Expected number of unpecked chicks - NYT article

Question

In this article, the winner of the math competition answered this question correctly:

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?

The answer was given as 25. I am interested to know the correct method for solving this problem as well as a general way to find this out for N chicks. I was thinking of trying to solve by figuring out the number of occurrences of (something?) inside a binary string of length 100 (or N), but I don't know if that is the right way to approach it.

Answer 1

For any individual chick, there is a $0.5$ chance that the one on its right won't peck it, and a $0.5$ chance that the one on its left won't peck it. So overall, it has a $0.25$ chance of not being pecked.

Since this is true for every chick, we can add up $0.25(100)=25$ to get the number of unpecked chicks.

"But wait," you might say, "the probabilities that chicks are unpecked aren't independent! If chick $n$ is unpecked, then chicks $n-2$ and $n+2$ will be pecked. Surely we have to take this into account!"

Fortunately, there's a very useful theorem called linearity of expectation. This tells us that for any random variables $X$ and $Y$, independent or not, $E[X+Y]=E[X]+E[Y]$, where $E$ is the expected value. What we're formally doing above is assigning a random variable $X_i$ to the $i$th chick that is equal to $0$ if the chick is pecked and $1$ if it's unpecked. Then the total number of unpecked chickens is $\sum X_i$, so the expected number of unpecked chickens is $$E\left[\sum X_i\right]=\sum E[X_i]$$ by linearity of expectation. But $E[X_i]$ is just the probability that the $i$th chicken is unpecked, thus justifying our reasoning above.

Answer 2

The probability that a chick is unpecked is the probability that its left neighbors pecks left and its right neighbor pecks right. Assuming that each chick chooses left or right with probability $0.5$ and that choices are independent, you get the result.

The number of chicks doesn't affect the $0.25$ probability as long as it is greater than $2$.

Answer 3

Consider the even numbered chicks, and ask the expectation of how many odd chicks are unpecked plus the number that are double pecked. If we create a random walk with step $k$ growing left if chick $2k$ pecks left, then we see that the number of unpecked or double-pecked odd chicks is the number of sign changes in a cyclical sequence of length 50. The expectation for that is 25. SO the number of unpecked chickes is half that, or 12.5. Adding, now the even chicks, we get the answer of $12.5\times 2=25$.

Answer 4

This is a simple problem. Look, ma, no equations!

Consider YOU are the only chicken that matters, and construct a table to say whether YOU get pecked. Your chance of being pecked comes down to only 4 outcomes. (1) YES - pecked twice. (2) YES - pecked from left wing only. (3) YES - pecked from right wing only. (4) NO - unpecked.

The table has 4 elements, all of equal probability, 1 of which is unpecked. YOU are therefore pecked 3:1 ratio or 3:4 opportunities or 75% of the time. For convenience, this needs to be conducted for 100 trials of YOU, and the answer is that 25 times YOU will NOT be pecked. The circular nature of the 100 chicks says that YOU are not unique, and your experience is the same as the others, so we extrapolate your experience of 100 trials to a single trial of 100 chicks just like YOU.

25 unpecked chicks, 50 get pecked once, 25 get double pecks.

This is the same table constructed for 100 women having two children and asking how many have no girls.

Answer 5

I tried to reduce complexity to adjust it to my intellectual competence, so I considered 4 chicken in an event matrix. That shows:

• the probability that 2 chicken (50 %) remain unpecked is 0.25
• the probability that 1 chicken (25 %) remains unpecked is 0.5
• the probability that no chicken (0 %) remains unpecked is 0.25

Answer 6

It is possible to have a state with $50$ unpecked chicks. It's also possible to have $0$ unpecked chicks and any integer between $0$ and $50$ is possible. $51$ unpecked chicks is impossible because then there would be no way to have $100$ pecks totally. If we can show that having $x$ unpecked chicks is equally likely as having $50-x$ unpecked chicks, then the average of both scenarios gives you $25$ unpecked chicks. So you number the chicks $1$ to $100$ such that consecutive numbers are adjacent to each other and $1$ is next to $100$. Change the directions of each chick that is divisible by $4$ and each chick that is congruent to $1 \mod 4$. I believe this involution gives us a one to one correspondence between states with $x$ unpecked chicks and states with $50-x$ unpecked chicks. Each state is equally likely and this means that $25$ has to be the expected number of unpecked chicks.

