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Relevant wiki: Expected Value

Since the price of shares never stays the same for $2$ or more consecutive days and the probability of growth is $\frac{7}{8}$ , then the probability of decline is $\frac{1}{8}$ .

Let $X_{n}$ be the random variable that represents the number of days required to decline for $n$ consecutive days occur.

In order price to decline for $n$ consecutive days, it must first decline for $n-1$ days. Then, the $n^{th}$ consecutive decline will occur on the next day with $\frac{1}{8}$ probability, or the process will start over on the next day with $\frac{7}{8}$ probability. Thus, we can write:

$E\left [ X_{n} \right ]=E\left [ \frac{1}{8}(X_{n-1}+1)+\frac{7}{8}(X_{n-1}+1+X_{n}) \right ]=E\left [ X_{n-1}+1+\frac{7}{8}X_{n} \right ]$

By linearity of expectations:

$E\left [ X_{n} \right ]=E\left [ X_{n-1}\right ]+1+\frac{7}{8}E\left [ X_{n} \right ]$

$\Rightarrow E\left [ X_{n} \right ]=8E\left [ X_{n-1}\right ]+ 8$

It's known that $E\left [ X_{1} \right ]=8$ , since that's number of days expected one day decline to occur. This result matches with logic and intuition. Then, by formula:

$E\left [ X_{2} \right ]=8\times 8+8=72$ , and finally

$E\left [ X_{3} \right ]=8\times 72+8=584$ .

The expected number of days until decline for 3 consecutive days occur is $\boxed{584}$ .