Expected rolls to get n consecutive Heads#
Expectation to get 1 Head = 2
E(2 consecutive H) = 6
E(3 consecutive H) = 14
clearly 2,, 6, 14, …
E(N heads) = 2+4+8+..+2^n = 2^(n+1) - 2
Method 1:#
For N = 2:
E = (1 + 1/2*( 1 + 1/2 * E) + 1/2 * E)
or E = 1 + 1/2 + 1/4E + 1/2E
or 1/4E = 3/2
E = 6
Method 2:#
1/2 chance gets a tail, E = E+1
1/4 chance gets a head then tail, E = E+2
1/4 chance gets 2 head, E = 2
E = 1/2 (E + 1) + 1/4(E + 2) + 1/4 * 2
or E = 3/4E + 3/2
E = 6
For N:#
\[
E = \frac{1}{2}(E + 1) + \frac{1}{4}(E + 2) + \frac{1}{8}(E + 3) + \cdots + \frac{1}{2^{N}}(E + N) + \frac{1}{2^N}N
\]
\[
E = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^{N}}\right)E + \left(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots + \frac{N-1}{2^{N-1}} + \frac{N}{2^N} + \frac{N}{2^N}\right)
\]
\[
E = \left(1 - \frac{1}{2^{N}}\right)E + \left(2 - \frac{1}{2^{N-1}}\right)
\]
\[
\frac{1}{2^{N}} E = 2 - \frac{1}{2^{N-1}}
\]
\[
E = 2^{N+1} - 2
\]