24

Chickens in Boxes

24.1 THE PROBLEM (AND SOME WARM-UPS, TOO)

This book opened with a lot of commentary (mostly unhappy, I’m afraid¹) on Marilyn vos Savant’s mathematics and, if only for symmetry reasons, it seems appropriate to end with another of her puzzles. So, the final solved problem in this book is from her Parade Magazine column of August 4, 2002, in which she printed the following question from a reader:

Suppose you’re on a game show. There are four boxes in an L-shaped configuration, like this:

The host tells you:

What are the chances that a chicken is in the corner box? In one way, the chances seem to be 1 in 3; but in another way, the chances seem to be 1 in 2. They can’t both be right!

To that vos Savant added a little challenge of her own, writing, “What’s the answer, readers? … Also answer this: Which box is most likely to contain a chicken? We’ll publish the first letter that correctly answers both questions.”

Well, who could resist that? I soon had the solution in hand, but I didn’t bother sending it in. I figured plenty of her other readers would. I simply stuck the solution in my files and forgot about it, recalling it only after I started writing this book. Apparently nobody else sent anything in either, however, as vos Savant never—so far as I know, and I looked pretty hard—wrote anything more on chickens in boxes.

This problem is one that can be solved in about two minutes with the honorable technique of counting. Don’t dismiss counting as only for kids; you’ll recall we used that approach in an earlier problem dealing with nontransitive bingo cards. If you can solve a good probability problem by simply counting, the very first things you should count are your blessings! There is, in fact, a large number of other interesting probability classics that are also nothing but counting puzzles. And so, before I show you the chickens-in-boxes solution, I’ll illustrate (let me count the ways—five times!) the counting approach, in increasing order of subtleness.

(1) Assume the birth of a boy or a girl is equally likely, and then consider a family with two children. If you are told that the older child is a boy, what is the probability that both children are boys? If you are told that at least one of the children is a boy, what is the probability that both children are boys? The mystery of this problem is that, at first glance, it is difficult to see the distinction between the two given conditions; they appear to be providing essentially the same information. And yet the two questions have quite different answers. Here’s how to see that. Make the following table showing the four possible, equally likely two-children families:

If we are told that the older child is a boy, then we know the only possibilities are the ones in the last two rows. Since only one of those rows (the last one) has both children as boys, then the probability of both children being boys (with the given condition) is 1/2. If we are told that at least one of the children is a boy, then we know the only possibilities are the ones in the last three rows. Since only one of those rows (the last one) has both children as boys, then the probability of both children being boys (with the given condition) is 1/3.

(2) A big jar contains an unknown number of pennies. Sixty of them are taken out, given a dab of red paint, and then returned to the jar. The jar is given a vigorous shaking, and then 100 pennies are drawn at random, without replacement. If 15 of the drawn pennies have a dab of red paint on them, about how many pennies were there originally in the jar? This may seem to be simply a silly game, but substitute “lake” for jar and “fish” for penny and you have the very clever, so-called “recapture method” used to estimate the size of the fish population in a lake. Now, how does it work? Getting 15 red pennies out of the total of 60 painted ones in the jar means we drew one-fourth of the red pennies. So, we expect that the 85 unpainted pennies we drew to also be about one-fourth of the unpainted pennies. That is, there were around 340 unpainted pennies. Combined with the 60 painted pennies, a reasonable estimate is that there were originally around 400 pennies in the jar.

(3) There are four cards on the table in front of you, each with a colored side face down. Two of those faces are black and two are white. If two of the cards are picked at random and turned face up, what is the probability they will have the same color? Most people instantly answer with 1/2, but, while intuitive, that answer is wrong. Here’s why. Turn the first selected card face up. It’s either black or white. Suppose it’s black. That means the three cards that are still face down consist of one black card and two white cards. The probability of picking the remaining black card is 1/3. If, on the other hand, the first selected card is white, that means the three cards that are still face down consist of one white card and two black cards. The probability of picking the remaining white card is 1/3. Since each of these two possibilities has probability 1/2 of occurring, our answer is (1/2)(1/3) + (1/2)(1/3) = 1/3.

(4) Recall the hat-matching problem I mentioned back in the preface. Suppose there are N hats in a box, where N = 2, 3, or 4. Find the exact probability for, in each of these three cases, that at least one man gets his hat back. To answer this question, all we need do is write down all N! permutations of the first N positive integers and then count how many have at least one instance of the jth integer in the jth position (starting at the left). For example, if N = 2 there are two permutations of 1,2 (that, and 2,1). The first has two matches, and the second has none. So, the probability for the N = 2 case is 1/2. For N = 3 we have six permutations and (I’ll let you confirm) four of them have at least one match. So, the probability for the N = 3 case is 4/6 = 2/3. If you do this for N = 4 you’ll find that of the 24 permutations there are 15 that have at least one match, and so the probability is 15/24 = 5/8. Beyond N = 4 this does get a bit tiresome. The exact theoretical answer for any N is 1 − 1/2! + 1/3! −… ± 1/N!. As N → ∞ (as in the example I used in the preface of a million men and their hats), this becomes the power series expansion for 1 − e⁻¹ ≈ 0.632, the value I gave in the preface.

