Counting Your Chickens Before They’re Pecked

It started with a brief story in the New York Times about Luke Robitaille, a 13-year-old student from Euless, Texas, who won the Raytheon Mathcounts National Competition by correctly answering the following question:

In a barn, 100 chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks?

Robitaille took less than a second to buzz in with the correct answer, according to the Times.

The next day, Jordan Ellenberg tweeted a followup problem:

Text of Ellenberg's tweet: 100 chicks in a circle. Each pecks R or L at random. Pecked chicks don't peck. Iterate until no two unpecked chicks adjacent. How many left?

Since I don’t have to squeeze this story into 140 characters, I’ll fill in some details of Ellenberg’s question, as I understand it. Where the original problem called for a single round of synchronized random pecking, we now have multiple rounds. During a round, each chick randomly turns either left or right and pecks one of its neighbors. However, once a chick has been pecked, it will never peck again, even if it continues to receive pecks. When two adjacent chicks peck each other in the same round, they both drop out of the pecking game for all future rounds. If an unpecked chick winds up sitting between two pecked neighbors, it can never be pecked and will therefore keep on pecking forever. The question is, what proportion of the flock will survive to become invulnerable peckers?

Spoilers below, so now’s the time to work out the answers for yourself. While you’re busy with that, I’m going to say a few words about chickens, and about the rhetoric and semiotics of mathematical “word problems.”

My only direct knowledge of poultry comes from boyhood visits to my Aunt Noretta’s farm in southern New Jersey. That’s not much of a claim to expertise, but for what it’s worth I never saw her chickens sit in a circle, and they didn’t peck randomly. (They had a pecking order!) Furthermore, nothing I observed in their social interactions resembled the turn-the-other-cheek behavior of the chickens described in this problem. Why does a pecked chick never peck again? This is a bigger riddle than the quantitative question we are asked to address. Has the chick suddenly discovered the wisdom and power of nonviolence? I can think of another explanation, but it’s not for the squeamish: Maybe pecked chicks don’t peck back because pecks are lethal.

I know it’s silly to demand narrative realism in a story like this one. Mathematical word problems belong to a genre where no one expects verisimilitude. They are set in a world where knaves always lie and knights always speak the truth, where shipwrecked sailors obsess about the divisibility properties of a pile of coconuts, where people don’t know the color of the hat on their own head. Even the laws of physics yield to mathematical necessity: A fly shuttling between oncoming locomotives instantaneously reverses direction. Those chicks sitting in a circle are not fluffly bundles of yellow plumage; they are mathematical abstractions. They have coordinates and state variables rather than feathers.

I’m okay with abstraction; by all means, let us strip away extraneous detail. Nevertheless, isn’t the point of word problems to connect the mathematics to some aspect of familiar experience? Consider the ancient and famous river-crossing problem, where the fox must not be left alone with the chicken, which must not be left alone with the bag of corn. These constraints are easy to understand when you know something about the dietary preferences of foxes and chickens. That kind of intuitive boost is not to be found in the pecking problem. On the contrary, a little knowledge of avian behavior actually makes the problem more perplexing.

But no matter. Onward! Have you come up with your answers?

The single-round problem from the Mathcounts Competition yields to the oldest trick in the probability book. A chick remains unpecked only if both of its neighbors turn away and peck in the other direction. On both the left and the right, the probability of escaping a peck is \(\frac{1}{2}\), and the two events are independent, so the probability of staying unpecked on both sides is \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). This argument applies identically to all the birds in the circle, so you can expect 25 percent of the chicks to come through unscathed.

Do you agree with this analysis? I came up with it pretty quickly when I read the Times article (though not nearly fast enough to beat Luke Robitaille to the buzzer). But then I began to have doubts. Is it strictly true that a chick’s left and right neighbors are totally independent? After all, they are connected by a chain of other chicks. Perhaps some influence can propagate around the circle, creating a correlation between left and right and altering the probability of survival.

Time for an experiment: Write the program, run the simulation. Set up a ring of 100 unpecked chickens and allow a single round of random simultaneous pecking. Repeat many times and calculate the mean number of unpecked birds remaining. (Some quick notation: Let \(N\) be the number of chicks in the ring and \(S\) be the number that survive unpecked. I’ll use \(\bar{S}\) for the mean value of \(S\) averaged over \(R\) repetitions of the experiment.) My results:

\(R\) \(\bar{S}\)
100 24.79
10,000 24.9881
1,000,000 25.000274
100,000,000 24.99991518

As expected, the mean is quite close to 25 survivors. Furthermore, each time the sample size increases by a factor of 100, the accuracy of the approximation improves about tenfold. This pattern conforms to a statistical rule of thumb—that the fluctuations in a random process are proportional to the square root of the sample size. Thus the slight departures from \(\bar{S} = 25\) appear to be innocent random noise, not some systematic bias.

So that settles it, right?

Well, the simulation looks pretty convincing for the specific case of \(N = 100\) chicks, but the result might differ for other values of N. In particular, perhaps there’s some finite-size effect that becomes apparent only when N is small. Consider a “circle” of just two chicks. In this situation the left neighbor and the right neighbor are one and the same chicken! No matter what random choices are made, the two chicks immediately peck each other, and the proportion of survivors is not 25 percent but zero.

The next-larger “circle” consists of three chicks arranged in a triangle. The two neighbors of a chick are distinct, but they are also neighbors of each other. What happens when the three chickens are set loose on one another? The system has \( 2^3 = 8\) possible pecking patterns, and we can easily examine all of them. In the diagram, the arrows indicate where the chicks choose to direct their pecking.

Three site enumeration

In two cases, where all the chicks peck left or all peck right, there are no survivors. In every other instance exactly one chick remains unpecked. Aggregating the eight patterns, we find six unpecked chicks out of 24 total chicks, for a proportion of \(\bar{S} = \frac{1}{4}\). Thus it appears the finite-size anomaly afflicts only the two-chick version of the problem.

But wait! There’s another possible confounding factor. Can we be sure of seeing the same outcome for both even and odd numbers of chicks? For any odd value of N there is just one way to annihilate all the chicks in a single round: They must all peck in the same direction. For even N, however, another pattern also leads to immediate extinction: Adjacent chicks can pair up, knocking each other out. Won’t this extra pathway slightly alter the overall probability of survival?

Let’s see what happens with N = 4. Now there are \(2^4 = 16\) possible outcomes:

Four site enumeration 16

As expected, four patterns leave no survivors at all. On the other hand, there are also four patterns that leave two chicks unpecked rather than just one. Miraculously, the extra losses and the extra gains balance exactly. In all we have 16 survivors out of 64 chicks, so the ratio is again \(\bar{S} = \frac{1}{4}\).

After that long and twisty detour through the combinatorics of chicken pecking, we are right back where we started. The probability of surviving unpecked after a single round of pecking is \(\frac{1}{4}\) for any \(N \gt 2\). All of my fretting about finite-size effects and odd-even disparities was a waste of time. So why have I inflicted it on you? Well, although those worries turn out to be unfounded, they are not farfetched. Making just a small change to the pecking protocol leads to a different outcome. Let the pecking be sequential rather than simultaneous. Some designated chick initiates the sequence of pecks, and then the birds take turns, proceeding clockwise around the circle. When a chick’s turn comes, if it has already been pecked, it does nothing. If it is unpecked, it pecks either its left or its right neighbor, choosing randomly. The round ends when every chick has had a turn.

For \(N = 2\) it’s easy to see that the first chick to peck always survives and the other chick always dies, for a survival rate of \(\frac{1}{2}\). With a little more pencil-and-paper chicken scratching, you can establish that the 50 percent survival rate also holds for \(N = 3\). Looking at very large values of \(N\), computer experiments indicate that the survival fraction again approaches \(\frac{1}{2}\) as N goes to infinity. Between these extremes, however, there’s some funny business:

Sequential pecking survival edit

At \(N = 4\) the survivor rate dips below 0.47. (The exact probability is \(\frac{30}{64} = 0.46875\).) This is a minimum. But as the rate recovers back toward 0.5, there is some telltale wiggling in the curve that reveals an odd-even bias: The survival probability is depressed further for even N than for odd N. This is just the kind of behavior I was looking for (but not finding) in the original Mathcounts version of the problem.

Let us now take up Ellenberg’s problem of iterated pecking (using the simultaneous rather than the sequential protocol). We already know that after the first round we can expect to find about one-fourth of the chicks still unpecked. Clearly, the unpecked fraction cannot increase after multiple rounds. Thus in the final state the expected surviving fraction \(\bar{S}\) must lie somewhere between zero and \(\frac{1}{4}\).

It’s helpful to look at a typical configuration of pecked (●) and unpecked (○) chicks after a single round of synchronized pecking:


(You’ll have to use your imagination to connect the left and right ends of this array and thereby form a ring.) Notice that there are long strings of pecked chicks, but the unpecked chicks appear in only two configurations. They are either singletons (●○●) or pairs (●○○●). The cause of this pattern is not hard to understand. After a round of pecking, a group of three consecutive unpecked chicks (●○○○●) is impossible. The middle chick must have pecked either left or right, and so it cannot have two unpecked neighbors.

