How many consecutive heads do we observe in a run of coin tossing of length n? Although the problem seems to be easy to answer, this would be actually a little bit tough when we try to prove it straightforwardly. The expected number of consecutive heads in a run is 3n-2/8 (n≧2) using the recursive formula.

However, if we define a solitary head coin such that a head coin is isolated by neighboring tail coin(s) in a run, the problem of how many solitary heads we observe in a run can be solved easily. The expected number of solitary heads in a run is n+2/8 (n≧2). Since the problem of solitary head coin becomes a dual problem of the above, the consequence of the problem of the consecutive heads is derived easily by considering the probability of a solitary coin appearance.