How can I formulate and determine the overall probability that the consecutive data are overlapping temporally? Suppose we have the data $A, B, C, D, E,$ and $F$ that are expected to arrive at the destination at $t_A, t_B, t_C,t_D,t_E,$ and $t_F$, respectively. However, the channel in which these data propagate induces randomness such that the data arrive at random times $t'_A, t'_B, t'_C,t'_D,t'_E,$ and $t'_F$, respectively. $T$ is some constant time separation to avoided overlapping and ideally all data should arrive within $t_i$ and $t_i+T$. But, $T$ cannot be too large as it can degrade performance (smaller $T$ is preferred). As an example, $C$ is arriving early and overlaps temporally (that is, in time) with $B$, $D$ is arriving late and overlaps temporally with $E$, and $F$ is arriving early and overlaps temporally with $E$.

I want to know the probability that two consecutive data overlap temporally with one another (illustrated by the darker regions). In other words, the time at which two consecutive data arrive is less than $\tau\ (\tau<T)$, which is the duration of a data and is the same for all data.
Let the random arrival time of a data follow the Normal distribution, such that $t'_i \sim N(\mu_i,\sigma^2_i)$, where $i \in (A,B,C,D,E,F)$ and $\mu_i=t_i$.
Then, from here, that probability is
$$P(\text{Two consecutive data are overlapping})=P(Z<\tau)=P\left(\frac{Z-\mu_Z}{\sigma_Z}<\frac{\tau-\mu_Z}{\sigma_Z}\right) \\ =\Phi\left(\frac{\tau-\mu_Z}{\sigma_Z}\right)\  (1)$$
where $Z=t_j-t_k, j \neq k$ and $j \in (F,E,D,C,B), k \in(E,D,C,B,A)$.
Eq. (1) allows me to find the probability that two data overlap.
How can I formulate and determine the overall probability that the consecutive data are overlapping temporally?
I may be mistaken, but what I gather is that the overall probability is $P(A \text{ and } B \text{ are overlapping})$ and $P(B \text{ and } C \text{ are overlapping})$ and $P(C \text{ and } D \text{ are overlapping})$ and $P(D \text{ and } E \text{ are overlapping})$ and $P(E \text{ and } F \text{ are overlapping})$.
How can I proceed further?
Additionally, since $E$ is overlapping temporally with $D$, we consider them to be destroyed. Then $F$ will not be overlapping temporally with $E$. So, how can we incorporate this condition in the overall probability?
Thank you in advance.
 A: The situation is a set of data packets, each of fixed width $\tau$ in time, that ideally start at times $T,2T,3T,...$. The start time of each packet, however, is normally distributed around its ideal starting time, with variance $\sigma^2$.
In the terminology of the question, $Z$ represents the actual difference in start times between 2 consecutive packets.* So by construction, $\mu_Z = T$. If the packet arrival times are independent (except for their defined ideal arrival times), the variance of the difference in arrival times, $\sigma_Z^2$, is $2\sigma^2$. So the probability of 2 consecutive events overlapping can be put a bit more directly as:**
$$\Phi\left(\frac{\tau-T}{\sqrt2\sigma}\right)$$
For concreteness, if you wanted this probability to be 1% or less, you would need approximately $\left(\frac{\tau-T}{\sqrt2\sigma}\right) < -2.326,$ or $T> \tau +3.29 \sigma$.

How can I formulate and determine the overall probability that the consecutive data are overlapping temporally?

If by this you mean the probability that none of the packets overlap versus at least one pair overlapping, the usual interest in a case like this, then you don't want to use the "and" operator with respect to the individual probabilities of overlap, as you do in the question:

what I gather is that the overall probability is $P(A \text{ and } B \text{ are overlapping})$ and $P(B \text{ and } C \text{ are overlapping})$ and $P(C \text{ and } D \text{ are overlapping})$ and $P(D \text{ and } E \text{ are overlapping})$ and $P(E \text{ and } F \text{ are overlapping})$.

That would be close to the probability that all of the packets overlap. (The assumption that the second of an overlapping set of packets is destroyed and thus doesn't overlap with the next packet complicates things a bit.)
If you want to know the probability that all packets were received correctly without overlap, you want to use the "and" operator on the individual probabilities of non-overlap. For each potential overlap, the probability of non-overlap is
$$1- \Phi\left(\frac{\tau-T}{\sqrt2\sigma}\right).$$
Then use the "and" operator on these probabilities of non-overlap. So for non-overlap of 3 packets (2 possible overlaps) you have the square of this probability, for 4 packets the cube, etc. Your example is for 6 packets, with 5 potential overlaps.
Once you have thus determined the probability that no packets overlapped, the probability that some of the packets overlapped (which I think is what this question is getting at) is 1 minus that probability of no overlap.
This type of moving back and forth between probabilities of events and their complements often helps to simplify analysis of problems like this.

*The question currently shows $Z=t_j-t_k$ where the $t_i$ represent the ideal arrival times. Based on the context, I take that to be a typo, with the intent being for $Z$ to represent the actual difference in arrival time, $Z=t_j'-t_k'$.
**One potentially useful trick would be to re-define the time scale in terms of $\sigma$. In particular, if you let one unit of time equal $\sqrt2\sigma$ then this would be just $\Phi\left(\tau-T\right)$. Some find working in such dimensionless units to be simpler.
