Throw a coin $10$ times without knowing the mass distribution and get $10$ heads and $0$ tail, what is the probability of head in the $11$th time? Throw coin A $10$ times without knowing the mass distribution and get $10$ heads and $0$ tails, what is the probability of facing up for the $11$th time?
Throw coin B $100$ times without knowing the mass distribution and get $99$ heads and $1$ tails, what is the probability of facing up for the $101$th time?
Which coin is more likely to face up in the next toss?

I already know that parameter estimation methods such as maximum likelihood estimation can be used to estimate the most likely mass distribution of this coin;
I already know that Laplace smoothing can help me better compare the difference between Coin A and Coin B;
But how should we calculate their probability of heading up in the next toss?
I just want a percentage.
Thanks for your help.
 A: There are various approaches to this problem depending
on assumptions and statistical philosophy. One possibility is to take a Bayesian approach (as suggested in comments by @little), @Quillo, and @BrianTung.)
Let's say you have a chance to look at the coin before
you start tossing. If it 'looks like a fair coin' to
you, then you might start with a prior distribution on the true Heads probability
$\theta \sim \mathsf{Beta}(2,2)$ with density proportional to $$f(\theta) \propto \theta(1-\theta) = \theta^{2-1}(1-\theta)^{2-1},$$
for $0 < \theta < 1.$ [The symbol $\propto$ is read as "proportional to".]
There are various ways to characterize what you might
have in mind, making this particular choice of a prior distribution. One of them is that
you think there's a 95% chance that the true Heads
probability $\theta$ is in the interval $(0.0843, 0.9057)$ and that $E(\theta) = 0.5.$
q = qbeta(c(.025,.975) ,2 ,2);  q
[1] 0.09429932 0.90570068

After getting 10 Heads and 0 Tails in ten tosses, your
binomial likelihood function would be proportional to
$$g(x|\theta) \propto \theta^{10}(1-\theta)^0.$$
Using Bayes' Theorem to obtain the posterior distribution on $\theta$ you would have the posterior density function
proportional to
$$h(\theta|x) = f(\theta)\times g(x|\theta)\\  \propto \theta^{2-1}(1-\theta)^{2-1} \times 
\theta^{10}(1-\theta)^0 \\ = \theta^{12-1}(1-\theta)^{2-1},$$
which we recognize at the kernel (density function without norming factor) of $\mathsf{Beta}(12,2),$ which has
$E(\theta) = 12/14 = 6/7 = 0.8571$ and
$P(0.6397 <\theta< 0.9808) = 0.95.$ Also, taking the
mean as your guide, you might say a likely value
of $P(X_{11} = 1) - P(\mathrm{H\;on\;11th}) = 6/7.$
Or if you wanted to give an interval estimate,
you could say that probability is in the interval
$(0.6387,0.9808).$
qbeta(c(.025,.975), 12, 2)
[1] 0.6397026 0.9807933

The answer does depend in a small way on your choice
of the prior beta distribution, but it would not
have been much different if you had chosen $\mathsf{Beta}(1,1),$ or $\mathsf{Beta}(.5,.5)$ or some other beta
distribution with small and equal shape parameters.
Similar arguments could be used for Coin B with 99 Heads and 1 Tails in 100 tosses.
Notes: If you were to take a frequentist probability approach
and consider it a Law of the Universe that this
particular coin is fair, then you would not be much impressed
by the evidence of ten Heads in a row and you would
say that the coin has had an 'anomalous lucky streak,'
and that its probability of Heads on the 11th toss
is still $1/2.$ This is not much different from
a Bayesian starting with a prior distribution something
like $\mathsf{Beta}(10^6, 10^6),$ which would
overwhelm any results from only ten tosses and give
essentially a probability $1/2$ on the 11th toss.
A traditional frequentist statistician might wonder if we should update our estimate of the probability of Heads after 10 Heads in a row, but
it is not clear what procedure should be used for that. It may be too much to say the coin must
be two Headed.
