If you play games to $4$ points, where you have to win by $2$, you can assume the players play 6 points. If no player wins by $2$, then the score is tied $3-3$, and then you play pairs of points until one player wins both. This means the the chance to win a game to $4$ points, when your chance to win each point is $p$, is
$$p^6 + 6p^5(1-p) + 15p^4(1-p)^2 + 20 p^3(1-p)^3 \frac{p^2}{p^2 + (1-p)^2}$$.
In top level men's play, $p$ might be about $0.65$ for the server. (It would be $0.66$ if men didn't ease off on the second serve.) According to this formula, the chance to hold serve is about $82.96\%$.
Suppose you are playing a tiebreaker to $7$ points. You can assume that the points are played in pairs where each player serves one of each pair. Who serves first doesn't matter. You can assume the players play $12$ points. If they are tied at that point, then they play pair until one player wins both of a pair, which means the conditional chance to win is $p_sp_r/(p_sp_r + (1-p_s)(1-p_r))$. If I calculate correctly, the chance to win a tiebreaker to $7$ points is
$$ 6 p_r^6 ps + 90 p_r^5 p_s^2 - 105 p_r^6 p_s^2 + 300 p_r^4 p_s^3 -
840 p_r^5 p_s^3 + 560 p_r^6 p_s^3 + 300 p_r^3 p_s^4 - 1575 p_r^4 p_s^4 +
2520 p_r^5 p_s^4 - 1260 p_r^6 p_s^4 + 90 p_r^2 p_s^5 - 840 p_r^3 p_s^5 +
2520 p_r^4 p_s^5 - 3024 p_r^5 p_s^5 + 1260 p_r^6 p_s^5 + 6 p_r p_s^6 -
105 p_r^2 p_s^6 + 560 p_r^3 p_s^6 - 1260 p_r^4 p_s^6 + 1260 p_r^5 p_s^6 -
462 p_r^6 p_s^6 + \frac{p_r p_s}{p_r p_s + (1-p_r)(1-p_s)}(p_r^6 + 36 p_r^5 p_s - 42 p_r^6 p_s + 225 p_r^4 p_s^2 - 630 p_r^5 p_s^2 +
420 p_r^6 p_s^2 + 400 p_r^3 p_s^3 - 2100 p_r^4 p_s^3 + 3360 p_r^5 p_s^3 -
1680 p_r^6 p_s^3 + 225 p_r^2 p_s^4 - 2100 p_r^3 p_s^4 + 6300 p_r^4 p_s^4 -
7560 p_r^5 p_s^4 + 3150 p_r^6 p_s^4 + 36 p_r p_s^5 - 630 p_r^2 p_s^5 +
3360 p_r^3 p_s^5 - 7560 p_r^4 p_s^5 + 7560 p_r^5 p_s^5 -
2772 p_r^6 p_s^5 + p_s^6 - 42 p_r p_s^6 + 420 p_r^2 p_s^6 -
1680 p_r^3 p_s^6 + 3150 p_r^4 p_s^6 - 2772 p_r^5 p_s^6 + 924 p_r^6 p_s^6)$$
If $p_s=0.65, p_r=0.36$ then the chance to win the tie-breaker is about $51.67\%$.
Next, consider a set. It doesn't matter who serves first, which is convenient because otherwise we would have to consider winning the set while having the serve next versys winning the set without keeping the serve. To win a set to $6$ games, you can imagine that $10$ games are played first. If the score is tied $5-5$ then play $2$ more games. If those do not determine the winner, then play a tie-breaker, or in the fifth set just repeat playing pairs of games. Let $p_h$ be the probability of holding serve, and let $p_b$ be the probability of breaking your opponent's serve, which may be calculated above from the probability to win a game. The chance to win a set without a tiebreak follows the same basic formula as the chance to win a tie-breaker, except that we are playing to $6$ games instead of to $7$ points, and we replace $p_s$ by $p_h$ and $p_r$ by $p_b$.
The conditional chance to win a fifth set (a set with no tie-breaker) with $p_s=0.65$ and $p_r=0.36$ is $53.59\%$.