These constraints simplify the analysis of subsequent rounds. The singletons are essentially immortal and unchangeable: The unpecked chick in the middle can never be pecked, and the pecked neighbors can never be unpecked. For the pairs, there are four possible fates, corresponding to the four ways the two active chicks could choose to peck:

Fates of pairs

In any one round, all four of these events have the same probability, namely \(\frac{1}{4}\). The first three result states are terminal, in the sense that further rounds of pecking will leave them unchanged. In the fourth case we are left with an adjacent pair again, which will therefore face the same set of choices in the next round. Eventually, as the number of rounds goes to infinity, the fourth case must yield one of the other outcomes, and thus in the long run we can consider the fourth case to have probability zero and each of the other three cases to have probability \(\frac{1}{3}\).

And now it’s time to bring all these contingent events together and work out a chicken’s long-term probability of survival. The diagram below presents the scheme. In the first round of pecking, three-fourths of the chicks are eliminated immediately. Of the remaining one-fourth, half are singletons, which survive indefinitely. The other surviving chicks are members of pairs, with another pecking chick as either a right neighbor or a left neighbor.

Probability calculation

The lower part of the diagram summarizes the effect of all subsequent rounds, which are assumed to continue until all pairs have been either annihilated or reduced to singletons. (I call this pecking to completion.) For each pathway that leads to a surviving singleton, the probability is the product of the individual probabilities encountered along that pathway. There are three such pathways, with probabilities \(\frac{1}{8}, \frac{1}{48}\), and \(\frac{1}{48}\), for a sum of \(\frac{1}{6}\).

I have to confess that I did not come up with this analysis—or with the correct answer—on my first try. I was able to work it out only after I had run a simulation and thus knew what I was looking for. Even then I had trouble with double counting.

Here are the simulation results:

\(R\) \(\bar{S}\)
100 16.53
10,000 16.6835
1,000,000 16.664404
100,000,000 16.66701664

Again note that accuracy seems to improve as the square root of the sample size, although the variance here is larger than in the single-round experiment.

What about finite-size effects? In circles with only two or three members, the fate of the chicks is fully decided after a single round of pecking: \(\bar{S}\) is 0 and \(\frac{1}{4}\) respectively. Thus these smallest rings escape the \(\frac{1}{6}\) rule, but it appears that circles of all larger sizes converge to \(\frac{1}{6}\). There’s no evidence of even-odd discrepancies.

Another approach to understanding the iterated chicken-pecking problem is through the theory of Markov chains. For a ring of \(N\) chicks we list all \(2^N\) states of the flock and assign a probability to each transition between states. Consider a ring of four chicks, which has 16 states. Symmetries allow us to consolidate some sets of states, and other states can be ignored because they are unreachable from the starting state of four unpecked chicks (Dots0000).

Markov symmetry classes

Only the four states in the red box need to be retained in the model. The transitions between them are recorded in a directed graph, where each arrow is labeled with the corresponding probability. Note that the starting state Dots0000 has only outgoing arrows; there is no way to re-enter the state once you leave. The states Dots1000 and Dots1111 are absorbing: The only outgoing arrow leads directly back to the same state; thus, once you reach one of those states, you never escape it.

Markov state diagram

The essential information from the directed graph can be captured in a \(4 \times 4\) matrix, where the rows and columns are labeled with the four states, and the matrix entries represent the probability of a transition from the row state to the column state. The entries in each row sum to 1, as they must if they are to represent probabilities.

Markov matrix

The pattern of zero entries in the transition matrix implies that certain states can’t be reached from other states, even by an indirect route. For this reason the Markov model is said to be irregular. That’s a bit awkward, because regular Markov models are easier to analyze and understand. In a regular model, when you take successive powers of the transition matrix, it converges to a steady state, where all the rows are identical and every column consists of a single, repeated value. This fixed point reveals the system’s long-term probability distribution. An irregular Markov model may not even have a stable limiting distribution, but this one does, and it seems to offer some insight. Every ring of four chickens must wind up in one of the two absorbing states. With probability two-thirds that terminal state will be Dots1000 and with probability one-third Dots1111. This result is consistent with the finding that one-sixth of the chickens survive unpecked.

Markov matrix limit

So, finally, that wraps it up, right? Both the contest problem and Ellenberg’s iterative extension asked for the expected number of surviving chickens, and we have supplied the answers: for a circle of \(N\) chickens, the expected number of survivors \(\bar{S}\) is \(\frac{N}{4}\) after a single round of pecking and \(\frac{N}{6}\) upon pecking to completion. Ironically, though, the expected value of a probabilistic process doesn’t necessarily tell you what to expect. Consider a simpler problem: When you flip a fair coin 100 times, how many heads do you expect to see? The obvious answer is 50, and it’s correct in the sense that no other number has a higher likelihood of correctly predicting the outcome of the experiment. However, the probability of seeing exactly 50 heads is only about 0.08, and thus some other number will turn up more than 90 percent of the time.

Instead of looking only at the expected value, let’s examine the range of possible \(S\) values in the pecking game. We’ve already established that zero survivors is a possible outcome, so that forms a lower bound. What is the upper bound—the maximum number of survivors? In the single-round process, every chick pecks, and so after that round every chick must have at least one pecked neighbor. On the basis of this fact I claim that the surviving population can never be greater than \(\frac{N}{2}\). (Do you agree? It took me a while to persuade myself it’s true.)

If \(S\) can never be greater than \(\frac{N}{2}\), the next question is whether it can ever attain that bound. And if we can have equal numbers of pecked (●) and unpecked (○) chicks, how are they arranged in the ring? It’s tempting to propose the following configuration:


This is a stable state: The unpecked chicks can never be pecked, so no further changes are possible. And the fraction of survivors is \(\frac{1}{2}\). But there’s a problem with this pattern: It cannot be reached from the starting state. Look at any of the black pecked chicks and ask yourself: Which of its neighbors did it peck? Neither of them, evidently, since they are both unpecked. But that’s not possible, given that every chicken must peck in the first round.

Although the alternating black and white arrangement is ruled out, we’re on the right track. There’s another configuration that also leaves one-half of the chicks unpecked after a single round, and that pattern is achievable from the starting state:


When you join the ends to form a ring, every chick, whether pecked or not, has one pecked neighbor. It turns out this is the only way—after allowing for some obvious symmetries—to reach 50 percent survivorship. (Strictly speaking, 50 percent is attainable only when \(N\) is divisible by 4, but \(S\) is never less than \(\frac{N-2}{2}\).)

When the pecking continues to completion, the upper bound of \(S = \frac{N}{2}\) is no longer reachable. Suppose we tried to maintain \(\frac{N}{2}\) over multiple rounds of pecking. Clearly we would have to start in the first round with the maximal-survivor state ●●○○●●○○●●○○. However, at least half of the unpecked chicks in this configuration must succumb in subsequent rounds, leaving no more than \(\frac{N}{4}\) survivors.

Does this argument mean that \(S = \frac{N}{4}\) is the greatest possible after pecking to completion? No, it doesn’t. There’s another pattern where one of every three chicks survives:


This configuration is reachable in a single round and stable indefinitely, since none of the pecking chicks has any pecking neighbors. No other arrangement has a higher density of survivors once the pecking process goes to completion.

To summarize: After one round of pecking the number of surviving chicks must lie somewhere between zero and \(\frac{N}{2}\), and the expected number \(\bar{S}\) is right in the middle at \(\frac{N}{4}\). After all further rounds of pecking are completed, the count of unpecked chicks is between zero and \(\frac{N}{3}\), with the expected value again in the middle, at \(\bar{S} = \frac{N}{6}\).

“How many chickens survive?” is a question that seems to call for a numeric answer, but in truth the most informative response is not a number at all; it is a distribution:

Fig x100r1000000 wide

Each curve records the results of a million experiments with a ring of 100 chicks, giving the frequency of each possible value of \(S\). As expected, the one-round distribution has a peak at 25 survivors, and the iterated curve peaks at 17 (the closest integer to \(\frac{100}{6}\). Note that the red curve is not only shifted to the left but is also slightly taller and narrower.

To get a better view of the details, let’s zoom in. For the sake of smoother curves, I’m going to switch to experiments with \(N = 10{,}000\) chickens. First the green single-round curve, then the red one for the iterated pecking experiment:

Fig x10000r1000000 single zoom

Fig x10000r1000000 zoom

With the larger value of \(N\), the curves now peak at 2500 and at 1666.67—exactly the positions expected for \(\frac{N}{4}\) and \(\frac{N}{6}\). Finding the peaks at these positions is no surprise, but what governs the width and the overall shape of the curve? In other words, what is the mathematical nature of the distributions?

One guess that’s always worth a try is the normal (or Gaussian) distribution. For the pecking problem, a normal distribution defines \(P(S)\), the probability of observing \(S\) survivors, as follows:

\[P(S) = \frac{1}{\sigma\sqrt{2 \pi}} \exp -\frac{1}{2}\left(\frac{S - \mu}{\sigma}\right)^2.\]

That’s a pretty messy equation for such a familiar concept, but it’s possible to tease out the basic meaning. The equation defines a symmatric curve with a peak where \(S\) is equal to \(\mu\), the mean of the distribution. The width of the peak depends on \(\sigma\), the standard deviation. Because the area under the curve is a constant, \(\sigma\) also effectively determines the height: A narrower peak has to be taller.

