packing is employed. Including this constraint in the question makes for good mathematics—and a brainteaser that is incomprehensible to the nonmathematicians for whom brainteasers are meant.
Answer for the fireman-brakeman-engineer problem: The brakeman, who lives halfway between Detroit and Chicago, also lives nearest Mr. X. Mr. X earns exactly three times as much as the brakeman. Mr. X cannot be Mr. Robinson because Mr. Robinson lives in Detroit (and therefore could not be the brakeman’s nearest neighbor). Mr. X cannot be Mr. Jones because Mr. Jones earns $20,000 a year, which could not be exactly three times the brakeman’s salary.
Because Mr. X cannot be Mr. Robinson or Mr. Jones, he must be Mr. Smith.
The passenger whose name is the same as the brakeman’s lives in Chicago. Therefore, he cannot be Mr. Robinson, who lives in Detroit. He cannot be Mr. Smith because Mr. Smith lives nearest the brakeman, who lives halfway between Detroit and Chicago.
Because the passenger whose name is the same as the brakeman’s cannot be Mr. Robinson or Mr. Smith, he must be Mr. Jones. Because his name is the same as the brakeman’s, the brakeman must be Jones.
Smith beats the fireman at billiards, so the fireman cannot be Smith. The fireman cannot be Jones because Jones is the brakeman. So Smith must be the engineer.
If the brakeman is Jones and the engineer is Smith, the fireman must be Robinson.
1.“Knowing” and “not knowing” are mutual (i.e., if one person knows a second, then the second knows the first, and if one person does not know a second, then the second does not know the first).
2.“Three people all know each other” means that each of the three must know the other two. It is not sufficient for A to know B and B to know C. C must also know A. Likewise, “three people all do not know each other” means that none of the three know any other of the three.
3.Games magazine (January–February, 1978).
4.This assumes a flat surface. When we talk about a surface that is not flat, it is a whole different ball game. Here we are dealing with a flat surface because the circumference of the earth or baseball is a circle on a plane. If we were to draw a circle on the earth or the baseball, then plane geometry would not apply.