The chance to win a set with a tie-breaker with $p_s=0.65$ and $p_r=0.36$ is $53.30\%$.
The chance to win a best of $5$ sets match, with no tie-breaker in the fifth set, with $p_s=0.65$ and $p_r=0.36$ is $56.28\%$.
So, for these win rates, how many games would there have to be in one set for it to have the same discriminatory power? With $p_s=0.65, p_r=0.36$, you win a set to $24$ games with the usual tiebreaker $56.22\%$, and you win a set to $25$ game with a tie-breaker possible $56.34\%$ of the time. With no tie-breaker, the chance to win a normal match is between sets of length $23$ and $24$. If you simply play one big tie-breaker, the chance to win a tie-breaker of length $113$ is $56.27\%$ and of length $114$ is $56.29\%$.
This suggests that playing one giant set is not more efficient than a best of 5 matches, but playing one giant tie-breaker would be more efficient, at least for closely matched competitors who have an advantage serving.
Here is an excerpt from my March 2013 GammonVillage column, "Game, Set, and Match." I considered coin flips with a fixed advantage ($51\%$) and asked whether it is more efficient to play one big match or a series of shorter matches:
... If a best of three is less efficient than a single long match, we
might expect a best of five to be worse. You win a best of five $13$
point matches with probability $57.51\%$, very close to the chance to win
a single match to $45$. The average number of matches in a best of five
is $4.115$, so the average number of games is $4.115 \times 21.96 = 90.37$. Of
course this is more than the maximum number of games possible in a
match to $45$, and the average is $82.35$. It looks like a longer series
of matches is even less efficient.
How about another level, a best of three series of best of three
matches to $13$? Since each series would be like a match to $29$, this
series of series would be like a best of three matches to $29$, only
less efficient, and one long match would be better than that. So, one
long match would be more efficient than a series of series.
What makes a series of matches less efficient than one long match?
Consider these as statistical tests for collecting evidence to decide
which player is stronger. In a best of three matches, you can lose a
series with scores of $13-7 ~~ 12-13 ~~ 11-13$. This means you would win $36$
games to your opponent's $33$, but your opponent would win the series.
If you toss a coin and get $36$ heads and $33$ tails, you have evidence
that heads is more likely than tails, not that tails is more likely
than heads. So, a best of three matches is inefficient because it
wastes information. A series of matches requires more data on average
because it sometimes awards victory to the player who has won fewer
games.
Best Answer
The analysis is complicated by the prospect that the game goes into "overtime" in order to win by a margin of at least two points. (Otherwise it would be as simple as the solution shown at https://stats.stackexchange.com/a/327015/919.) I will show how to visualize the problem and use that to break it down into readily-computed contributions to the answer. The result, although a bit messy, is manageable. A simulation bears out its correctness.
Let $p$ be your probability of winning a point. Assume all points are independent. The chance that you win a game can be broken down into (nonoverlapping) events according to how many points your opponent has at the end assuming you don't go into overtime ($0,1,\ldots, 19$) or you go into overtime. In the latter case it is (or will become) obvious that at some stage the score was 20-20.
There is a nice visualization. Let scores during the game be plotted as points $(x,y)$ where $x$ is your score and $y$ is your opponent's score. As the game unfolds, the scores move along the integer lattice in the first quadrant beginning at $(0,0)$, creating a game path. It ends the first time one of you has scored at least $21$ and has a margin of at least $2$. Such winning points form two sets of points, the "absorbing boundary" of this process, whereat the game path must terminate.
This figure shows part of the absorbing boundary (it extends infinitely up and to the right) along with the path of a game that went into overtime (with a loss for you, alas).
