Confirmation on how to calculate this (simple?) probability

Assume that $$n$$ birds $$0,1,...,n − 1$$ sit around a circle. Suddenly, each bird randomly pecks the birds immediately to its left or right each with probability $$1/2$$. What is the expected number of unpecked birds?

I don't know if I'm thinking of this from a simple point of view, was wondering if someone can let me know if I'm correct or not.

We have n number of birds. If it pecks left or right, it's a probability of $$1/2$$ Unpecked would just mean they were the part of the $$1/2$$ probability.

So the answer would be $$\textbf{(1/2)}^\textbf{n}$$

Can someone let me know if my approach is correct?

• You can't just multiply probabilities to get an expected number. As the number of birds $n$ goes up, do you expect the number of unpecked birds to go up or down? As $n$ goes up, do you expect $(1/2)^n$ to go up or down? Under what circumstances does a bird go unpecked? Commented Oct 9, 2018 at 17:07
• You are asked to find the expected number of unpecked birds, but your answer has the looks of a probability. Commented Oct 9, 2018 at 17:07
• You may find it useful to break this down into odd and even $n$. Commented Oct 9, 2018 at 17:10
• While you could try to come up with an exact probability distribution finding the probability that exactly 0 birds remain unpecked, exactly 1 bird remains unpecked, exactly 2 birds etc... that is a terribly inefficient and difficult approach. Note, your question asks you to calculate the expected number of unpecked birds. It does not ask you to find the pdf for the number of unpecked birds. As it turns out, you can calculate the expected number without ever needing to refer to the exact pdf. Commented Oct 9, 2018 at 17:13

Hint: Use indicator functions for each bird which are $$1$$ if the bird is unpecked and $$0$$ otherwise. Then use linearity of expectation.