We can fit a normal distribution to the pecking data using a procedure that finds the optimal values of \(\mu\) and \(\sigma\)—those that minimize the discrepancy between the data points and the mathematical model. In the two graphs below the fitted models are superimposed on the two data plots, first for one round of pecking and then for pecking to completion:

Fig x10000r1000000 single zoom with normal

Fig x10000r1000000 iterated zoom with normal

The fits appear to be quite close indeed, with the theoretical curves splitting the experimental ones from end to end. In some sense this result has to be counted a success, and yet I don’t find this approach to the problem fully satisfying. The normal curve provides a very good descriptive model of the pecking process, but not a predictive or explanatory one. Remember, the curve is fitted to the data, not the other way around. I see no obvious way to construct a specific normal distribution from what I know about the underlying interactions of pecking chickens. In particular, where do the values of \(\sigma\) in the two models come from? Why is \(\sigma \approx 25\) in the one-round model and \(\sigma \approx 23.6\) in the iterated model? These values look like free parameters, which we have to tune to suit the data. Moreover, they will differ for every value of \(N\). Another issue: the normal curve is a continuous distribution, defined over the entire real number line. The pecking function is discrete; it makes sense only for integer numbers of chickens.

Let’s set aside the normal curve and consider another plausible model: the binomial distribution, which is discrete, and which turns up in many probabilistic contexts. Suppose you roll 10,000 dice and count how many of them come to rest with a 1 showing on the upper face. When you repeat this experiment many times, the expected number of 1s is one-sixth of 10,000, the same as the expected number of survivors in the iterated chicken-pecking experiment. With dice, there’s a well-known mathematical expression that defines not just the expected value but also the form of the entire distribution. Assume that every die has probability \(p\) of showing a \(1\). We are going to roll \(N\) dice and we want to know the probability of seeing \(k\) \(1\)s for any \(k\) between \(0\) and \(N\). The formula that supplies this information is:

\[P(k) = {N \choose k} p^k (1 - p)^{N - k}.\]

Here \(p^k (1 - p)^{N - k}\) gives the probability of any specific arrangement of \(k\) \(1\)s among \(N\) dice. The binomial coefficient \(N \choose k\), equal to \(N! / k! (N-k)!\), counts the number of such arrangements.

With \(N = 10000\) and \(p = \frac{1}{6}\) we get a curve showing the outcome of the dice-rolling experiment mentioned above. Perhaps the same curve also describes what happens to the iterated pecking model, which has the same expected value? Alas no.

Fig x10000 and binom 10000

The binomial curve is wider and flatter than the distribution of iterated pecking survivors. What has gone wrong? When I first saw the graph, I had an inkling. As noted above, the binomial coefficient \(N \choose k\) counts all the ways of choosing \(k\) items from a set of size \(N\). This is appropriate for an experiment with dice, since all possible arrangementds of \(k\) successes among \(N\) trials are equally likely. In particular, when you roll \(10{,}000\) dice, you could conceivably see no \(1\)s at all, or all \(10{,}000\) dice could land with a \(1\) showing face up; the entire range of outcomes has probability greater than zero.

The pecking problem is different. It’s not possible for 100 percent of the chickens to remain unpecked. Thus only a subset of the \(N \choose k\) arrangements are attainable. If the binomial distribution is going to work in this context, we need to adjust it somehow to include only the feasible outcomes.

With the thought that it’s easier to solve a problem if you already know the answer, I tried fiddling with the parameters of the distribution to see how the graph responded. My goal was to squeeze the curve into a narrower and taller profile while keeping it centered at the same mean. The mean is equal to \(Np\), so if we decrease \(N\) we have to increase \(p\) by the same factor. Here are the results of some experiments:

Fig x10000r1000000 withBinom

The dark green curve is the one we’ve already seen, for a binomial distribution with \(N = 10000\) and \(p = \frac{1}{6}\). Going to \(N = 5000\) and \(p = \frac{1}{3}\) appears to be a step in the right direction, and \(N = 3333\) and \(p = \frac{1}{2}\) is even better. Then, with \(N = 2500\) and \(p = \frac{2}{3} \ldots\) Bingo! The yellow curve is an excellent match to the pecking data. Thus it appears we can predict the survivorship of an \(N\)-member pecking ring by constructing a binomial distribution with parameters \(N^\prime = \frac{N}{4}\) and \(p^\prime = 4p\).

I can pull the same trick to find a binomial distribution that matches the single-round pecking data. This time the magic numbers that bend the curve to the correct trajectory are \(N’ = \frac{N}{3} = 3333\) and \(p’ = 3p = \frac{3}{4}\).

Fig x10000r1000000 single zoom with binomial

Unlike the normal distribution, the binomial model is constructive, or predictive. From the two parameters \(N’\) and \(p’\) we can calculate both the mean of the distribution and the standard deviation. The mean is simply \(N’ p’\); the standard deviation is \(\sqrt{N’ p’ (1 - p’)}\). For the example of the \(10{,}000\) chickens pecking to completion, the mean \(\mu\) works out to \(1{,}666.666 \dots\) (as expected), and the standard deviation \(\sigma\) is \(23.570226\). (The fitted normal distribution had \(\sigma = 23.567337\).) For the single-round case, \(\mu\) is exactly \(2500\) and \(\sigma\) is \(25\). (To avoid roundoff errors, I am taking \(N\) to be \(9999\) instead of \(10{,}000\).)

Hooray, eh? At last we have a formula for calculating the shape and location of the chicken-pecking distribution, based on a few simple parameters—\(N’\) and \(p’\). But I’m still grumpy, indeed more perplexed and frustrated than ever. Maybe the model explains the data, but what explains the model? With \(10{,}000\) chickens and a first-round survivor probability of \(\frac{1}{4}\), why does the formula call for \(N’ = 3333\) and \(p’ = \frac{3}{4}\)? Where do those numbers come from? And why \(N’ = 2500\) and \(p’ = \frac{2}{3}\) for the iterated case?

I am embarrassed to admit how long I have spent helplessly flailing and thrashing in the bogs of probability theory, trying to solve these mysteries. (I even turned to a recent book called The Probability Lifesaver, which I highly recommend—but it didn’t save my life.) In the search for answers I have investigated the multinomial extensions of binomials. I have looked into convolutions of distributions and computed contingent probabilities. I have filled whole pads of scratch-paper with soldierly rows of ●s and ○s, searching for patterns that would explain those enigmatic fractions \(\frac{N}{3}\) paired with \(\frac{3}{4}\), and \(\frac{N}{4}\) with \(\frac{2}{3}\). Night after night I’ve gone to bed with a promising idea, only to awaken and recognize a fatal flaw.

Now I believe I do have a correct explanation. It has passed the overnight test several nights in a row. I’m going to reveal it, but not until the end of this essay. Perhaps you’ll figure it out on your own before then. In the meantime, I’m going to widen the horizons of the chicken problem.

Our cozy circle of chickens is a one-dimensional structure. You can go clockwise or counterclockwise around the ring; there are no other meaningful directions in this little universe. Now suppose that instead of getting all our chickens in a row, we arrange them in a grid, an array of columns and rows, covering a region of a two-dimensional surface. To avoid leaving a subset of chickens on the exposed edges of a rectangular array, we can mate the left edge with the right edge and the top edge with the bottom edge. (Topologically, this turns the rectangle into a torus.) Getting real chickens to cooperate in this experiment would be even harder than in the one-dimensional version, but no matter; we’ve long since lost all touch with barnyard reality.

The most important fact about the two-dimensional flock is that each chicken has four neighbors instead of two. With twice as many hostile neighbors, one might well guess that a chicken would be more vulnerable to a pecking attack. On the other hand, each of those neighbors spreads its pecking over twice as many potential targets. How do these competing effects balance out?

For a single round of pecking, we can calculate the survival probability in the same way we did for the one-dimensional system. A chick remains unpecked only if all of its neighbors turn elsewhere to peck. Each neighbor does so with probability \(\frac{3}{4}\), and so the probability that all of them turn away is \(\left(\frac{3}{4}\right)^4\). Numerically, this works out to about 0.3164, compared with 0.25 in the circle. Thus the fraction surviving is greater in two dimensions than in one; the distraction of having more targets outweighs the danger of having more attackers. The distribution observed in computer experiments confirms this finding.

Fig 1D and 2D single

Here’s what a \(40 \times 40\) lattice of chicks looks like after a single round of pecking.

2d 40x40 lattice one round

There are 1,600 chicks in the two-dimensional array. If you count the unpecked ○s, you’ll find there are 501, for a survival fraction of 0.3131, close to the theoretical value of 0.3164. Simulations confirm the expected survival rate of \(\left(\frac{3}{4}\right)^4\) for \(N \times N\) lattices with any value of \(N\) greater than \(2\). (For the \(2 \times 2\) grid, the survival rate is \(\frac{1}{4}\), as in the one-dimensional system. There’s a reason why!)

When I stare at the pattern above, I notice a certain stringy or loopy texture, with chains of ○s separating blobs of ●s. This might be a trick of the eye and mind, but I think not. In two dimensions the no-three-in-a-row restriction is lifted; the array includes rows and columns with as many as six consecutive unpecked chicks, as well as diagonal lines. But you will not see a solid \(3 \times 3\) block () of unpecked chicks, or even a \(3 \times 3\) cross (). Such patterns cannot exist because the chick in the middle of the block must have pecked one of its four neighbors. More generally, the system is still bound by the rule that every chick, whether pecked or unpecked, must have at least one pecked neighbor.

