# Expected value of number of heads in n flips

We flip a coin n times and calculate the number of heads. for the obtained number of heads of the previous n tosses we will toss the coin again. what is the expected value of heads in this tossing process?

• What have you tried? Where have you gotten stuck? – WaveX Oct 6 '18 at 6:16
• How we can calculate this expected value after tossing for the obtained number of heads of the previous n tosses? – lighting Oct 6 '18 at 6:21

Let $$H_1$$ be the number of head in the first $$n$$ tosses, and $$H_2$$ be the number of head in the second. We are looking to compute $$E(T)$$, where $$T=H_1 + H_2$$. Note that $$E (T)=E(H_1)+E(H_2)$$. $$E(H_1)$$ is simply $$n*0.5 = n/2$$. To compute $$H_2$$, we will use the law of total expectation, \begin{align} E(H_2) = E(E(H_2|H_1)) &=\sum_{i=0}^n E(H_2|H_1=i)P(H_1=i) \\ &= \sum_{i=0}^n i/2 \binom{n}{i}(0.5)^n \\ &= (0.5)^{n+1}n2^{n-1}\\ &= n/4 \end{align} Thus, $$E(T)=n/2+n/4 = 3n/4$$
• +1 More concise: $\mathbb E[\mathbb E[H_2\mid H_1]]=\mathbb E\frac12H_1=\frac14n$ – drhab Oct 6 '18 at 8:03