The previous subsection presented some examples to test your intuition around data. This chapter presents a few problems and puzzles to flex your quantitative intuition, logic reasoning, and thinking muscles.
Some of these are well-known problems and their variations that you might encounter during a job interview for a technical/analytical role. That is a terrible practice. Data scientists are not hired to solve brain teaser problems on the spot. They are hired to be problem solvers; you are allowed to think deeply about a problem, you are not required to shoot a probability result from the hip. The point about the birthday problem (Section 6.1.1) is not that the probability of at least one match in a group of 23 people is 0.5—that is the number everyone remembers. The point is how you approach such a problem and find a general solution—what is the pattern?
Alas, questions like the ones here do show up in interviews. Maybe this chapter helps you prepare for that.
The problems are interesting because they make you think about probabilities and patterns. And they are fun.
Solutions are provided in the call out boxes. Try to solve the puzzles before taking a look at the solutions.
This is a classical problem in probability, and a popular one because it is relatable yet somewhat counterintuitive. The probability is higher than what most people expect. It goes like this:
What is the probability that in a group of \(n\) randomly chosen people, at least two share the same birthday?
“Birthday” is meant as one of 365 days of the year, not adjusting for leap years. Also, we are not taking the birth year into account. A birthday for the purpose of this problem is April 10, or August 15, etc.
The standard version of the problem uses \(n=23\), because you can imagine yourself in a group of that size—a classroom, for example—and the probability of at least two shared birthdays is also relatable.
How likely do you think it is that at least two people share a birthday in a group of 23? How about a group of 35?
The probability of at least two shared birthdays in a group of 23 is about 0.5; it is 0.05073, to be more exact. How do you interpret that? If you were to assemble groups of 23 randomly chosen people, than half of those groups would have at least two shared birthdays. Pretty high, eh?
What happens to the probability of a shared birthday when the groups get larger? How about in a group of 35 people? The probability of a shared birthday increases to 0.814. In a group of 50 people, the probability is 0.97. In a group of 100, it is virtually certain that there are at least two identical birthdays, \(p=0.999999\). With only 10 people in a group, it would be surprising to have identical birthdays, but it is not a rare event, \(p=0.117\).
How do you calculate those probabilities? First, whenever you see the expression “at least” in a probability statement, it is probably easier to calculate the probability of the complement event and subtract that from 1. \[ \Pr(\text{at least two identical birthdays}) = 1 - \Pr(\text{no matching birthdays}) \]
What is the probability that no birthdays match in a group of \(n\)? You can compute this by considering the possible choices as people enter the group. The birthday of the first person can be chosen from 365 days, but the birthday for the second person has only 364 choices, otherwise we would have a match. Since the members of the group are chosen at random, the birthdays are independent and the probability of no matches is the product \[ \Pr(\text{no matches}) = \frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{365-n+1}{365} \]
You can write this in terms of factorials as \[ \Pr(\text{no matches}) = \frac{1}{365^n} \frac{365!}{(365-n)!} \] Finally, the probability of at least two shared birthdays is \[ \Pr(\text{at least two shared birthdays}) = 1 - \frac{1}{365^n} \frac{365!}{(365-n)!} \]
If you were to compute this, you’d run into problems because the factorials are larger than what a finite precision computer can represent. The following R function uses two tricks to compute the birthday probability efficiently:
The lgamma function in R computes the log of the Gamma function, and that gives us access to an efficient way to compute the components of the probability on the log scale.
birthday_prob <- function(n) {
log_p <- lgamma(365+1) - lgamma(365-n+1) - n*log(365)
return (1-exp(log_p))
}
birthday_prob(10)[1] 0.1169482birthday_prob(23)[1] 0.5072972birthday_prob(35)[1] 0.8143832birthday_prob(50)[1] 0.9703736birthday_prob(100)[1] 0.9999997A family has two children and we know that one of them is a boy. What is the probability that the family has two boys?
