To make this more clear, I will generalize to sequences of $N$ words drawn from an bank with $A$ words, and you want to find the probability there exists a repeated subsequence of length $L$. In your case, $A=65536$, $L=50$, and $N=400\,000\,000$.
I cannot exactly compute the probability of two repeated sequences of length $L$ existing. However, I can prove the probability is at most $10^{-200}$. This is small enough such that, if you repeated this experiment every second until the sun died out, you could still be confident that none of the experiments had two repeated subsequences of length $L$.
For each $i,j\in \{1,2,\dots,N-L+1\}$, let $E_{i,j}$ be the event that the $i^\text{th}$ subsequence of length $L$ is equal to the $j^\text{th}$ subsequence of length $L$. If call the sequence of random words $W_1,W_2,\dots,W_N$, then $E_{i,j}$ is the event that
$$
(W_i,W_{i+1},\dots,W_{i+L-1})=(W_{j},W_{j+1},\dots,W_{j+L-1})
$$
Claim: For all $i,j\in \{1,\dots,N-L+1\}$, with $i<j$, $P(E_{i,j})= A^{-L}$.
Note: This claim applies to overlapping sequences, as well as non-overlapping ones.
Proof: In order for $E_{i,j}$ to occur, we need $W_{i+k}=W_{j+k}$ for each $k\in \{0,1,\dots,L-1\}$.
The probability that $W_j=W_i$ is $1/A$. This is because, no matter word $W_i$ is, the probability that $W_j$ equals that same word is $1/A$.
Conditional on $\{W_j=W_i\}$ occurring, the probability that $W_{j+1}=W_{i+1}$ is also $1/A$. This follows by the same logic, together with the fact that $W_{j+1}$ is independent of $W_i$ and $W_j$, so conditioning on $\{W_j=W_i\}$ does not affect the probability distribution of $W_{j+1}$.
Continuing this argument, for all $k\in \{1,\dots,L-1\}$, the probability that $W_{j+k}=W_{i+k}$, conditional on $W_{j+m}=W_{i+m}$ for all $m\in \{1,\dots,k-1\}$, is equal to $1/A$.
Finally, to compute the probability that $\{W_{i+k}=W_{j+k}\}$ occurs for each $k$, we multiply together all of these conditional probabilities, resulting in $A^{-L}$.$\tag*{$\blacksquare$}$
To conclude, there are $\binom{N-L+1}2$ events of the form $E_{i,j}$, and each occurs with probability $A^{-L}$. To see two repeated words, we need at least one of the $E_{i,j}$ to occur. By the union bound, the probability of some $E_{i,j}$ occurring bounded above as follows:
$$
P(\text{two repeated words of length $L$})\le \binom {N-L+1}2 A^{-L}
$$
Plugging back in your specific numbers (see WA computation), you get the probability of this collision is at most $10^{-200}$.
