I have the following problem where, given $X_n$ is a random variable that equals the number of tails minus the number of heads when n fair coins are flipped, what is the expected value of $X_n$?

I am having a difficulty getting started on this problem. Could someone offer a suggestion as to how this problem should be modeled?


2 Answers 2


We can start and finish by saying that by symmetry the expected number is $0$.

Or let $X$ be the number of tails. So $n-X$ is the number of heads. Thus tails minus heads is $2X-n$. This has mean $2E(X)-n$, which is $0$. This second argument will also give the answer for a coin that is biased, having probability $p$ of tails. Then the expected value of the difference is $2pn-n$.

  • $\begingroup$ Why does 2E(X)-n equal 0? $\endgroup$
    – shachna
    Jul 27, 2016 at 4:21
  • 1
    $\begingroup$ @shachna: The number $X$ of tails has binomial distribution $p=1/2$. So $E(X)=(1/2)n$. Equivalently, $2E(X)-n=0$. $\endgroup$ Jul 27, 2016 at 4:25

The key is linearity of expectation, and I suggest finding a way to write $X_n$ as a sum of simple 'indicators' corresponding to individual coin flips. So you would be thinking, "if I see a heads, then value to add is __.... and tails ......"


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