Since more chicks survive the first round of pecking in a two-dimensional world, it seems plausible there might also be a greater proportion of survivors when the pecking continues to completion. Let’s try the experiment:

2d 40x40 lattice final

In this \(40 \times 40\) array there are 238 survivors out of 1,600 chicks, which is less than the one-sixth survival rate seen in one dimension. In a sample of a million such pecking grids, I found that the mean survival rate \(\bar{S}\) is about 0.1533. Compare the distributions for one- and two-dimensional systems:

Fig 1D and 2D mult

In going from 1D to 2D the peak shifts to the left, with the mean moving from 0.1667 to 0.1533. The 2D hump is also a little taller and skinnier, thus showing reduced variance.

Why stop at two dimensions? Let us ask our ever-accommodating chickens to roost in a three-dimensional lattice, again with opposite boundaries joined to create the 3D equivalent of a toroidal surface. It’s not hard to guess how this experiment is going to turn out. Back in one dimension, where every chick had two neighbors, the fraction of survivors after a single round of pecking was \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). In two dimensions, with four neighbors, the corresponding number was \(\left(\frac{3}{4}\right)^4 = \frac{81}{256}\). In the three-dimensional pecking party each chick has six neighbors, so the obvious extrapolation is \(\left(\frac{5}{6}\right)^6 = \frac{15625}{46656}\), with a value of \(\approx 0.3349\). Running the simulation supports this surmise, and shows a clear trend when we construct chicken lattices with still higher numbers of dimensions.

Fig 1D to 7D single

From this series of results we can boldly generalize: When every chick has \(n\) neighbors, the fraction expected to survive a single round of pecking is:

\[\left(\frac{n - 1}{n}\right)^n.\]

As \(n\) increases, this expression converges on a value of approximately \(0.36787944\). Does that number look familiar? It is \(\frac{1}{e}\). (Changing the minus sign to a plus generates \(e\) itself, \(2.71828\).) When I stumbled upon this formula, the sudden appearance of \(\frac{1}{e}\) took me by surprise, but it shouldn’t have. The constant turns up in the same way in a model of rumor spreading that I wrote about some years ago.

What about the iterated pecking process in higher dimensions? The fraction of survivors shows a steady decline as the number of dimensions increases:

Fig 1D to 7D mult

The proportion of chicks that never get pecked falls from 16.7 percent in one dimension to about half that when we embed our intrepid chickens in seven-dimensional space. In other words, a higher-dimensional space raises the initial survival rate (after one round of pecking), but depresses long-term survival (after pecking to completion). Here’s another way of showing the effect of dimension—tracking the mean number of survivors remaining after each round of pecking in one dimension through seven dimensions.

Fig progressive 1D to 7D

I can offer a rough, hand-wavy rationale for this trend. If you are a chick in a one-dimensional ring, your chance of surviving the first round of pecking is only \(\frac{1}{4}\), but if you make it through that round, your chance of avoiding a peck in the second round is at least \(\frac{1}{2}\). Why the improvement? It’s because of your own actions: Your pecking in the first round eliminated the threat from one of your two neighbors. Your odds continue improving in subsequent rounds: The longer you last, the greater the chance that you will hang on until all your neighbors are pacified.

The same trend holds in higher dimensions, but the magnitude of the effect tapers off. In four dimensions, for example, you have eight neighbors, and your chance of surviving the first round is \(\left(\frac{7}{8}\right)^8\), or about 0.34. Because you peck one of those neighbors, your probability of making it through the second round is better, but only slightly so: \(\left(\frac{7}{8}\right)^7\), or 0.39.

Looking at the graphs above, one might surmise that as the dimension \(D\) goes to infinity, the number of survivors (after pecking to completion) will drop to zero. To explore this idea, we don’t actually need infinite-dimensional space. What matters most is not the geometric arrangement of the chickens but the number of neighbors, and we can approximate an infinite-dimensional lattice just by declaring that all chickens are nearest neighbors. In other words, the who-pecks-whom graph becomes complete, with an arc from every chick to every other chick. This does seem to be a recipe for annihilation; you can’t be safe as long as even one other chicken continues to peck. But the details of the end game allow a little room for variation. Will there be one survivor or none?

Peter Winkler discusses a similar problem, “Group Russian Roulette,” in Mathematical Puzzles: A Connoisseur’s Collection (p. 33). The actors in his version are not chickens but “armed and angry people,” who engage in rounds of simultaneously shooting random neighbors. Winkler observes that the probability of a survivor does not approach a limit as \(N\) increases. I don’t see this effect in the chicken problem: There is almost always a last chicken standing. What makes the difference, I believe, is that Winkler’s roulette players don’t waste their ammunition on players who have already been shot, whereas the chickens continue to peck at neighbors who don’t peck back.

Finally, I return to the narrow confines of one dimension and to the mysterious binomial distributions that seem to predict the statistics of chicken pecking in this system. To review: If you roll 10,000 dice and count those that show a \(1\), you can expect to find about 1667. If you put 10,000 chicks in a circle and wait until all the pecking is done, you can expect about 1667 unpecked survivors. The dice experiment is described by a binomial distribution with parameters \(N = 10000\) and \(p = \frac{1}{6}\). The same model doesn’t work for the chickens: The predicted distribution is much broader than the observed one. But that’s not the weird part. The real puzzler is why a different binomial model, with parameters \(N’ = 2500\) and \(p’ = \frac{2}{3}\), does seem to match the experimental results.

The dice model’s failure to work for chicken pecking is not really a surprise. A key assumption underlying the binomial distribution is that the events or objects being counted are independent. That’s true for dice; one die doesn’t care what the others do. But the circle of pecking chickens is all about interactions between neighbors. If you have already been pecked, that alters the odds that your neighbors will eventually be pecked. Independence enters the binomial distribution through the coefficient \(N \choose k\). Given \(N\) dice with \(k\) of them showing \(1\)s, all possible interleavings of the \(1\)s among the other dice are equally likely; the binomial coefficient counts those arrangements. But given \(N\) chicks with \(k\) of them unpecked, it’s not true that all arrangements are equally likely. Indeed, many patterns, such as ○○○, are impossible.

If neighbor interactions spoil the binomial model with \(N = 10{,}000\) and \(p = \frac{1}{6}\), how are those interactions overcome in the model with \(N’ = 2500\) and \(p’ = \frac{2}{3}\)? For the longest time I was beguiled by the observation that 2500 is the expected number of survivors after a single round of pecking, and two-thirds of those individuals can be expected to survive all subsequent rounds. Surely, having those two numbers turn up in the binomial distribution cannot be a meangingless coincidence. Maybe not, but I was able to make sense of the situation only when I gave up on that line of inquiry.

What’s needed is a model in which we count the arrangements of 2500 objects, where two-thirds of the objects can be considered successes or survivors. I have found such a model. The objects are not individual chickens. They are groups of four chickens. Consider this set of 4-tuples:

a = ○●●●
b = ●○●●
c = ●●●●

If you select elements from this set at random and string them together, any sequence you create could be an output of the iterated pecking process. A typical result looks like this:


Note that this sequence satisfies all the rules for flock of chickens that has pecked to completion. All unpecked ○s are singletons, surrounded by pecked neighbors. At least two ●s separate every pair of ○s, and this ensures that every element of the sequence has at least one ● neighbor. There is no way of con­catenating any selection of a, b, and c elements that violates these rules. Furthermore, if a, b, and c are chosen with equal probability, the expected proportion of ○s in the sequence is \(\frac{1}{6}\).

I am deeply ambivalent about this discovery. On the one hand, it’s always a relief to get to the bottom of a problem that has stumped you. On the other hand, what we have here is a recipe for creating a sequence with the same structure and statistics as the product of the pecking process, but it offers no insight into the nature of that process. There’s no connection with the behavior of the chickens. Worse, it’s not even a true or exact model. Although the curve appears to coincide with the data, it’s only an approximation. The proof of this fact is simple. The binomial distribution with \(N’ = 2500\) and \(p’ = \frac{2}{3}\) has an absolute cutoff at \(2500\). For any number of survivors greater than \(2500\), the model assigns a probability of zero. Yet the flock of \(10{,}000\) pecking chickens can in fact leave up to \(3333\) survivors.

The defect becomes visible in a smaller model, such as this one with \(N = 24\):

Fig x24r1000000 withBinom

The predicted and observed curves exhibit slight mismatches everywhere, but pay particular attention to the right tail of the distribution, where the binomial curve (purple) dives to zero for all survivor numbers greater than six, whereas the experimental data (red) include 6718 instances with seven survivors and 49 instances with eight survivors.

A similar model for the one-round pecking process uses a set of four 3-tuples:

a = ○●●
b = ○●●
c = ●●○
d = ●●●

Again it generates a sequence that looks very much like the outcome of a pecking experiment, but fails to reproduce the tail of the distribution. In the model the highest possible density of survivors is \(\frac{1}{3}\) whereas it should be \(\frac{1}{2}\).

Perhaps you’re thinking that a cute high school problem about chicks pecking their neighbors doesn’t really merit an 8,000-word screed on Markov chains and probability distributions, with tables and equations and 25 graphs and diagrams. That thought has crossed my mind, too. However, I want to add just a few more words to argue that the exercise is not totally frivolous.