This is a problem about a conditional probability, since we are told that one of the children is a boy. The sample space of possible configurations—assuming two genders—is \(\{ (B,B), (B, G), (G, B), (G,G)\}\), where \(B\) stands for a boy and \(G\) stands for a girl.
The conditional probability in question is \[ \Pr(\text{two boys} | \text{one is a boy}) = \frac{\Pr(\text{two boys and one is a boy})}{\Pr(\text{one is a boy})} = frac{1}{3} \] If you answered 1/4, then you were thinking about the unconditional (the marginal) probability of two boys in a family of 2 children. The conditioning removes the event \(\{(G,G)\}\) from the set of possibilities. And only one of the remaining events meets the two boy criterion.
An airline has a single seat open on a flight, but \(n=100\) standby passengers hoping to get on the flight. You are one of the passengers on standby. To be fair to all standby passengers, the airline decides to drop 100 equal-sized pieces of paper into a bucket. 99 of them are blank, one says “Last Seat”. The papers are folded and shuffled in the bucket.
The standby passengers queue and each passenger gets to pick one piece of paper without replacement—that is, they keep the slip and do not return it to the bucket. Also they cannot unfold and look at the slip until all of them re drawn. After the last slip is drawn the standby passengers announce who is the lucky person that drew the “Last Seat” by checking their slip.
Here is the question: if you have your choice to pick first, second, last, or at any particular position in the queue, which position would you choose?
It does not matter when you draw the paper if the pieces were properly shuffled. This is a completely random sample even if the sampling is done sequentially. Your chance of drawing the “Last Seat” slip is 1/100, whether you draw first, last, or at any other position in the queue.
Note that this would be different if passengers would announce the result of their draws before the next draw. The conditional probability of choosing the “Last Seat” slip on the next draw increases with every bank slip that preceded.
You have two stacks of cards. The first is a regular 52-card deck. The second stack contains two regular 52-card decks, thus has 104 cards. Both stacks are shuffled well. You choose two cards in sequence and you win if they are both red. Would you prefer to choose from the 52-card stack or the 104-card stack?
You want to choose from the larger stack. The probability to draw two red cards in sequence from a stack of \(n\) cards (with \(n/2\) red ones) is \[ \frac{n/2}{n} \times \frac{n/2-1}{n-1} \] For the first draw the probabilities are identical: \(26/52\) and \(52/104\). But for the second draw the probabilities are \[ \frac{51}{103}=0.495 > \frac{25}{51}=0.49 \]
There is a slightly higher chance to draw two red cards from the larger stack.
There are many versions of this basic puzzle, using ants, camels, and other animals. We use robots here, the puzzle goes like this. Three robots are placed at the corners of a triangle. A robot can choose to move along either side of the triangle that meet at its corner (Figure 6.1). What is the probability that any two robots will collide?
Each robot has two possible movements, so there are a total of 2 x 2 x 2 = 8 possible moves on the triangle. There are two ways in which there won’t be any collisions, if all choose to go clockwise or counter-clockwise. In those cases they will follow each other around the triangle (Figure 6.2).
Any other choice of movements will result in at least one collision. So the probability of any collision if the robots choose their movements at random is 6/8 = 3/4. There is a 75% chance that any two robots will collide.
You are engaged in a game of Russian Roulette with a six shooter. One bullet is placed in a chamber, the chamber is spun, and your opponent pulls the trigger. The gun does not fire. Now it is your turn and before you pull the trigger you are given the choice to spin the chamber one more time or leave the gun as is.
How do you decide?
You should spin the chamber again, because it will give a 1/6 chance to fire the bullet. If you leave the gun as is, because the gun did not previously fire, you have a 1/5 chance to fire the bullet.
By spinning the chamber you have a higher likelihood to live after the shot.
This is the same setup as the previous problem. But now there are two bullets in consecutive chambers of the gun. The first shot does not go off and it is your turn to pull the trigger. Should you spin the chamber first?
