Call your game Game $1$. Game $2$ is almost the same, except that we continue for the full $13$ rounds, and we win Game $2$ if we get one or more match. The probability of winning Game $2$ is the same as the probability of winning Game $1$. So to answer your question, it is enough to find the probability of winning Game $2$.
The probability of a match at position $i$ is $\frac{4}{52}$. Now sum over all $i$ from $1$ to $13$. We get $\binom{13}{1}\frac{4}{52}$. This is $1$, and is the expected (mean) number of matches in Game $2$. But of course it is not the probability of winning Game $2$.
The problem is that we have counted twice every situation in which we have a match at $i$ and also a match at $j$. For clarity of thought, though in fact it doesn't matter, assume that $i\lt j$. The probability of a match at $i$ is $\frac{4}{52}$. Given that we have a match at $i$, the probability of a match at $j$ is $\frac{4}{51}$. So the probability of a match at $i$ and $j$ is $\frac{4}{52}\cdot \frac{4}{51}$. There are $\binom{13}{2}$ ways of picking $i$ and $j$. So from $\binom{13}{1} \frac{4}{52}$ we subtract $\binom{13}{2}\frac{4}{52}\cdot\frac{4}{51}$.
But we have subtracted too much! For we have subtracted one too many times all situations in which we had three matches. Let $i\lt j\lt k$. The probability of matches at $i$, $j$, and $k$ is $\frac{4}{52}\cdot\frac{4}{51}\cdot \frac{4}{50}$. There are $\binom{13}{3}$ ways of choosing the triple $(i,j,k)$. So we will add back $\binom{13}{3} \frac{4}{52}\cdot\frac{4}{51}\cdot \frac{4}{50}$.
But we have added back too much, for we have added back too many times the situations in which we have $4$ matches. So we must subtract $\binom{13}{4}\frac{4}{52}\cdot\frac{4}{51}\cdot \frac{4}{50}\cdot\frac{4}{49}$.
Continue. We are using the Method of Inclusion/Exclusion.
For practical work, we could probably stop. Already the term $\binom{13}{4}\frac{4}{52}\cdot\frac{4}{51}\cdot \frac{4}{50}\cdot\frac{4}{49}$ is fairly small. But it is not hard to continue to the end and get an exact answer.
Remark: Your simulation gave an answer quite close to the truth. The terms I mentioned in the post give roughly $0.639$, and the next term in the Inclusion/Exclusion is an "add" term.
For $i,j\in\{1,2,3\}$, let $a_{i,j}$ denote the number of $i$ cards being dealt with number $j$ spoken.
We have $\sum_j a_{i,j}=4$ and for a winning game $a_{i,i}=0$.
The number of winning positions for a given $(a_{i,j})$ is
$$\frac{18!}{a_{2,1}!a_{3,1}!(18-a_{2,1}-a_{3,1})!}\cdot\frac{17!}{a_{1,2}!a_{3,2}!(17-a_{1,2}-a_{3,2})!}\cdot\frac{17!}{a_{1,3}!a_{2,3}!(17-a_{1,3}-a_{2,3})!}. $$
We need to sum this over all $(a_{i,j})$ and divide by the total count $$ \frac{52!}{4!4!4!40!}.$$
(Actually, we need just let $a_{1,2}, a_{2,3}, a_{3,1}$ run from $0$ to $4$ and this determines $a_{1,3}=4-a_{1,2}$ etc.)
The final result is
$$p=\frac{58388462678560}{7151046448045500}=\frac{24532967512}{3004641364725}\approx 0.008165 $$
(I just noted that Harold has performed a Monte Carlo simulation with matching result)
Best Answer
For $0\le k\le n$, let $p(n,k)$ be the probability to win the game if it starts with $n+1$ cards $0,1,\ldots, n$ and the first card shown is card $k$. By symmetry, $p(n,k)=p(n,n-k)$. If $k\ge \frac n2$, then the first guess will be "lower" (where both possible guesses a just as good in case $k=\frac n2$) and succeed with $\frac{k}{n}$, namely if the next card $l$ is one of $0,\ldots,k-1$. This gives us the recursion $$ p(n,k)=\frac1n\sum_{l=0}^{\max(k,n-k)-1}p(n-1,l)$$ (with the bases case $p(0,0)=1$). If we start with $n$ cards and have not seen the first card yet, the winning probability is then $$p(n)=\frac1n\sum_{k=0}^{n-1} p(n-1,k)$$
I doubt that there is a simple closed formula for $p(n)$. One can combute by hand that with a five card deck the probability is $p(5)=\frac{31}{60}$, which is nicely close to $50:50$. Just for fun, I let my computer evaulate this for a deck of $32$, where the overall winning probability is $$p(32)=\frac{593119019981747761359995677819}{4111419327088961408862781440000000}\approx 0.00014 \approx0.7585^{32}$$ as well as for for a deck of $52$, where it is $$p(52)=\\\frac{2502415780357808863930924231118727082667638137165232707577843}{10082271896367984821457579607050470871911188180110409728000000000000}\\\approx 2.48\cdot10^{-7}\approx 0.7454^{52} $$ This may be surprising because the first guess will be correct with at least $\frac 34$. So how come that the $0.7585$ per round decrease to $0.7454$ per round when the deck is enlarged from $32$ to $52$ cards? The problem is that after a correct guess of "lower", say, all cards above the previuos card are excluded. Thus the next card is a bit more likely to be of medium size - but that's the situation where guessing right is harder!