A group of $n\geq 3$ people is sitting at a round table, so that each person has two neighbors,one clockwise neighbour and one counter clockwise neighbour. Each person flips a fair and independent coin. A person starts singing if and only if (i) his coin comes up heads, (ii) the coin of his clockwise neighbour comes up tails, and (iii) the coin of his counter clockwise neighbour comes up tails. Let X be the random variable whose value is the number of people that are singing. What is the expected value $E(X)$ of X?

I'm unsure of how exactly to figure out the $E(X)$, however I have a guess that it's $\frac{n}{8}$ since only the you, or the one sitting counter-clockwise, and his counter-clockwise can sing. Therefore there is 3 people that could possibly sing. However two people sing if it's landed tails twice after a single heads. Therefore the possibility is:

$$\frac{n}{2^3} = \frac{n}{8}$$

This this reasonable?


2 Answers 2


Let $Y_i$ be a Boolean variable that is 1 if the $i$th person sings and 0 otherwise. Then, $X = \sum_i Y_i$. And $E(Y_i) = Pr\{Y_i = 1\} = 1/8$.

By linearity of expectation, $E[X] = E[\sum_i Y_i] = \sum_i E[Y_i] = n/8$.


Your answer is absolutely right; but, the explanation could use some work. Let me give you one way that you could explain it!

The key here is that if $\xi_i$, $i=1,2,\ldots,n$, is the indicator variable of the event "person $i$ starts singing", then $X=\xi_1+\xi_2+\cdots+\xi_n$, and therefore $$ \mathbb{E}[X]=\sum_{i=1}^{n}\mathbb{E}[\xi_i]=\sum_{i=1}^{n}P(i\text{ starts singing}). $$ Now, for any $i$, person $i$ starts singing if and only if three coin tosses turn out in one specific way, and those coin tosses are fair and independent; hence $P(i\text{ starts singing}0=\frac{1}{2^3}=\frac{1}{8}$, and as a result we have \begin{equation*} \mathbb{E}[X]=\sum_{i=1}^{n}\frac{1}{8}=\frac{n}{8}. \end{equation*}


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