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I am studying for an exam and found the following problem.

Let $L = l_0, l_1, ... , l_{n+1}$ be a list of items. For each item from 0 to $n+1$, we flip a coin (fairly). We add item $l_i$ (with $1 \le i \le n$) to a set if $l_i$'s toss came up heads, and both of its neighbours ($l_{i - 1}$ and $l_{i + 1}$) came up tails. What is the probability that $l_i$ is included in the set?

My initial thought is that it is simply 1/8. The probability that the coin is heads is 1/2, the one before it being tails has probability 1/2, and the same idea for the toss after it. Am I missing something, or is it really that simple? They are obviously dependent on one another.

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    @Alon Amit: The role of items are as a generic 'term'. We could think of them themselves as coin tosses (rather than having a coin toss associated with it) and if toss i-1 and i+1 are tails, while toss i is heads, then we write that toss number down on a piece of paper. What is the probability that it is written on the paper? (and following from that, what are the expected number of tosses written on the page)2011-04-11

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If that's the whole question, yes, it really is that simple. Perhaps there's a part (b)...

On the other hand, maybe this is intended to test the student's ability to filter out irrelevant aspects of the problem. This is actually a significant issue with many probability students, who might insist on using a sample space involving all $n+2$ coin flips.

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    @shoes, @Henry: Sorry, I've deleted my braindead comment about the expectation value being influenced by the correlations.2011-04-11