If the chamber is spun, the probability to fire a bullet is 2/6 = 1/3. Since the previous shot did not go off, you know that the chamber is in one of four possible positions (Figure 6.3). One of those positions will be followed by a bullet in the chamber—the probability of a shot going off is 1/4.
Because 1/4 is less than 1/3 you do not want to spin the chamber again.
Figure 6.4 shows a logic reasoning puzzle. The first row makes sense if the strange operator is addition, but that does not work for the next rows. You have to find the meaning of that operator, then apply the pattern to solve the last equation.
We need to find a pattern that expresses the operations in terms of familiar algebra. If the operator in Figure 6.4 is interpreted as multiplication then we get 4, 10, 18, all smaller than the values on the right hand side. How much smaller? Exactly by the left-most number. The pattern that seems to apply to the first three rows is
Applying this pattern to the last row yields 96 as the solution (Figure 6.5).
This, by the way, is not the only solution. There are other patterns that will lead to a different result for the last row. Those patterns are equally valid. Can you find another pattern that yields a solution?
Here is a sequence of numbers.
\[ \begin{array}{c} 1 \\ 11 \\ 21 \\ 1211 \\ 111221 \\ 312211 \\ ?? \end{array} \]
What is the next number in the sequence?
What is the pattern in the sequence of numbers?
The pattern is that the numbers for the following row are obtained by spelling out the numbers in the current row, then replacing the words with the numbers they represent. For example, take 1211 in the fourth row. Spelling it out gives “one one one two two ones”. Now replace the words with numbers: “111221”.
The missing entry at the end of the sequence is thus \[ 13112221 \]
This could be called a math puzzle but it is just as much—maybe more—a pattern recognition puzzle. Consider the 4 statements:
\[ \begin{align*} 5^a &= 6\\ 6^b &= 7\\ 7^c &= 8\\ 8^d &= 625 \end{align*} \] Given these statements, can you find \[ \sqrt{abcd} = ?? \]
Trying to solve this by brute force is going to be tricky,
\(a \log(5) = \log(6)\), \(b \log(6) = \log(7)\), and so forth.
But by simply substituting the left hand side of the previous equation into the next equation we get: \[ 6^b = (5^a)^b = 5^{a\times b}=7 \] Continuing along those lines we arrive at \[ 8^d = 5^{a \times b \times c \times d} = 625 \] Now we know \(abcd = 4\) and \(\sqrt{abcd} = 2\).
Figure 6.6 shows a matrix with 9 cells and eight of the cell values. Can you determine the relationship between the cells and their values to determine the value in the missing cell?
Start by looking at the numbers in rows and columns, focusing on complete rows and columns. In the first colunm the numbers 7, 49, 46 do not stand in a discernible relationship. It is easier for the numbers in the first row, going left to right: \[ 7 + 9 = 16 \qquad 16 + 8 = 24 \] In the third column we see a similar pattern \[ 24 + 7 = 31 \qquad 31 + 6 = 37 \]
In the last row, going left to right, the pattern is \[ 46 - 4 = 42 \qquad 42 - 5 = 37 \] If the left-to-right pattern uses subtraction, the right-to-left pattern in the row uses addition: \[ 37 + 5 = 42 \qquad 42 + 4 = 46 \] Figure 6.7 shows the numbers added to the previous cell when you navigate the matrix in a clock-wise pattern. The missing value is thus 51.
Suppose you are guarding \(n\) criminals in an open field. You have one gun with a single bullet. You are a good shot and being fired at means death—the criminals know that. Their behavior is governed by the following rules:
How do you guard the criminals and stop them from escaping?
Imagine there is only a single criminal, \(n=1\). Since he/she would definitely be shot at during an escape, they would face certain death and not escape.
What happens if there are two criminals? If they both try to escape, there is a 50:50 chance to survive, hence they will both try to escape. To prevent that from happening you would tell one of the two (you do not need to tell both!) that you would shoot them, should they both attempt to escape. That criminal now faces certain death and will not escape. That brings you back to the situation with a single criminal.
