Should we assume that the 58% chance of winning is fixed and that points are independent?
I believe that Whuber's answer is a good one, and beautifully written and explained, when the consideration is that every point is independent from the next one. However I believe that, in practice it is only an interesting starting point (theoretic/idealized). I imagine that in reality the points are not independent from each other, and this might make it more or less likely that your co-worker opponent gets to a win at least once out of 50.
At first I imagined that the dependence of the points would be a random process, ie not controlled by the players (e.g. when one is winning or loosing playing differently), and this should create a greater dispersion of the results benefiting the lesser player to get this one point out of fifty.
A second thought however might suggest the opposite: The fact that you already "achieved" something with a 9.7% of chance may give some (but only slight) benefit, from a Bayesian point of view, to ideas about favouring mechanisms that get you to win more than 85% probability to win a game (or at least make it less likely that your opponent has a much higher probability than 15% as argued in the previous two paragraphs). For instance, it could be that you score better when your position is less good (it is not strange for people scoring much more different on match points, in favor or against, than on regular points). You can improve estimates of the 85% by taking these dynamics into account and possibly you have more than 85% probability to win a game.
Anyway, it might be very wrong to use this simple points statistic to provide an answer. Yes you can do it, but it won't be right since the premises (independency of points) are not necessarily correct and highly influence the answer. The 42/58 statistic is more information but we do not know very well how to use it (the correctness of the model) and using the information might provide answers with high precision that it actually does not have.
Example
Example: an equally reasonable model with a completely different result
So the hypothetical question (assuming independent points and known, theoretical, probabilities for these points) is in itself interesting and can be answered, But just to be annoying and skeptical/cynical; an answer to the hypothetical case does not relate that much to your underlying/original problem, and might be why the statisticians/data-scientists at your company are reluctant to provide a straight answer.
Just to give an alternative example (not neccesarily better) that provides a confusing (counter-) statement 'Q: what is the probability to win all of the total of 50 games if I already won 15?' If we do not start to think that 'the point scores 42/58 are relevant or give us better predictions' then we would start to make predictions of your probability to win the game and predictions to win another 35 games solely based on your previously won 15 games:
- with a Bayesian technique for your probability to win a game this would mean: $p(\text{win another 35 | after already 15}) = \frac{\int_0^1 f(p) p^{50}}{\int_0^1 f(p) p^{15}}$ which is roughly 31% for a uniform prior f(x) = 1, although that might be a bit too optimistic. But still if you consider a beta distribution with $\beta=\alpha$ between 1 and 5 then you get to:

which means that I would not be so pessimistic as the straightforward 0.432% prediction The fact that you already won 15 games should elevate the probability that you win the next 35 games.
Note based on the new data
Based on your data for the 18 games I tried fitting a beta-binomial model. Varying $\alpha=\mu\nu$ and $\beta=(1-\mu)\nu$ and calculating the probabilities to get to a score i,21 (via i,20) or a score 20,20 and then sum their logs to a log-likelihood score.
It shows that a very high $\nu$ parameter (little dispersion in the underlying beta distribution) has a higher likelihood and thus there is probably little over-dispersion. That means that the data does not suggest that it is better to use a variable parameter for your probability of winning a point, instead of your fixed 58% chance of winning. This new data is providing extra support for Whuber's analysis, which assumes scores based on a binomial distribution. But of course, this still assumes that the model is static and also that you and your co-worker behave according to a random model (in which every game and point are independent).
Maximum likelihood estimation for parameters of beta distribution in place of fixed 58% winning chance:

Q: how do I read the "LogLikelihood for parameters mu and nu" graph?
A:
- 1) Maximum likelihood estimate (MLE) is a way to fit a model. Likelihood means the probability of the data given the parameters of the model and then we look for the model that maximizes this. There is a lot of philosophy and mathematics behind it.
- 2) The plot is a lazy computational method to get to the optimum MLE. I just compute all possible values on a grid and see what the valeu is. If you need to be faster you can either use a computational iterative method/algorithm that seeks the optimum, or possibly there might be a direct analytical solution.