Let's count. The number of ways the game can end with $y$ points for your opponent is the number of distinct paths in the integer lattice of $(x,y)$ scores beginning at the initial score $(0,0)$ and ending at the penultimate score $(20,y)$. Such paths are determined by which of the $20+y$ points in the game you won. They correspond therefore to the subsets of size $20$ of the numbers $1,2,\ldots, 20+y$, and there are $\binom{20+y}{20}$ of them. Since in each such path you won $21$ points (with independent probabilities $p$ each time, counting the final point) and your opponent won $y$ points (with independent probabilities $1-p$ each time), the paths associated with $y$ account for a total chance of
$$f(y) = \binom{20+y}{20}p^{21}(1-p)^y.$$
Similarly, there are $\binom{20+20}{20}$ ways to arrive at $(20,20)$ representing the 20-20 tie. In this situation you don't have a definite win. We may compute the chance of your win by adopting a common convention: forget how many points have been scored so far and start tracking the point differential. The game is at a differential of $0$ and will end when it first reaches $+2$ or $-2$, necessarily passing through $\pm 1$ along the way. Let $g(i)$ be the chance you win when the differential is $i\in\{-1,0,1\}$.
Since your chance of winning in any situation is $p$, we have
$$\eqalign{ g(0) &= p g(1) + (1-p)g(-1), \\ g(1) &= p + (1-p)g(0),\\ g(-1) &= pg(0). }$$
The unique solution to this system of linear equations for the vector $(g(-1),g(0),g(1))$ implies
$$g(0) = \frac{p^2}{1-2p+2p^2}.$$
This, therefore, is your chance of winning once $(20,20)$ is reached (which occurs with a chance of $\binom{20+20}{20}p^{20}(1-p)^{20}$).
Consequently your chance of winning is the sum of all these disjoint possibilities, equal to
$$\eqalign{ &\sum_{y=0}^{19}f(y) + g(0)p^{20}(1-p)^{20} \binom{20+20}{20} \\ = &\sum_{y=0}^{19}\binom{20+y}{20}p^{21}(1-p)^y + \frac{p^2}{1-2p+2p^2}p^{20}(1-p)^{20} \binom{20+20}{20}\\ = &\frac{p^{21}}{1-2p+2p^2}\left(\sum_{y=0}^{19}\binom{20+y}{20}(1-2p+2p^2)(1-p)^y + \binom{20+20}{20}p(1-p)^{20} \right). }$$
The stuff inside the parentheses on the right is a polynomial in $p$. (It looks like its degree is $21$, but the leading terms all cancel: its degree is $20$.)
When $p=0.58$, the chance of a win is close to $0.855913992.$
You should have no trouble generalizing this analysis to games that terminate with any numbers of points. When the required margin is greater than $2$ the result gets more complicated but is just as straightforward.
Incidentally, with these chances of winning, you had a $(0.8559\ldots)^{15}\approx 9.7\%$ chance of winning the first $15$ games. That's not inconsistent with what you report, which might encourage us to continue supposing the outcomes of each point are independent. We would thereby project that you have a chance of
$$(0.8559\ldots)^{35}\approx 0.432\%$$
of winning all the remaining $35$ games, assuming they proceed according to all these assumptions. It doesn't sound like a good bet to make unless the payoff is large!
I like to check work like this with a quick simulation. Here is
Rcode to generate tens of thousands of games in a second. It assumes the game will be over within 126 points (extremely few games need to continue that long, so this assumption has no material effect on the results).When I ran this, you won in 8,570 cases out of the 10,000 iterations. A Z-score (with approximately a Normal distribution) can be computed to test such results:
The value of $0.31$ in this simulation is perfectly consistent with the foregoing theoretical computation.
Appendix 1
In light of the update to the question, which lists the outcomes of the first 18 games, here are reconstructions of game paths consistent with these data. You can see that two or three of the games were perilously close to losses. (Any path ending on a light gray square is a loss for you.)
Potential uses of this figure include observing:
The paths concentrate around a slope given by the ratio 267:380 of total scores, equal approximately to 58.7%.
The scatter of the paths around that slope shows the variation expected when points are independent.
If points are made in streaks, then individual paths would tend to have long vertical and horizontal stretches.
In a longer set of similar games, expect to see paths that tend to stay within the colored range, but also expect a few to extend beyond it.
The prospect of a game or two whose path lies generally above this spread indicates the possibility that your opponent will eventually win a game, probably sooner rather than later.
Appendix 2
The code to create the figure was requested. Here it is (cleaned up to produce a slightly nicer graphic).