How does this generalize to larger groups of criminals? Assign a number from 1 to \(n\) to the criminals and tell them that should any subgroup of them try to escape, the one with the highest number in the group will be shot. Or you can tell all of them that the first criminal who tries to escape will be shot. Nobody can afford to be first because it would mean certain death. Since noone will be first, noone can be second and have a non-zero probability of surviving.
Three opaque jars are sitting on a table. The jars are labeled “Apples”, “Oranges”, and “Apples & Oranges”. Unfortunately, all three are labeled incorrectly.
Your task is to assign the labels correctly to the jars. What is the smallest number of fruit you have to choose in order to correctly label the three jars?
You choose one fruit from the jar that is labeled incorrectly as “Apples & Oranges”. If you pull an apple, you know this is the jar with the apples, otherwise it is the jar with the oranges. Now you have two jars left whose labels just need to be flipped since you were told that all three jars are labeled incorrectly.
This puzzle has many variations:
The trick is that each door or fork in the road is guarded by a separate guard. One of the guards is a liar who never tells the truth, the other guard always tells the truth. And, you get to ask only one question. What question do you ask to reveal the right door (assuming you want to get to a good outcome)?
You should ask the following question of either of the guards:
If I would ask the other guard which road leads to my destination, what would they say?
A variation for other setups would be “If I would ask the other guard which door leads to freedom, …”
Then, when you hear the answer, choose the opposite road or door.
Why does this work? Regardless of which guard you ask you get the wrong answer, pointing out the fork in the road or the door you want to avoid.
If you ask the guard who never lies, he knows that the liar would point you to the wrong road. Since he speaks the truth, he will tell you that the other guard will lead you to the wrong road.
If you ask the guard who always lies, he knows the other guard would point you to the correct road. But since he always lies, he will tell you that the other guard will point you at the wrong road.
Either way, the answer given is be the road or door to avoid.
This puzzle is about logic reasoning and not about probability. Surprisingly, it is related to the robot movement puzzle in Section 6.1.5.
Consider the following three statements:
Who is telling the truth and who is lying?
This puzzle is related to the robot movement in that there are \(2^3 = 8\) possible choices, each of the three characters could either be truthful or lying. It is different from the robot movement in that it is not a question of probability. While robots choose one of the two directions at random, our characters are either lying or telling the truth. We have to reason which one it is.
With 8 possible choices you can go about it by finding combinations that are inconsistent, a process of elimination.
Suppose that Jenn tells the truth. Then Gavin and Brian are liars. According to Gavin’s statement, that would mean Brian is telling the truth. But Brian’s statement contradicts the assumption that Jenn tells the truth. Jenn must be a liar.
If Jenn is not telling the truth, there are three possibilities:
Let’s look at the first option. If Gavin tells the truth than Brian is lying, which means Jenn would be truthful. We already ruled out this possibility. But if Gavin is not truthful, then 3. cannot be the case either.
We are down to the second option: Brian speaks the truth and the other two are liars. Let’s see if everything makes sense in this scenario: If Gavin does not speak the truth, then Brian is not a liar. Brian’s statement that Jenn is a liar is consistent with what we already found.
Conclusion: Only Brian is truthful.
Figure 6.8 shows a triangle made from 10 coins. Can you change this into an upside-down triangle by moving only 3 coins?
The solution is shown in Figure 6.9. First, focus on the seven coins in the center of the triangle. The original and the inverted triangle share these; they do not need to move at all. We can focus on the three coins at the edges.
You have ten coins in front of you, but you cannot see whether heads or tails are up on any of them. You are being told that five of them show heads, five of them show tails (Figure 6.10).
Without looking at the face of the coins, your task is to divide the coins into two groups, each contains the same number of heads. The number of heads in each group does not have to be five. For example, you can end up with two groups, each of which contains 2 heads. Or two groups, each of which contains 3 heads. You cannot look at the coins but you can flip any coin as many times as you like.
