To lump or not to lump

Suppose I have a very simple asset whose price takes only three possible values: $X_t\in \{-1,0,1\}$. I also got some discrete time series $X = (X_t)_{t\geq 0}$ and I would like to come up with a trading rule based on these observations. Let's focus on the following naive approach: given the current level of the asset, I would like to estimate of what the next change would be. Thus, I am fitting this time series into Markov Chain where I disregard transition that do not change the state. For example, based on the following sample from the time series: $$\dots0,0,0,0,0,0,0,1,1,1,0,0,0,-1,-1\dots \tag{1}$$ and suppose that there are no appearances of $0$ anymore. I can conclude that out of $10$ appearances of $0$, $7$ are followed by $1$ and $3$ are followed by $-1$, hence a naive algorithm would say that transition probabilities are $p(1|0) = 0.7$ and $p(-1|0) = 0.3$.

Now, I wanted to make it faster, so I preprocessed the data to get rid of repetitions since I thought that it shall not affect the end result. For example, the sample $(1)$ transforms into $$\dots0,1,0,-1\dots \tag{2}$$ but now $p(1|0) = 0.5$ and $p(-1|0) = 0.5$ which is quite different from the previous estimation. Of course, that's just a simple example, but it gives a general impression: the lumping procedure $(1)\to(2)$ changes the estimates of transition probabilities. It surprised me first, but now it seems very natural: in $(2)$ when sampling I give equal weight to each interval of consecutive $0$'s whereas in $(1)$ more weight goes to a longer interval of consecutive $0$'s.

The question is: given my purpose, what would be the correct method to estimate probabilities? Note that I am trying to predict the price move without taking into account for how long have I stayed at the current price of the asset, only the price level itself. From that perspective, I guess the second approach is more appropriate: if I observe new prices and I see the price is $0$, I'd say I'd rather rely on that the next price change happens with probabilities $(0.5,0.5)$ rather than $(0.7, 0.3)$. At the same time, here I do not feel confident since my background in statistics is rather weak, so any feedback on this topic is highly appreciated. Namely: what is formally more correct, to use probabilities from $(1)$ or from $(2)$, or if both are correct depending on how they are used, what is a proper way of using them. Practical comment on that approach is also welcome.

Now, let's ask what the probability of getting a one is, assuming that we won't stay at zero, i.e. $P(X_1=1 | X_0=0, X_1 \not = 0)$. We recall that $$P(A|B)=P(A,B) / P(B)$$ and therefore get
\begin{align*} P(X_1=1 | X_0=0, X_1 \not = 0) &=\frac{P(X_1=1, X_1 \not = 0 | X_0=0)}{P(X_1 \not = 0| X_0=0)}\\ & =\frac{P(X_1=1 | X_0=0)}{P(X_1 = 1| X_0=0) + P(X_1 = -1| X_0=0)} \\ & =\frac{P(1|0)}{P(1|0) + P(-1|0)} \end{align*}
Which is just your second definition where you found it to be $(.5,.5)$ as indeed half the times it transitioned away from zero it transitioned to one.
Importantly, this is not $P(1|0)$! I think this is where you were getting confused before.