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My teacher gave the following question as a practice question for the exam... I was just wondering if someone could check if my answer is correct:

A group of n ≥ 3 people is sitting at a round table, so that each person has two neighbors, one clockwise neighbor and one counter clockwise neighbor. 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 neighbor comes up tails, and (iii) the coin of his counter clockwise neighbor 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?

So the chance of any of the given n people getting heads with both of their neighbours getting tails is $\frac{1}{8}$. So $X_i = \frac{1}{8}$

Then, using linearity of expectation:$$\mathbb{E}\big(X\big) = \mathbb{E}\bigg(\sum\limits_{i=1}^n X_i \bigg) = \sum\limits_{i=1}^n \frac{1}{8} = \frac{n}{8}$$ Is this correct?

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  • $\begingroup$ Looks good to me. $\endgroup$ – Gerry Myerson Apr 8 '15 at 3:50
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I would be more careful with your notation. As you have defined it, $X_i$ is an indicator random variable (e.g., a Bernoulli random variable) that equals $1$ if person $i$ gets heads while persons $i \pm 1$ get tails (using cyclical numbering), and $0$ otherwise; thus $$\operatorname{E}[X_i] = \Pr[X_i = 1] = \frac{1}{8},$$ and linearity of expectation (note independence is not required, nor does it hold in this case) gives $$\operatorname{E}[X] = \sum_{i=1}^n \operatorname{E}[X_i] = \sum_{i=1}^n \frac{1}{8} = \frac{n}{8}.$$

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  • $\begingroup$ Oh ok thank you. I suppose I can't just assume random random indicator values like that. $\endgroup$ – user137720 Apr 8 '15 at 14:09

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