A stock is prices at $ \$100$ and follows a one-period binomial process with an up move that equals 1.05 and a down move that equals 0.97. If one million Bernoulli trials are performed and the average terminal stock price is $ \$102$, the probability of an up move is closest to ____?
My book gives the solution as follows: $$ p \cdot 105 + (1-p) \cdot 97 = 102$$and thus computes the value of $ \text p$. However. I can't understand how a first step average of the process equals to the average terminal stock price that comes after one million Bernoulli trials. Can someone help me with this question?