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I recently watched this 3Blue1Brown video that has a small problem at the end (9:41), which I haven't been able to solve. The problem is the following (my own phrasing):

Suppose that you have $n$ students sitting in a circle taking a test. It's a hard test, so each student tries to cheat off of his neighbour, choosing randomly which neighbour to cheat from. What is the expected number of students that does not have a neighbour cheating from them?

I am aware that it is stated in the video that a link to a solution can be found in the description. I have not been able to find a solution following this link, however. I have found the following values: $$\begin{array}{c|ccccc} n&3&4&5&6&7\\ \hline E(n)&\frac{3}{4}&\frac{4}{4}&\frac{5}{4}&\frac{6}{4}&\frac{7}{4} \end{array},$$ which suggests that $E(n)=\frac{n}{4}$ for $n>2$, but I do not see how to prove this. I keep on running into the problem of the probabilities not being independent. I've also tried to phrase the problem in terms of 'paths' formed by following which student is watching which, but this has not lead to anything so far. Any help or hint is welcome.

Jos
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  • $n/4$ is correct, by linearity of expectation. You don't need independence for this to be true. – saulspatz Aug 30 '18 at 00:47
  • How does this follow from linearity of expectation? It's not even clear to me precisely how the distributions we get for different values of $n$ are related... – Jos Aug 30 '18 at 00:48
  • Let $X_i$ be a random variable that is $1$ if no one is cheating off student $i$ and $0$ otherwise, so that the expected number of students with no one cheating from them is $E(\sum X_i) = \sum E(X_i)$ by linearity. – saulspatz Aug 30 '18 at 00:54
  • @saulspatz So simple yet so brilliant. Thanks! – Jos Aug 30 '18 at 00:56
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    This is isomorphic to https://math.stackexchange.com/questions/2282622/expected-number-of-unpecked-chicks-nyt-article/2282639 – Misha Lavrov Oct 12 '18 at 00:18

1 Answers1

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Let $X_i$ be a random variable that is $1$ if no one is cheating off student $i$ and $0$ otherwise, so that the expected number of students with no one cheating from them is $E(∑Xi)=∑E(Xi)={n\over4}$ by linearity of expectation.

saulspatz
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