Mathematics does not owe us a tidy, closed-form, one-line solution to every problem, but we’d be foolish to give up the quest too easily. In this case, computer simulations are easy and productive. By running a program for five minutes I can get answers to a multitude of detailed questions, and I don’t have serious doubts about the correctness of those answers. But they don’t help me make the connection between the microscopic mechanisms (a chicken pecks left or right at random) and macroscopic observations (the distribution has \(\mu = \frac{1}{6}\) and \(\sigma = 23.56\)). Richard Hamming’s old chestnut says the purpose of computing is insight, not numbers, but insight is just what I’m missing.

Second, this is not really a problem about chickens, whether real or abstract. It is a gateway to a collection of other many-body problems in statistical physics and dynamical systems and cellular automata.

Finally, I’ve had fun, and what’s the harm in that? Maybe the fun’s not over. What about zombie chickens, whose pecks bring other chickens back to life?

Update 2017-07-11: Carl Witty has worked out the correct probability distribution for the single-round case. See his comment below.

Posted in biology, computing, featured, mathematics, problems and puzzles, statistics | 8 Comments

The short arm of coincidence

James Tanton tosses off number theory problems the way John D. Rockefeller handed out dimes. I wrote about one of Tanton’s problems back in January. Then a few weeks ago this tweet about factorials and squares snagged my attention, and it hasn’t let go:

Tweet reads: 4!+1 = 25, a square number. 5!+1 = 121, a square number. Another example? Two more examples?

With pencil and paper it’s easy to show that \(6!\) doesn’t work. The factorial of \(6\) is \(1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720\); adding \(1\) brings us to \(721\), which is not a square. (It factors as \(7 \times 103\).) On the other hand, \(7!\) is \(5040\), and adding \(1\) yields \(5041\), which is equal to \(71^2\). This makes for a very cute equation:

\[7! + 1 = 71^2.\]

Continuing on, you can establish that \(8! + 1\), \(9! +1\) and \(10! + 1\) are not square numbers. But to extend the search much further, we need mechanized assistance. Here’s a Julia function that does the obvious thing, generating successive factorials and checking each one to see if it is \(1\) less than a perfect square:

function search_fac_sqr(maxn)
  fac = big(1)                      # bigints needed for n > 20
  for n in 1:maxn
    fac *= n                        # incremental factorial
    r = isqrt(fac + 1)              # floor of sqrt
    if r * r == fac + 1
      println(n, "! + 1 = ", r, "^2 = ", r^2)
  println("That's all folks!")

With this tool in hand, let’s check out \(n! + 1\) for all \(n\) between \(1\) and \(100\). Here’s what the program reports:

4! + 1 = 5^2 = 25
5! + 1 = 11^2 = 121
7! + 1 = 71^2 = 5041
That's all folks!

Those are the three cases we’ve already discovered with pencil and paper—and no more are listed. In other words, among all values of \(n! + 1\) up to \(n = 100\), only \(n = 4\), \(n = 5\), and \(n = 7\) yield squares. When I continued the search up to \(n = 1{,}000\), I got exactly the same result: no more squares. Likewise \(n = 10{,}000\) and \(n = 100{,}000\). Allow me to mention that the factorial of \(100{,}000\) is a rather large number, with \(456{,}574\) decimal digits. At this point in the search, I began to grow weary; furthermore, I began to lose hope. When \(99{,}993\) successive values of \(n\) fail to produce a single square, it’s hard to sustain faith that success might be just around the corner. Nevertheless, I persisted. I got as far as \(n = 500{,}000\), which has \(2{,}632{,}341\) decimal digits. Not one more perfect square in the whole lot.

What can we learn from this evidence—or lack of evidence? Are 4, 5, and 7 the only values of \(n!\) that lie \(1\) short of a perfect square? Or are there more such cases somewhere out there along the number line, maybe just beyond my reach, waiting to be found? Could there be infinitely many? If so, where are they? If not, why not?

To my taste, the most satisfying way to resolve these questions would be to find some number-theoretical principle ensuring that \(n! + 1 \ne m^2\) for \(n \gt 7\). I have not discovered any such principle, but in a dreamy sort of way I can imagine what a proof might look like. Suppose we eliminate the “\(+1\)” part of the formula, and search for integers such that \(n! = m^2\). It turns out there is just one solution to this equation, with \(n = m = 1\). You needn’t bother lathering up your laptop in the quest for larger examples; there’s a simple proof they don’t exist. In any square number, all the prime factors must be present an even number of times, as in \(36 = 2 \times 2 \times 3\times 3\). In a factorial, at least one prime factor—the largest one—always appears just once. (If you’re not sure why, check out Bertrand’s postulate/Chebyshev’s theorem.)

Of course when we put the “\(+1\)” back into the formula, this whole line of reasoning falls to pieces. In general, the factorization of \(n!\) and of \(n! + 1\) are totally different. But maybe there’s some other property of \(n! + 1\) that conflicts with squareness. It might have something to do with congruence classes, or quadratic residues. From the definition of a factorial, we know that \(n!\) is divisible by all positive integers less than or equal to \(n\), which means that \(n! + 1\) cannot be divisible by any of those numbers (except \(1\)). This observation rules out certain kinds of squares, namely those that have small primes in their factorization. But for all \(n \gt 4\) the square root of \(n!\) greatly exceeds \(n\), so there’s plenty of room for larger factors, as in the case of \(7! + 1 = 71^2\).

Here’s another avenue that might be worth exploring. The decimal representation of any large factorial ends with a string of \(0\)s, formed as the products of \(5\)s and \(2\)s among the factors of the number. Thus \(n! + 1\) must look like

\[XXXXX \ldots XXXXX00000 \ldots 00001,\]

where \(X\) represents any decimal digit, and the trailing sequence of \(0\)s now ends with a single terminal \(1\). Can we figure out a way to prove that a number of this form is never a square? Well, if the final digit were anything other than \(1, 4,\) or \(9\), the proof would be easy, but lots of squares end in \(\ldots 01\), such as \(10{,}201 = 101^2\) and \(62{,}001 = 249^2\). If there’s some algebraic argument along these lines showing that \(n! + 1\) can’t be a square, it will have to be something subtler.

All of the above is make-believe mathematics. I have stirred up some ingredients that look like they might make a tasty confection, but I have no idea how to bake the cake. Perhaps someone else will supply the recipe. In the meantime, I want to entertain an alternative hypothesis: that nothing prevents \(n! + 1\) from being a square except improbability.

The pattern observed in the \(n! + 1 = m^2\) problem—a few matches among the smallest elements of the sequences, and then nothing more for many thousands of terms—is not unique to factorials and squares. Other pairs of sequences exhibit similar behavior. For example, I have tried matching factorials with triangular numbers. The triangulars, beginning \(1, 3, 6, 10, 15, 21, \ldots\), are defined by the formula \(T(m) = m(m + 1)/2\). If we look for factorials that are also triangular, we get \(1! = T(1) = 1\), then \(3! = T(3) = 6\), and finally \(5! = T(15) = 120\). No more examples appear through \(n = 100{,}000\).

What about factorials that are \(1\) less than a triangular, satisfying the equation \(n! + 1 = T(m)\)? I know of only one case: \(2! + 1 = 3\). Broadening the search a little, I found that \(n! + 4\) is triangular for \(n \in {2, 3, 4}\), again with no more hits up to \(100{,}000\).

For another experiment we can bring back the square numbers and swap out the factorials, replacing them with the ever-popular Fibonacci sequence, \(1, 1, 2, 3, 5, 8, 13, \ldots\), defined by the recurrence \(F(n) = F(n - 1) + F(n - 2)\), with \(F(1) = F(2) = 1\). It’s been known since the 1960s that \(1\) and \(144\) are the only positive integers that are both Fibonacci numbers and perfect squares. Looking for Fibonacci numbers that are \(1\) less than a square, I found that \(F(4) + 1 = 4\) and \(F(6) + 1 = 9\), with no other instances up to \(F(500{,}000)\).

We can do the same sort of thing with the Catalan numbers, \(1, 1, 2, 5, 14, 42, 132 \ldots\), another sequence with a huge fan club. I find no squares other than \(1\) among the Catalan numbers up to \(n = 100{,}000\); I don’t know if anyone has proved that none exist. A search for cases where \(C(n) + 1 = m^2\) also comes up empty, but there are a few low-lying matches for \(C(n) + k = m^2\) for \(k \in {2, 3, 4}\).

Finding similar behavior in all of these diverse sequences changes the complexion of the problem, in my view. If we discover some obscure, special property of \(n! + 1\) that explains why it never lands on a square (for large values of \(n\)), do we then have to invent another mechanism for Fibonacci numbers and still another for Catalan numbers? Isn’t it more plausible that some single, generic cause lies behind all the observations?

But the cause can’t be too generic. It’s not the case that you can take any two numeric sequences and expect to see the same kind of pattern in their intersections. Consider the factorials and the prime numbers. By the very nature of a factorial, none of them except 2! = 2 can possibly be prime, but there’s no obvious reason that \(n! + 1\) can’t be a prime. And, indeed, for \(n \le 100\) nine values of \(n! + 1\) are prime. Extending the search to \(n \le 1000\) turns up another seven. Here is the full set of known numbers for which \(n! + 1\) is prime:

\[1, 2, 3, 11, 27, 37, 41, 73, 77, 116, 154, 320, 340, 399, 427, 872, 1477, \\ 6380, 26951, 110059, 150209\]

They get rare as \(n\) increases, but there’s no hint of a sharp cutoff, as there is in the other cases explored above. Does the sequence continue indefinitely? That seems a reasonable conjecture. (For more on this sequence, including references, see Chris K. Caldwell’s factorial prime page.)