Make two groups of five coins each. Whatever the number of heads or tails within the two groups, you know that originally five of the coins showed head. Now take the coins in one of the groups and turn them all over (Figure 6.11). Voilà.
A farmer is on his way back from the market, with him he has a fox, a chicken and some grain. To get home he needs to cross a river using a small boat that can accommodate only him and one of the other items. Unfortunately, if the fox is left alone with the chicken it will eat it. If the chicken is left alone with the grain, it will eat it. How can the farmer cross the river and bring home the fox, the chicken, and the grain?
This will take several trips across the river:
The trick is to take one item—here, the chicken—back and forth to make sure it is not alone with the item it would destroy.
A teacher writes six words on a board:
cat \(\quad\) dog \(\quad\) has \(\quad\) max \(\quad\) dim \(\quad\) tag
The teacher chooses one of the words and writes each letter of the word on a separate piece of paper. Three students, Tobias, Julian, and Amelie receive one of the papers each. These students are known to do well on their logic exams. The teacher asks
“Tobias, do you know the word?”
Tobias replies immediately, “Yes, I do.”
The teacher then asks
“Julian, do you know the word?”
He thinks for a moment and replies, “Yes, I do.”
The teacher then asks
“Amelie, do you know the word?”
She thinks for a moment and replies, “Yes, I do.”
Which of the six words did the teacher choose?
The word is dog.
Since Tobias knew immediately what the word was he must have received one of the unique letters that identifies a word:
c o h s x i
This eliminates tag as a possibility. Having heard Tobias’ answer, Julian now looks at the remaining unique letters that are left among the words
cat \(\qquad\) dog \(\qquad\) has \(\qquad\) max \(\qquad\) dim
This list is
t g h s
because “a”, “d”, and “m” appear more than once and the other letters were eliminated because Tobias knew immediately what the word was. Notice that “h” and “s” are still in the list because if Tobias had received “s”, “h” would still be a possible letter and vice versa.
This eliminates max and dim from consideration. Based on the remaining unique letters and the piece of paper he received Julian can now figure out the word.
Amelie narrows it down the same way. Among the remaining words,
c a t \(\qquad\) d o g \(\qquad\) has
there are only two letters left to consider, “a”, and “d”. The letter “a” appears in multiple words so it cannot be it; the only unique letter left is “d”. The other letters are either not unique or were in the initial list based on which Tobias knew immediately what the word was and in the list eliminated by Julian.
Tobias had received the letter “o”, Julian had received the letter “g”, and Amelie had received the letter “d”.
Imagine that you hire a consultant to work for you for five days. At the end of each day you need to pay them 1/5th of a gold bar. You have a single gold bar (worth 5 fifths) and need to cut it up so you can pay the consultant at the end of each day.
What is the minimum number of cuts that allow you to pay the consultant every day?
You need only two cuts to cut the gold bar into three pieces of sizes 1/5, 1/5, and 3/5.
Then you pay the consultant as follows:
You enjoy your job but find out that your position is slated for elimination during a round of budget cuts. The HR representative offers to keep the position from the elimination list and gives you the following chance:
How should you divide the marbles between the two bowls if you want to keep your job? (How would you divide the marbles if you do not care about the job?)
This problem is in the analytic category because it is not just about probabilities, it is about expected values. Let \(\pi_1\) denote the probability to choose a blue marble from the first bowl and \(\pi_2\) the probability to choose a blue marble from the second bowl. Suppose you want to keep your job, then you want to maximize the likelihood that a blue marble is chosen. Since bowls are chosen at random, we are trying to maximize \[ q = \frac12 \pi_1 + \frac12 \pi_2 \] Suppose you divide the marbles equally between the two bowls, then \(\pi_1 = \pi_2 = 0.5\) and \(q = \frac12 0.5 + \frac12 0.5 = 0.5\). That makes sense, it is equally likely in this scenario to end up with a blue or red marble.
