Understanding Stochastic Processes

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I think I have an understanding of how a stochastic process is defined technically, but I've recently come across an example of one given in my notes that doesn't seem to fit with how I understood them. I would like to ask whether my understanding of stochastic processes is incorrect, or if the example given doesn't fit the definition.

Definition: A stochastic process in discrete time is a sequence of random variables $X_{1},X_{2},...$ all on the $same$ probability space $(\Omega,F,P)$.

The example here will show you what my understanding of this definition is: For a simple random walk each outcome is a sequence of "H" or "T" so we can write $\omega=\omega_{1}\omega_{2},...$ where $\omega_{i}\in{H,T}$. Then $(Z_{n})_{n\geq0}$ is defined by $Z_{i}(\omega)=Z_{i}(\omega_{1}\omega_{2},...\omega_{i},...)$$=$ \begin{cases} 1 & \omega_{i}=H \\ 0 & \omega_{i}=T & \end{cases} Then $(X_{n})_{n\geq0}$ where $X_{n}=X_{0} +\sum_{i=1}^{n} Z_{i}$ is the simple random walk starting at $X_{0}$.

My understanding breaks down when the $Z_{i}$ are described to be independent and identically distributed. I think I can see how these random variables could be independent, however, to be identically distributed would mean if $Z_{1}(\omega)=1$ then $Z_{i}(\omega)=1$ $\forall i\geq0$. To me this doesn't make sense. What am I missing here?

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The fact that $Z_1$ and $Z_i$ are identically distributed does not tell you that $Z_1=1$ implies $Z_i=1$. It only tells you that $P(Z_1=1)=P(Z_i=1)$.

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$Z_1(\omega)=1$ implies $Z_i(\omega)=1$ for all $i$ is definitely false.

What is true is that $\mathbb{P}(Z_1=1)=\mathbb{P}(Z_i=1)$ for all $i$.