My question is this: Can we understand these curious patterns in terms of mere chance coincidence? The values of \(n! + 1\) form an infinite sequence of integers spread over the number line, dense near the origin but becoming extremely sparse as \(n\) increases. The values of \(m^2\) form another infinite sequence, again with diminishing density, although the dropoff is not as steep. Maybe factorials bump into squares among the smallest integers because there just aren’t enough of those integers to go around, and some of them have to do double duty. But in the vast open spaces out in the farther reaches of the number line, a factorial can wander around for years—maybe forever—and not meet a square.

Let me try to state this idea more precisely. Since \(n!\) cannot be a square, we know that it must lie somewhere between two square numbers; the arrangement on the number line is \((m - 1)^2 \lt n! \lt m^2\). The distance between the end points of this interval is \(m^2 - (m - 1)^2 = 2m - 1\). Now choose a number \(k\) at random from the interval, and ask whether \(n! + k = m^2\). Exactly one value of \(k\) must satisfy this condition, and so the probability of success is \(1/(2m - 1)\), or roughly \(1 / (2 \sqrt{n!})\). Because \(\sqrt{n!}\) increases very rapidly, this probability takes a nosedive toward zero as \(n\) increases. It is represented by the red curve in the graph below. Note that by \(n = 100\) the red curve has already reached \(10^{-80}\).

Plot of probability of coincidence of factorial and square, fibonacci and square, factorial and prime

The green curve gives the probability of a collision between Fibonacci numbers and squares; the shape is similar, though it dives off the precipice a little later. The Fibonacci-square curve approximates a negative exponential: The probability is proportional to \(\phi^{-\sqrt{F(n)}}\), where \(\phi = (\sqrt{5} + 1) / 2 \approx 1.618\). The factorial-square curve is even steeper because the factorial function is superexponential: \(n!\) grows faster than \(c^n\) for any fixed \(c\).

The blue curve, recording the probability of coincidences between factorials and primes, has a very different shape. In the neighborhood of \(n!\) the average distance between consecutive primes is approximately \(\log n!\), which grows just a little faster than \(n\) itself and very much slower than \(n!\). The probability of collision between factorials and primes is roughly \(1 / \log n!\). The continuous blue curve corresponds to this smooth approximation. The blue dots sprinkled near that line give the probability based on actual distances between consecutive primes.

What to make of those curves? Is it legitimate to apply probability theory to these totally deterministic sequences of numbers? I’m not quite sure. Before confronting the question directly, I’d like to retreat a few steps and look at a simpler model where probability is clearly entitled to a seat at the table.

Let us borrow one of Jacob Bernouilli’s famous urns, which have room to hold an infinite number of ping pong balls. Start with one black ball and one white ball in the urn, then reach in and take a ball at random. Clearly, the probability of choosing black is \(1/2\). Put the chosen ball back in the urn, and also add another white ball. Now there are three balls and only one is black, so the probability of drawing black is \(1/3\). Add a fourth ball, and the probability of black falls to \(1/4\). Continuing in this way, the probability of black on the \(n\)th draw must be \(\frac{1}{n + 1}\).

If we go on with this protocol forever—always choosing a ball at random, putting it back, and adding an extra white ball—what is the probability of eventually seeing the black ball at least once? It’s easier to answer the complement of this question, calculating the probability of never seeing the black ball. This is the infinite product \(\frac{1}{2} \times \frac{2}{3} \times\frac{3}{4} \times\frac{4}{5} \ldots\), or:

\[P(\textrm{never black}) = \prod_{n = 1}^{\infty} 1 - \frac{1}{n+1}\]

The product goes to zero as \(n\) goes to infinity. In other words, in an endless series of trials, the probability of never drawing black is \(0\), which means the probability of seeing black at least once must be \(1\). (“Probability \(1\)” is not exactly the same thing as “certain,” but it’s mighty close.)

Now let’s try a different experiment. Again start with one black ball and one white ball, but after the first draw-and-replace cycle add two white balls, then four white balls, and so on, so that the total number of balls in the urn at stage \(n\) is \(2^n\); throughout the process all of the balls but one are white. Now the probability of never seeing the black ball is \(\frac{1}{2} \times \frac{3}{4} \times\frac{7}{8} \times\frac{15}{16} \ldots\), or:

\[P(\textrm{never black}) = \prod_{n = 1}^{\infty} 1 - \frac{1}{2^n}\]

This product does not go to zero, no matter how large \(n\) becomes. Neither does it go to \(1\). The product converges to a constant with the approximate value \(0.288788095\). Strange, isn’t it? Even in an infinite series of draws from the urn, you can’t be sure whether the black ball will turn up or not.

These two urn experiments do not correspond directly to any of the sequence coincidence problems described above; they simply illustrate a range of possible outcomes. But we can rig up an urn process that mimics the probabilistic treatment of the factorials-and-squares problem. At the \(n\)th stage, the urn holds \(1 + 2 \sqrt{n!}\) balls, only one of which is black. The probability of never seeing the black ball, even in an infinite series of trials, is

\[\prod_{n = 1}^{\infty} 1 - \frac{1}{1 + 2 \sqrt{n!}}.\]

This expression converges to a value of approximately \(0.2921426977\). It follows that the probability of seeing black at least once is \(1 - 0.2921426977\), or \(0.7078573023\). (No, that number is not \(1/\sqrt{2}\), although it’s close.)

An urn process resembling the factorials-and-primes problem gives a somewhat different result. Here the number of balls in the urn at stage \(n\) is \(\log n!\), again with just one black ball. The infinite product governing the cumulative probability is

\[\prod_{n = 2}^{\infty} 1 - \frac{1}{\log n!}.\]

On numerical evidence this expression seems to dwindle away to zero as \(n\) goes to infinity (although I’m not \(100\) percent sure of that). If it does go to \(0\), then the complementary probability that the black ball will eventually appear must be \(1\).

Some of these results leave me feeling befuddled, and even a little grumpy. Call me old-fashioned, but I always thought that rolling the dice infinitely many times ought to be enough to settle beyond doubt whether a pattern appears or not. In the harsh light of eternity, I would have said, everything is either forbidden or mandatory; as \(n\) goes to infinity, probability goes to \(0\) or it goes to \(1\). But apparently that’s not so. In the factorial urn model the probability of never seeing a black ball is neither \(0\) nor \(1\) but lies somewhere in the neighborhood of \(0.2921426977\). What does that mean, exactly? How am I supposed to verify the number, or even check its first few digits? Running an infinite series of trials is not enough; you need to collect a statistically significant sample of infinite experiments. For an exact result, try an infinite series of infinite experiments. Sigh.

The urn model corresponds in a natural way to the randomized version of the factorial-square problem, where we look at \(n! + k = m^2\) and choose \(k\) at random from an appropriate range of values. But what about the original problem of \(n! + 1 = m^2\)? In this case there’s no random variable, and hence there’s no point in running multiple trials for each value of \(n\). The system is deterministic. For each \(n\) the factorial of \(n\) has a definite value, and either it is or it isn’t adjacent to a perfect square. There’s no maybe.

Nevertheless, there might be a way to sneak probabilities in through the back door. To do so we have to assume that factorials and squares form a kind of ergodic system, where observing one chain of events for a long period is equivalent to watching many shorter chains. Suppose that factorials and squares are uncorrelated in their positions on the number line—that when a factorial lands between two squares, its distance from the larger square can be treated as a random variable, with every possible distance being equally likely. If this assumption holds, then instead of looking at one value of \(n!\) and trying many random values of \(k\), we can adopt a single value of \(k\) (namely \(k = 1\)) and look at \(n!\) for many values of \(n\).

Is the ergodic assumption defensible? Not entirely. Some distances between \(n!\) and \(m^2\) are known to be more likely than others, and indeed some distances are impossible. However, the empirical evidence suggests that the deviations must be slight. The histogram below shows the distribution of distances between a factorial and the next larger square for the first \(100{,}000\) values of \(n!\). The distances have all been normalized to the range \((0, 1)\) and classified in \(100\) bins. There is no obvious sign of bias. Calculating the mean and standard deviation of the same \(100{,}000\) relative distances yields values within \(1\) percent of those expected for a uniform random distribution. (The expected values are \(\mu = 1/2\) and \(\sigma = 1/12\).)

relative distance of n! from m^2, in 100 bins, for n ranging from 2 to 100,000

If this probabilistic approach can be taken seriously, I can make some quantitative statements about the prospects for ever finding a large factorial adjacent to a perfect square. As mentioned above, the overall probability that one or more values of \(n! + 1\) are equal to squares is about \(0.7078573023\). Thus we should not be too surprised that three such cases are already known, namely the examples with \(n = 4, 5,\) and \(7\). Now we can apply the same method to calculate the probability of finding at least one more case with \(n \gt 7\). Let’s make the question more general: “Whether or not I have seen any squares among the first \(C\) values of \(n! + 1\), what are the chances I’ll see any thereafter?” To answer this question, we can just remove the first \(C\) elements from the infinite product:

\[\prod_{n = C+1}^{\infty} 1 - \frac{1}{1 + 2\sqrt{n!}}.\]

For \(C = 7\), the answer is about \(0.0037\). For \(C = 100\), it’s about \(5.7 \times 10^{-80}\). We are sliding down the steep slope of the red curve.