- 3) The parameters $\mu$ and $\nu$ relate to the beta distribution https://en.wikipedia.org/wiki/Beta_distribution which is used as a model for the p=0.58 (to make it not fixed but instead vary from time to time). This modeled 'beta-p' is than combined with a binomial model to get to predictions of probabilities to reach certain scores. It is almost the same as the beta-binomial distribution. You can see that the optimum is around $\mu \simeq 0.6$ which is not surprising. The $\nu$ value is high (meaning low dispersion). I had imagined/expected at least some over-dispersion.
code/computation for graph 1
posterior <- sapply(seq(1,5,0.1), function(x) {
integrate(function(p) dbeta(p,x,x)*p^50,0,1)[1]$value/
integrate(function(p) dbeta(p,x,x)*p^15,0,1)[1]$value
}
)
prior <- sapply(seq(1,5,0.1), function(x) {
integrate(function(p) dbeta(p,x,x)*p^35,0,1)[1]$value
}
)
layout(t(c(1,2)))
plot( seq(1,5,0.1), posterior,
ylim = c(0,0.32),
xlab = expression(paste(alpha, " and ", beta ," values for prior beta-distribution")),
ylab = "P(win another 35| after already 15)"
)
title("posterior probability assuming beta-distribution")
plot( seq(1,5,0.1), prior,
ylim = c(0,0.32),
xlab = expression(paste(alpha, " and ", beta ," values for prior beta-distribution")),
ylab = "P(win 35)"
)
title("prior probability assuming beta-distribution")
code/computation for graph 2
library("shape")
# probability that you win and opponent has kl points
Pwl <- function(a,b,kl,kw=21) {
kt <- kl+kw-1
Pwl <- choose(kt,kw-1) * beta(kw+a,kl+b)/beta(a,b)
Pwl
}
# probability to end in the 20-20 score
Pww <- function(a,b,kl=20,kw=20) {
kt <- kl+kw
Pww <- choose(kt,kw) * beta(kw+a,kl+b)/beta(a,b)
Pww
}
# probability that you lin with kw points
Plw <- function(a,b,kl=21,kw) {
kt <- kl+kw-1
Plw <- choose(kt,kw) * beta(kw+a,kl+b)/beta(a,b)
Plw
}
# calculation of log likelihood for data consisting of 17 opponent scores and 1 tie-position
# parametezation change from mu (mean) and nu to a and b
loglike <- function(mu,nu) {
a <- mu*nu
b <- (1-mu)*nu
scores <- c(18, 17, 11, 13, 15, 15, 16, 9, 17, 17, 13, 8, 17, 11, 17, 13, 19)
ps <- sapply(scores, function(x) log(Pwl(a,b,x)))
loglike <- sum(ps,log(Pww(a,b)))
loglike
}
#vectors and matrices for plotting contour
mu <- c(1:199)/200
nu <- 2^(c(0:400)/40)
z <- matrix(rep(0,length(nu)*length(mu)),length(mu))
for (i in 1:length(mu)) {
for(j in 1:length(nu)) {
z[i,j] <- loglike(mu[i],nu[j])
}
}
#plotting
levs <- c(-900,-800,-700,-600,-500,-400,-300,-200,-100,-90,-80,-70,-60,-55,-52.5,-50,-47.5)
# contour plot
filled.contour(mu,log(nu),z,
xlab="mu",ylab="log(nu)",
#levels=c(-500,-400,-300,-200,-100,-10:-1),
color.palette=function(n) {hsv(c(seq(0.15,0.7,length.out=n),0),
c(seq(0.7,0.2,length.out=n),0),
c(seq(1,0.7,length.out=n),0.9))},
levels=levs,
plot.axes= c({
contour(mu,log(nu),z,add=1, levels=levs)
title("loglikelihood for parameters mu and nu")
axis(1)
axis(2)
},""),
xlim=range(mu)+c(-0.05,0.05),
ylim=range(log(nu))+c(-0.05,0.05)
)