(5) My final counting example is an old one, dating from a problem in Lewis Carroll’s 1890s book, Pillow Problems and a Tangled Tale. (Carroll’s real name, of course, was Charles Dodgson [1832–1898], and he was the author of Alice in Wonderland and Through the Looking-Glass.) Imagine you have an urn with a single ball in it, either black or white with equal probability. Which color it is you don’t know. You drop a white ball into the urn, and so now there are two balls in the urn. Then, at random, you reach in and take a ball out, which you see is white. What is the probability that the ball remaining in the urn is also white? Carroll gave the following faulty “proof” that it is 1/2 (he knew it was faulty, but he was just being funny). As he put it, at the start there is a white ball in the urn with probability 1/2. You then put a white ball in, and then take a white ball out. So, nothing has changed and the probability a white ball is in the urn is still 1/2. It is astonishing how many people actually think that is okay! Here’s the right way to do it. Let’s define the state of our problem to be a description of what is both inside and outside the urn. Initially, there are two possible, equally likely states:

where, in this notation, w_o is the original outside white ball and b and w_i are the possible black or white inside balls, respectively. Then, we drop w_o into the urn, which gives us two equally likely possible states:

Finally, we draw out a ball (which we are told is white), creating the possibility of three equally likely states:

Of these three equally likely possible states, two have a white ball remaining in the urn, and so the probability of a white ball remaining in the urn is 2/3.

Using this same approach, you should now be able to answer this generalization of Dodgson’s problem. An urn contains, with equal probability, either n white balls or n black balls. You first drop a white ball into the urn, then randomly withdraw a ball and find that it is white. Show that the probability the original n balls in the urn were white is (n + 1)/(n + 2).

The equally likely nature of the various possible states in Dodgson’s problem (and its generalization) is crucial to the analysis of the problem. Sometimes, however, making a mistake about “equally likely” can be a trap for the unwary. For example, recall my mention back in the preface of Todhunter’s criticism of D’Alembert. The error that tripped up D’Alembert was in the calculation of the probability of throwing a head in two tosses of a fair coin. He said it was 2/3, while a modern student of probability would instead calculate the answer to be 3/4, as follows. If we write H for heads and T for tails, then there are four possible outcomes to the two tosses: HH, HT, TH, and TT, with each possibility having equal probability (for a fair coin) of 1/4. These four possibilities are the sample points in the sample space of the experiment of tossing a coin twice. The first three of these outcomes have a head (or two), and so the probability of the event “we see a head” is the ratio of the number of favorable outcomes (3) to the total number of outcomes (4).

How did D’Alembert get 2/3? He asserted that if H appears on the first toss, then the experiment immediately ceases, and the second toss simply doesn’t occur. Thus, there are only three possible outcomes: H, TH, and TT. Since there are two “favorable” outcomes out of three total outcomes, the answer is 2/3. The error in this is that the sample points in this sample space are not equally likely, with the first one (H) having a probability of 1/2 and each of the other two having a probability of 1/4. What D’Alembert should have written for the answer is Prob(H) + Prob(TH) = 1/2 + 1/4 = 3/4.

Let’s now apply the counting approach to vos Savant’s chickens-in-boxes problem.

24.2 THEORETICAL ANALYSIS

Before we apply any special constraints, there are 16 possible states for the four boxes, each of which, independently, may or may not contain a chicken. (I’ll assume the “may” and “may not” are equally likely, and so each possibility has probability 1/2.) In the following table I’ve listed all of these states, where a 0 means an empty box and a 1 means a box with a chicken in it. Now, let’s apply the given constraints. Only rows 3, 4, 5, 6, 9, and 10 satisfy the constraint that there is one chicken in the vertical boxes (Boxes 1, 2, and 3). Only rows 2, 3, 6, 7, 10, 11, 14, and 15 satisfy the constraint that there is one chicken in the horizontal boxes (Boxes 3 and 4). Only rows 3, 6, and 10 satisfy both constraints.

Looking at those three rows, we see that only one of them (row 3) has a chicken in the corner box (Box 3), and so the probability of a chicken in the corner box is 1/3. As for vos Savant’s extra-credit question, to identify the box most likely to contain a chicken, in those three rows we see that Boxes 1, 2, and 3 each have a chicken in just one row, while Box 4 has a chicken in two of the three rows. So, Box 4 is the box most likely to have a chicken.

And with this dramatic calculation finished, so is this book of puzzles with solutions. The next puzzle, the last one in this book, comes with no solution because it remains an open challenge. If you solve it, fame is guaranteed. But be forewarned: while it is easy to understand, it has stumped all who have tried to solve it before you.

NOTE

1. This is a probability book, and so I’ve limited myself to commenting on vos Savant’s often curious approach to probability mathematics. But I have occasionally found her mathematical physics to be a bit shaky, as well. The very next question in her chickens-in-boxes column, for example, was from another reader asking about the reality (or not) of a triple rainbow. Vos Savant correctly said such a phenomenon does indeed exist, but then put it in the wrong part of the sky! She puts it just above the primary/secondary rainbows, when in fact a ground observer would have to turn around and look away from the primary/secondary rainbows to be looking in the right direction. You can find the mathematical physics behind the primary and higher-order rainbows in my book When Least Is Best (Princeton 2007), pp. 179–198.