Your chances of retaining the job are maximized by placing a single blue marble in one bowl (\(\pi_1 = 1\)) and the remaining 99 marbles into the other bowl (\(\pi_2 = 49/99\)). Then \[ q = \frac12 + \frac12 \frac{44}{99} = 0.7474 \] With this configuration you have about a 3 in 4 chance to get a blue marble.
A chess board is made up of eight rows and columns of black and white positions (Figure 6.14). How many squares are on a board?
The quick answer is \(8 \times 8 = 64\) squares. However, that is only part of the story. The entire board is a single square as well, made up of the 64 individual squares. And we could place all kinds of \(2 \times 2\) squares inside the larger frame.
If you think about it for a bit there are \(8^2\) squares of size \(1 \times 1\), \(7^2\) squares of size \(2 \times 2\), \(6^2\) squares of size \(3 \times 3\) and so on. The total number of squares on a chess board is
\[ 8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2 = 204 \]
Suppose you are working in a library and are sorting books from a box that contains 32 fiction (F) and 17 non-fiction (NF) books. A steady supply of new books is available to add to the box. Your sorting algorithm goes as follows:
You randomly choose 2 books from the box.
Based on the types of books chosen you add another book from the supply to the box:
The entire procedure is depicted in Figure 6.15.
Since you add only one book to the box for every two books you remove, the box will eventually be empty. What is the type of the last book in the box? Is it a fiction or a non-fiction book?
The number of books in the bin goes down by one with each cycle: two books are removed from the bin, one book is added. How does this affect the number of fiction and non-fiction books that remain?
Let’s see how the number of non-fiction books in the bin changes in cases 1.–3. In the first case, there is no change. In the second case, the number of non-fiction books goes down by 2. In the third case, the number of non-fiction books also does not change: one is removed, one is added.
Since the number of NF books initially is an odd number, 17, we can conclude that after each cycle the number of NF books remains an odd number. It can never be an even number. Which leads to the conclusion that if there is only one book left in the bin it must be a non-fiction book.
Your car has four tires mounted to the wheels and a spare tire (S). That gives you five tires to work with. Each of the tires lasts at most 30,000 miles. If you can exchange tires among the five as many times as you wish, what is the furthest distance you can travel before you need to purchase a new tire?
Figure 6.16 depicts the initial tire life prior to driving the first mile. All tires, including the spare (S) have the same life expectancy of 30,000 miles.
The maximum total distance the five tires could travel before they are all worn out is 30,000 x 5 = 150,000 miles. The minimum distance of travel before you have to buy at least one new tire is 30,000 miles; it is achieved if you do not use the spare tire and run down the four tires currently mounted.
By optimizing how you use the spare tire, there must be an achievable distance between 30,000 and 150,000 miles. The optimal strategy is to wear all tires equally and to use the spare tire as much as possible. But we cannot use the spare for more than 30,000 miles, same as with the other four tires.
If the four tires on the car are equally worn, we can go at most 150,000/4 = 37,500 miles. The strategy is to get 30,000 miles from each of the tires on the car and 4 times 7,500 = 30,000 miles from the spare tire. In other words, the spare will have to give each of the four tires a 7,500 mile break.
Figure 6.17 shows how the spare tire is rotated for another tire after each leg of 7,500 miles. The right rear tire comes off after the fourth leg, it is worn out. The other tires still have 7,500 miles of life to go.
You have two wooden sticks and a box of matches. When a sticks is lit it will burn completely in exactly one hour. How do you use these ingredients to measure exactly 45 minutes?
Light the first stick on one end and light the second stick on both ends. Since an entire stick burns in one hour, the stick lit on both ends will burn down in 30 minutes (Figure 6.18).
At that point light the first stick on the other end. This will double the speed with which that stick, now reduced to 30 minutes burn time, will burn.
When the first stick is completely burned down, 45 minutes will have passed (Figure 6.19).