My explanation only works when the total number of chicks is positive and divisible by $4$. Luckily, $100$ is divisible by $4$.

Answer 7

I thought of pairs. Only scenario where they're unpecked is if two chicks next to each other peck away from each other. 25% probability - either both to left, both to right, both away!, or both towards each other. Pretty much the same thing as above.

Answer 8

It is overkill for this, but you can write a simulation to estimate the expected number. Such a simulation doesn't prove anything per se, but sometimes it is useful to write a quick simulation to confirm that a probability calculation is correct. Here is one in R:

count.unpecked <- function(n){
  chicks <- 0:(n-1)
  pecks <- sample(c(-1,1),n,replace = TRUE)
  pecked <- (chicks + pecks) %% n
  n - length(unique(pecked))
}

print(mean(replicate(100000,count.unpecked(100))))

This simulates the experiment 100,000 times. Output from my last run: 25.00679

Answer 9

I shall solve the problem generally, let the total number of chicks be k, such that k > 3 (the reason behind this shall become apparent soon). Consider the set of all possible orientations of the chicks, essentially, all the different ways the chicks can peck each other. Let's focus our attention on one particular chick, say chick A. Consider all possible configurations of chicks if chick A and the two adjacent chicks to A were removed. Let the number of such configurations be n. For any one of these n configurations, we can construct four new configurations including the three previously removed chicks, three in which chick A is pecked, and one in which it isn't. Obviously, the total number of configurations of chicks is 4n. Now in all 4n configurations, chick A is not pecked n times. This is true for any of the k chicks, so the total number of times a chick is not pecked throughout all 4n configurations is kn. The expected value of the number of chicks not pecked is the total number of chicks that are unpecked divided by the total number of possible configurations, which is kn/4n = k/4.

Answer 10

Represent chickens as variables ($a_{1}, a_{2}, \dots, a_{100})$. Assign it a $1$ if unpecked, $0$ otherwise. We want to find the expected value of the sum $a_{1} + \dots + a_{100}$. Notice that the expected value of each of these is $\frac{1}{4}$, so by linearity of expectation, the expected value of the sum is $100 \cdot \frac{1}{4} = 25$

Answer 11

I suspect the original answer to the competition question is not correct. I am not a mathematician and I apologize if I am wrong. Putting aside "how does the chick feel about pecking left and right" because it's completely random as stated, there are in total 6 possible sets of outcome, 4 that were mentioned and 2 additional ones. The last 2 sets represent another "chance" or another possible set of outcomes. The problem here is that we have 100 chicks and not 4.

1. all chicks peck to the left. 0% of chicks are unpecked.
2. all chicks peck to the right. 0% of chicks are unpecked.
3. all chicks, divided in pairs, peck each other, 0% of chicks are unpecked.
4. chicks are divided in groups of 4, where the pair in the middle pecks each other, while chicks on the edge peck this pair in the middle. 50% of chicks are unpecked.

There are however other possible probability distributions. Two new patterns emerge from this.

5. chicks are divided into groups of 3, where a pair of chicks pecks each other and one chick from this pair is double pecked by a chick on the LEFT. This chick on the left is unpecked which makes a total of 33%. The last, 100th chick can peck randomly left or right but remains unpecked itself as in example 4). This gives 34 unpecked chicks.
6. chicks are divided into groups of 3, where a pair of chicks pecks each other and one chick from this pair is double pecked by a chick on the RIGHT. The last, 100th chick can peck randomly left or right but remains unpecked itself as in example 4). This gives 34 unpecked chicks.

A mixture of all those sets of outcomes are, of course, also possible. Now we can calculate the median of these 6 possible outcomes: (0+0+0+50+34+34)/6=19.666 somewhat lower than the original solution.