Streak!

Actually, I almost solved it the following way: tossing a coin 16 times in a row for black stone has the probability $\left(\frac{1}{2} \right) ^{16}$ , for each toss. There are $2^{16}$ tosses. Expected value = $\mu p$ . This time, it's the expected value for the probability of tossing 16 coins in a row for black, in $2^{16})$ tosses. Thus, $\left(\frac{1}{2} \right) ^{16}(2^{16})=1$ . Hence, this 'adds up.' But apparently 16 tosses is not the answer; 15 is. Why? Because if you do 15, you get $2$ - which is the reciprocal of the probability of getting a black or a white. So...

$\frac{1}{\text{probability per toss}} = \text{number of tosses * probability per toss}^{\text{ expected contiguous tosses}}$

$\Rightarrow \text{expected contiguous tosses} = -\log _{ \text{ probability per toss }}{\left( \text{ probability per toss * number of tosses} \right) }$

$\Rightarrow \boxed{ \text{expected contiguous tosses} = -1 - \log _{ \text{ probability per toss }}{\left( \text{ number of tosses} \right) }}$

Hence,

So, is this right, or is this just coincidence?

According to " The Longest Run of Heads ", Mark F. Schilling, if you flip a fair coin $n$ times, then the expected length of the longest run is approximately $\log_2 n + \frac{\gamma}{\log 2} - \frac{3}{2}$ .