Cowboy Billy carries a Colt single action 6 shooter revolver. When he fires all 6 shots in a row, the time between the first bullet and the last is 60 seconds. How long would it take him to fire 3 shots?
It will take him 24 seconds to fire three shots. Wait, what?
The relevant pattern here is about the time elapsed between shots. If the shots are fired at regular intervals, then Billy will take 12 seconds between the six shots. 12 seconds after the first shot he fires the second bullet, 12 seconds after that he fires the third bullet.
Another way of thinking about this is the distance at which fence posts are placed. In a fence with six posts, the first one is at 0/5th total distance, the second post is located 1/5th of the total distance, and so on.
If five cats catch five mice in five minutes, how long does it take one cat to catch a mouse?
Five minutes.
Two merchants quarrel about how much wine is left in a barrel that has no lid. The first merchant peers into the barrel and says “It is more than half full.” The second merchant looks into the barrel and says “Nonsense, it is more than half empty.”
How do you determine which merchant is correct?
Tip the barrel on its edge until the liquid comes up to the edge of the barrel. If you can see the bottom of the barrel, it is more than half empty. If you cannot see the bottom of the barrel, it is more than half full.
Suppose that Andy and Brie are arrested as members of a criminal gang and held separately by the police. They cannot communicate. There is enough evidence to convict them on a lesser charge, but not on the principal charge. The police offers the following deal:
How should Andy and Brie behave to optimize their positions, that is, look out after their own interest?
The result of such a game is typically displayed in a payoff matrix that shows in each cell the payoff for the two players.
| Brie remains silent | Brie testifies | |
|---|---|---|
| Andy remains silent | \((-1, -1)\) | \((-3, 0)\) |
| Andy testifies | \((0, -3)\) | \((-2, -2)\) |
The “payoffs” are shown in the matrix as negative numbers, as they represent a penalty, years of imprisonment. The goal is to maximize the payoff, a number as large as possible.
The best situation for Andy is to testify when Brie remains silent. He would go free in this case (and does not mind Brie spending three years behind bars). Similarly, the best situation for Brie is to testify when Andy remains silent. These are the two diagonal cells in Table 6.1.
The situation does not play out as well for them if one testifies and the other also testifies. What is the best strategy?
The Nash equilibrium is a concept in game theory. It applies to non-cooperative games where players compete against each other. In the equilibrium state, no player can gain an advantage by changing their strategy. This assumes that the other player’s strategies do not change.
Suppose players Andy and Brie have chosen strategies A and B, respectively. In the Nash equilibrium, there is no other strategy available to Andy that would increase his expected payoff if Brie stays with strategy B. Similarly, there is no other strategy available to Brie that would increase her expected payoff from the game if Andy stays with strategy A.
The Nash equilibrium tells us not to consider player’s action in isolation. Instead, we need to take into account what other players are expected to do in evaluating a player’s choices.
The best outcome for either Andy and Brie would be to go free. But they do not know how the other one will behave. So what is the best strategy to play this game? Let’s rephrase testifying and remaining silent in terms of defecting and collaborating players of a game.
| Brie collaborates | Brie defects | |
|---|---|---|
| Andy collaborates | \((-1, -1)\) | \((-3, 0)\) |
| Andy defects | \((0, -3)\) | \((-2, -2)\) |
If Andy defects, his penalty will be less, regardless of whether Brie is collaborative or not (0 or 2 years compared to 1 or 3 years). The same applies to Brie, if she defects her penalty will be less regardless of what Andy does. The Nash equilibrium is that both players defect although they suffer worse penalties than if they had both cooperated.
There are two ducks in front of a duck, two ducks behind a duck, and a duck in the middle. How many ducks are there?
There are three ducks (Figure 6.20).
Two ducks are in front of the last duck. Two ducks are behind the first duck. There is one duck in the middle.
What do you sit on, sleep on, and brush your teeth with?
Are you not?