As a practical matter, further searching for another factorial-square couple does not look like a promising way to spend time and CPU cycles. The probability of success soon falls into the realm of ridiculously small numbers like \(10^{-1{,}000{,}000}\). And yet, from the mathematical point of view, the probability never vanishes. Removing a finite number of terms from the front of an infinite product cannot change its convergence properties. If the original product converged to a nonzero value, then so will the truncated version. Thus we have wandered into the canyon of maximal frustration, where there’s no realistic hope of finding the prize, but the probabilities tell us it still might exist.

I am going to close this shambling essay by considering one more example—a cautionary one. Suppose we apply probabilistic reasoning to the search for a cube that is \(1\) less than a square. If we were looking for exact matches between cubes and squares, we’d find plenty of them: They are the sixth powers: \(1, 64, 729, \ldots\). But integer solutions to the equation \(n^3 + 1 = m^2\) are not so abundant. One low-lying example is easy to find: \(2^3 + 1 = 3^2\), but after 8 and 9 where can we expect to see the next consecutive cube and square?

The probabilistic approach suggests there might be reason for optimism. Compared with factorials and Fibonaccis, cubes grow quite slowly; the rate is polynomial rather than exponential or superexponential. As a result, the probability of finding a cube at a given distance from a square falls off much less steeply than it does for \(n!\) or \(F(n)\). In the graph below, \(P(n^3 + k = m^2)\) is the orange curve.

Plot of probability of coincidence of factorial and square, fibonacci and square, factorial and prime, with added curve for cubes and squares

Note that the orange curve lies just below the blue one, which represents the probability that \(n!\) lies near a prime. The proximity of the two curves suggests that the two problems—factorials adjacent to primes, cubes adjacent to squares—might belong to the same class. We already know that factorial primes do seem to go on and on, perhaps endlessly. The analogy leads to a surmise: Maybe cube-square coincidences are also unbounded. If we keep looking, we’ll find lots more besides \(8\) and \(9\).

The surmise is utterly wrong. The problem has a long history. In 1844 Eugène Catalan conjectured that \(8\) and \(9\) are the only consecutive perfect powers among the integers; the conjecture was finally proved in 2004 by Preda Mihăilescu. For the special case of squares and cubes, Euler had already settled the matter in the 18th century. Thus, probabilities are beside the point.

All of the questions considered here belong to the category of Diophantine analysis—the study of equations whose solutions are required to be integers. It is a field notorious for problems that are easy to state but hard to solve. Catalan’s conjecture is one of the most famous examples, along with Fermat’s Last Theorem. When Diophantine problems are ultimately resolved, the proofs tend to be non-elementary, drawing on sophisticated tools from distant realms of mathematics—algebraic geometry in the proof of Fermat’s Last Theorem by Andrew Wiles and Richard Taylor, cyclotomic fields in Mihăilescu’s proof of the Catalan conjecture. As far as I know, probability theory has not played a central role in any such proof.

When I started wrestling with these questions a few weeks ago, I did not expect to discover a definitive solution. I’ve certainly fulfilled my expectations! As a matter of fact, in my own head the situation is more muddled now than it was at the outset. The realization that even an infinite series of experiments would not necessarily resolve some of the questions is deeply unsettling, and makes me wonder how much I really understand about probability theory. But that’s hardly unprecedented in mathematics. I suppose I’ll just have to get used to it.

Update: Thanks to a further tip from Tanton, I have learned that the problem has an extensive history, and also a name: Brocard’s problem, after Henri Brocard, who published on it in 1876 and 1885. Ramanujan mentioned it in 1913. Erdos conjectured there are no more solutions. Marius Overholt connected it with the abc conjecture. Bruce C. Berndt and William F. Galway established that there are no more solutions up \(10^9\). All this comes from the Wikipedia entry on Brocard’s problem. That article also mentions (but does not explain) that the solutions are called Brown numbers.

I have some more reading to do.

Posted in computing, featured, mathematics, problems and puzzles | 14 Comments

The uniqueness constraint

For the past few weeks the Sunday New York Times has been publishing a puzzle called Capsules, devised by Wei-Hwa Huang. Here are the instructions:

Place numbers in the grid so that each outlined region contains the numbers 1 to n, where n is the number of squares in the region. The same number can never touch itself, not even diagonally.

Here is a partially completed example:

Capsules puzzle with numbers entered in some of the cells.

The black, pre-printed numbers are the “givens,” supplied by the puzzle creator. I filled in the pencil-written numbers in a sequence of “forced” moves dictated by two simple rules:

  1. A number can be placed in a square if no other number is allowed there. For example, the three singleton squares in the bottom row must each hold a 1, and these squares are the obvious place to start solving the puzzle. After the 1s are written in, the square outlined in yellow in the diagram below can also be filled in; its neighbors forbid any number other than 3.
  2. A number can be placed in a square if the number has no other possible home within a region. The blue-outlined 1 in the diagram below was determined by this rule. There must be a 1 somewhere in the region, but none of the other squares can accommodate it.

The same partially completed puzzle grid, with two squares marked by blue and yellow outlines.

At this point in the solution process, with the grid in the state shown above, I was unable to find any other blank squares whose contents could be decided by following these two rules and no others. But I did spot a move based on a different kind of reasoning. Consider the two pairs of open squares marked in color:

The same puzzle, with two blank squares colored.

The salmon-pink squares must hold the numbers 2 and 5, but it’s not immediately clear which number goes in which square. Likewise the lime-green squares must hold 2 and 4, in one order or the other. I submit that the numbers must have the following arrangement:

A correct extension of the partial solution.

How do I justify that choice? Suppose the green 2 and 4 were transposed:

An incorrect extension of the partial solution, allowing an amiguity in two squares.

Then the pink 2 and 5 could be placed in either permutation, and no later moves elsewhere in the puzzle would ever resolve the ambiguity. This outcome is not acceptable if we assume the puzzle must have a unique solution. The uniqueness constraint might be expressed as a third rule:

  1. A number can be placed in a square if it is needed to prevent other squares from having multiple legal configurations.

I have vague qualms about this mode of puzzle-solving. It’s surely not cheating, but the third rule has a different character from the others. It exploits an assumed global property of the solution, rather than relying on local interactions. We are not making a choice because it is forced on us; we are choosing a cofiguration that will force a choice elsewhere.

In this particular puzzle it’s not actually necessary to apply the uniqueness constraint. There is at least one other pathway to a solution—which I’ll leave to you to find. Can we devise a puzzle that requires rule 3? I’m not quite sure the question is even well-formed. All constraint-satisfaction problems can be solved by a mindless brute-force algorithm: Just write in some numbers at random until you reach a contradiction, then backtrack. So if we want to force the solver to use a specific tool, we somehow have to outlaw that universal jackhammer.

The uniqueness constraint is not unique to the Capsules puzzle. I’ve encountered it often in kenkens, and occasionally in sudokus. I even have a sense of deja lu as I write this. I feel sure I’ve read a discussion of this very issue, somewhere in recent years, but I haven’t been able to lay hands on it. Pointers to precedents are welcome.

Addendum 2017-03-19: Jim Propp reminds me of his marvelous Self Referential Aptitude Test. The instructions begin:

The solution to the following puzzle is unique; in some cases the
knowledge that the solution is unique may actually give you a short-cut
to finding the answer to a particular question.

I completed the 20-question puzzle when SRAT first went public some years ago. This morning I found I was able to do it again with no diminution in enjoyment—or effort. I remembered none of the answers or the sequence of deductions needed to find them.

Highly recommended. And while you’re at it, check out Propp’s Mathematical Enchantments blog and his Twitter feed: @JimPropp.

Posted in problems and puzzles | 17 Comments

A Tantonalizing Problem

In times like these one craves distraction, or maybe anaesthesia. On the whole, mathematics is better for you than ethanol, and you can even do it while driving. So in spare moments I’ve been noodling away at the following problem, tweeted a week ago by James Tanton:

Tanton tweet

The answer to Tanton’s question is surely No: The series will never again land on an integer. I leaped to that conclusion immediately after reading the definition of the series and glancing at the first few terms. But what makes me so sure? Can I prove it?

I wrote a quick program to generate more terms:


Overall, the trend visible in these results seemed to confirm my initial intuition. When the fractions are expressed in lowest terms, the denominator generally grows larger with each successive term. Looking at the terms more closely, it turns out that the denominators tend to be products of many small primes, whereas the numerators are either primes or products of a few comparatively large primes. For example:

\[\frac{9649}{2520} = \frac{9649}{2^3 \cdot 3^2 \cdot 5 \cdot 7} \qquad \textrm{and} \qquad \frac{18358381}{4084080} = \frac{59 \cdot 379 \cdot 821}{2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17}.\]

To produce an integer, we need to cancel all the primes in the factorization of the denominator by matching primes in the numerator; given the pattern of these numbers, that looks like an unlikely coincidence.

