Uncertainty in a sample Given some game between two players A and B, after 10 wins of player A and none of player B, how big is the chance of player A to win again? Certainly, one can assume that player A is the more skilled player, since he won all games, but it feels also wrong to assume that player A has a 100% chance of winning now. How can I factor in, that B still has a small chance of winning, and it just didn't occur in that measure? Are there different ways to calculate that?
EDIT
To clarify the matter further, I don't look for a calculation that uses a given winning probability for A. Instead I want to understand, what knowledge is included in the fact that I observed 10 wins of A already (e.g. a basketball team won 10 matches against another one). What I can clearly say by this observation is, that A is stronger than B, but I cannot say for sure, that A will win all future games to come.
 A: This is a situation in which reasonable people might have different subjective
probabilities.
So one reasonable way to answer is to use a Bayesian approach. Suppose at the start (before
any games were played) you had a very neutral opinion about the
probability $\theta$ that A will win any one game, expressed
by the prior distribution $\theta \sim Beta(1,1),$ which is also
$Unif(0,1).$
Then the ten games are played giving you a likelihood proportional
to $\theta^{10}.$ According to the version of Bayes' Theorem that
states
$$\text{POSTERIOR} \propto \text{PRIOR} \times \text{LIKELIHOOD},$$
the kernel of the posterior beta distribution is
$\theta^{10}(1-\theta)^0,$ so that the posterior distribution is
$\theta \sim Beta(11, 1).$
From there, various statements are possible, including a 95% posterior
probability interval of $(.715, .998),$ computed in R statistical software as follows:
 qbeta(c(.025,.975), 11, 1)
 ##0.7150858 0.9977010

This can be interpreted as saying A's chances of winning the next game
are between 71.5% and 99.8%. If you want a single value for the random variable
$\theta,$ you could use the median 93.9% or the mean $11/12$ of this beta distribution
for A's chances of winning; thus 6.1% or 8.3% for B's chances of winning.
The PDF of $Beta(11,1)$ is shown below.

The idea of this sort of Bayesian analysis is to combine the initial subjective view with empirical data to get a posterior distribution that reflects both. If you started out with the view that the opponents might be vaguely equally
matched, you might use the (parabolic) prior $Beta(2,2),$  obtaining the
posterior $Beta(12,2)$ and slightly different probability interval, median,
and mean.