But there is reason for caution. Note the seventh term in the sequence, where the denominator has decreased from \(60\) to \(20\). To understand how that happens, we can run through the calculation of the term, which starts by summing the six previous terms.

\[\frac{60}{60} + \frac{120}{60} + \frac{150}{60} + \frac{170}{60} + \frac{185}{60} + \frac{197}{60} = \frac{882}{60}.\]

Then we calculate the mean, and add 1 to get the seventh term:

\[\require{cancel}\frac{882}{60} \cdot \frac{1}{6} = \frac{882}{360} = \frac{\cancel{2} \cdot \cancel{3} \cdot \cancel{3} \cdot 7 \cdot 7}{\cancel{2} \cdot 2 \cdot 2 \cdot \cancel{3} \cdot \cancel{3} \cdot 5} = \frac{49}{20} + 1 = \frac{69}{20}\]

Cancelations reduce the numerator and denominator of the mean by a factor of 18. It seems possible that somewhere farther out in the sequence there might be a term where all the factors in the denominator cancel, leaving an integer.

Another point to keep in mind: For large \(n\), the value of the Tanton function grows very slowly. Thus if integer values are not absent but merely rare, we might have to compute a huge number of terms to get to the next one. Reaching the neighborhood of 100 would take more than \(10^{40}\) terms.

value of tanton(n) for n from 1 to 300

So what do you think? Can we prove that no further integers appear in Tanton’s sequence? Or, on the contrary, might my instant conviction that no such integers exist turn out to be an alternative fact?

I’ve had my fun with this problem. I know the answer now, but I’m not going to reveal it yet. Others also deserve a chance to be distracted, or anaesthetized. I’ll be back in a few days to follow up—unless commenters explain what’s going on so thoroughly there’s nothing left for me to say.

Update 2017-01-30: Okay, pencils down. Not that anyone needs more time. As usual, my readers are way ahead of me. (See comments below, if you haven’t read them already.)

My own slow and roundabout voyage of discovery went like this. I had written a little piece of code for printing out n terms of the series, directly implementing the definition given in James Tanton’s tweet:

from fractions import Fraction as F
from statistics import mean

def tanton (n):
    seq = [F(1)]
    for i in range(n):
        seq.append(mean(seq) + 1)

But this is criminally inefficient. On every pass through the loop we calculate the mean of the entire sequence, then throw that work away and do it all again the next time. Once you have the mean of \(n-1\) terms, isn’t there some way of updating it to incorporate the nth term? Well, yes, of course there is. You just have to appropriately weight the new term, dividing by n, before adding it to the mean. Here’s the improved code:

from fractions import Fraction as F

def faster_tanton (n):
    m = F(1)
    for i in range(1, n):
        m += F(1, i)

Tracing the execution of this function, we start out with 1, then add 1, then add 1/2, then 1/3, then 1/4, and so on. This is 1 plus the harmonic series. That series is defined as:

\[H_{n} = \sum_{i=1}^{n} \frac{1}{i} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\]

The first 10 partial sums are:


One fact about the harmonic series is very widely known: It diverges. Although \(H_{n}\) grows very slowly, that growth continues without bound as \(n\) goes to infinity. Another fact, not quite as well known but of prime importance here, is that no term of the series after the first is an integer. The simplest proof shows that when you factor the numerator and the denominator, the denominator always has more \(2\)s than the numerator; thus when the fraction is expressed in lowest terms, the numerator is odd and the denominator even. This proof can be found in various places on the internet, such as StackExchange. There’s also a good explanation in Julian Havil’s book Gamma: Exploring Euler’s Constant.

Neither of those sources mentions anything about the origin or author of the proof. When I scouted around for more information, I found more than a dozen sources that attribute the proof to “Taeisinger 1915,” but with no reference to an original publication. For example, a recent paper by Carlo Sanna (Journal of Number Theory, Vol. 166, September 2016, pp. 41–46) mentions Taeisinger and cites Eric Weisstein’s Concise Encyclopedia of Mathematics; consulting the online version of that work, Taeisinger is indeed credited with the theorem, but the only reference is to another secondary source, Paul Hoffman’s biography of Erdős, The Man Who Loved Only Numbers; there, on page 157, Hoffman writes, “In 1915, a man named Taeisinger proved. . .” and gives no reference or further identification. So who was this mysterious and oddly named Taeisinger? I have never heard of him, and neither has MathSciNet or the Zentralblatt or the MacTutor math biography pages. In Number Theory: A Historical Approach John J. Watkins gives a slender further clue: The first initial “L.”

After some further rummaging through bookshelves and online material, I finally stumbled on a reference to a 1915 publication I could actually track down. In the Comptes Rendus Mathematique (Vol. 349, February 2011, pp. 115–117) Rachid Aït Amranea and Hacène Belbachir include this item in their list of references:

L. Taeisinger, Bemerkung über die harmonische Reihe, Monatsch. Math. Phys. 26 (1915) 132–134.

When I got ahold of that paper, here’s what I found:

Opening lines of the Theisinger 1915 paper

Not Taeisinger but Theisinger!

I still don’t know much of anything about Theisinger. His first name was Leopold; he came from Stockerau, a small town in Austria that doesn’t seem to have a university; he wrote on geometry as well as number theory.

What I do know is that a lot of authors have been copying each other’s references, going back more than 20 years, without ever bothering to look at the original publication.

Posted in mathematics, problems and puzzles | 8 Comments

The Polite Apocalypse

On the Beach cover of 1957 editionI’ve been rereading On the Beach, Nevil Shute’s novel about humanity’s last gasp in the aftermath of a nuclear war. The book was published 60 years ago, in 1957. I first read it in 1963, which is roughly when the events in the story are supposed to take place. It was also just a few months after the Cuban missile crisis, when annihilation was in the air.

Spoiler alert: Everybody dies.

The setting is Melbourne, Australia, the southernmost major city on the planet. The entire population of the Northern Hemisphere was wiped out in the war, and airborne radioactivity is slowly creeping across the Equator. Darwin and Cairns, on Australia’s north coast, are already ghost towns, and the people of Melbourne are told they have less than a year to go.

A U.S. submarine takes refuge in Melbourne’s harbor. Over a period of weeks the captain of that vessel, Dwight, forms an attachment to a young woman named Moira. There’s affection on both sides, and maybe passion, but Dwight is determined to remain faithful to his wife and children back in Connecticut. He buys them presents: a diamond bracelet, a fishing rod, a pogo stick. He speaks of them in the present tense. Is Dwight delusional? Not exactly. He knows perfectly well that his family are all dead, and that he’ll never rejoin them except in the sense that he too will soon be dead. But those deaths are abtractions.

He had seen nothing of the destruction of the war . . . ; in thinking of his wife and of his home it was impossible for him to visualize them in any other circumstances than those in which he had left them. He had little imagination, and that formed a solid core for his contentment in Australia.

It’s not just Dwight who lacks imagination—or chooses to ignore the truths it reveals. Moira studies shorthand and typing for a future job that will never exist. Her father harrows fields for crops that will never grow. Another couple plant hundreds of daffodils whose blooms they will never see, and they invest in a lawn mower for grass they’ll never cut.

The author himself seems to share this selective connection to reality. Everyone in his doomed society is unfailingly polite, and usually cheerful. Civilization may be ending, but not civility. There’s not a single act of violence or malice or even selfishness in the entire story. Shute mentions no hoarding or profiteering, much less rape and pillage. No marauding bandits or desperate refugees from the contaminated north descend on this last haven. On the Beach is the antithesis of that other Australian vision of apocalypse: the Mad Max movies. (The first of the series, in 1979, was filmed near Melbourne.)

It’s also worth noting that no one in Shute’s world takes any steps to prolong life. The government is not hollowing out mountains to keep the human germ line going until the atmosphere clears. Families are not digging fallout shelters in the back yard. These last few representatives of Homo sapiens may indulge in a variety of follies, but hope isn’t one of them.

Why am I writing about this sad book just now? Well, obviously, it’s inauguration day. Which feels more like termination day.

The threat of nuclear disaster has continued to shadow us through all the years since Shute wrote his novel. The danger of a planet-scouring war seemed particularly urgent when I was 13 and reading On the Beach for the first time. I stood up in front of my eighth-grade English class to give an oral report on the book. My performance was not interrupted by a duck-and-cover drill, but it could have been.

Now we have handed control of 4,700 nuclear warheads to a petulant brat, and the danger seems greater than ever.

Revisiting that sense of menace is why I picked up the book, but it’s not what has made the strongest impression on me this second time around. I am both drawn to and appalled by the stoic acceptance of Shute’s fictional Melbournites. Given their circumstances, their reaction is not inappropriate. The worst has already happened, there’s nothing they can do to change it, they may as well make the best of it. In the face of certain extinction, what can you do but shrug your shoulders? Maybe the best way of muddling through is just to plant some daffodils.

Given the current mood of the nation and the world, I suddenly find it easier to understand Dwight’s behavior. The urge to pretend is powerful. I too want to believe that life can go on as normal, that I can continue to enjoy the private pleasures of family and friends, that I can retreat to a cozy office or library and lose myself in the world of ideas, in the “less fretful cosmos” of mathematics and science, or art and literature for that matter.

But we are not yet huddled on the beach, the last of the doomed. It’s late, but not yet too late. This is not the moment for resignation and acquiescence. Tomorrow we march!

Posted in off-topic | 2 Comments