Fair coin toss and Bayes You have a coin and your prior assumption is that its probability of heads $\theta$ is chosen from a uniform distribution on $[0, 1]$. You toss the coin 10 times and get 6 heads. What is the estimate of $\theta$?
I figured that it has to be $\frac{6}{10}$ but is there a theorem or rule that can upend my guess?
One way to solve this is by using Maximum Likelihood Estimation. From Wikipedia (emphasis mine):

From the vantage point of Bayesian inference, MLE is a special case of maximum a posteriori estimation (MAP) that assumes a uniform prior distribution of the parameters (in our case, $\theta$).

Let $x_i$ denote the outcome (head = 1 / tails = 0) in a coin flip and let $X_n = (x_1, x_2, \dots, x_n)$ be a sequence of $n$ flips. We have that $m = \sum_{i=1}^n x_i$ is the number of heads in $n$ flips.
By Bayes' Theorem, we have that:
$$
f(\theta|X_n) = \frac{f(X_n|\theta) \cdot f(\theta)}{f(X_n)} 
$$
Assuming $\theta \sim \mathsf{Uniform}(0,1)$ we get:
$$
f(\theta|X_n) \propto f(X_n|\theta)
$$
This means that the most probable $\theta$ (i.e. MAP) that explains the sequence $X_n$ can be found by maximizing the likelihood $f(X_n|\theta)$ (i.e. MLE).
The likelihood of observing sequence $X_n$ given that the coin is parametrized by $\theta$ is:
$$
f(X_n|\theta)=\theta^m \cdot (1-\theta)^{n - m}
$$
Now, the maximum likelihood estimator of $\theta$ is:
$$
\widehat{\theta}_{\text{MLE}}(X_n) = \arg\max_\theta \big( f(X_n | \theta) \big) = \arg\max_\theta \big( \log f(X_n | \theta) \big)
$$
Note that maximizing $f(X_n|\theta)$ is the same as maximizing $\log f(X_n|\theta)$ since $\log$ is monotonously increasing, and in this setting it's much more convenient to work with $\log$.
Therefore:
$$
\log f(X_n|\theta) = \log \big(\theta^m \cdot (1-\theta)^{n - m} \big) = m \cdot \log \theta + \left(n-m\right) \cdot \log(1-\theta)
$$
Now we can maximize:
$$
\frac{\partial}{\partial \theta} \log f(X_n|\theta) =
\frac{m}{\theta} - \frac{n-m}{1-\theta} =
\frac{m\cdot(1-\theta)-\theta \cdot (n-m)}{\theta \cdot (1-\theta)} = 0
$$
Which finally gives:
$$
\boxed{\theta = \frac{m}{n}}
$$
I don't know of any pertinent probability theorem; my knowledge of probability is limited.  Also, I have no opinion on whether Peter's comment is accurate.  I would attack the problem as follows:
Suppose that $n$ is a fixed integer (e.g. 10).  Further suppose that $x \,\in \{0, 1, 2, \cdots, n\}$ and assume that the chance of a heads on any one toss is $x/n.$  Calculate the probability of getting exactly 6 heads out of 10 tosses as a function of $x.$ Denote this probability as $P_n(x).$
Then, Bayes theorem should kick in.  That is, let $F_n(x)$ denote the (relative) probability that the chance of a heads on any one toss is $x/n.$  Then $F_n(x) = \frac{P_n(x)}{P_n(0) + P_n(1) + \cdots + P_n(n)}.$  This will give you the relative value of $F_n(x).$
I presume that the interpretation of &quot;estimated probability for chance of a heads&quot; is interpreted as $\sum_{x=0}^n \,\frac{x}{n}F_n(x).$
The above is a key point, where I may well be mistaken.  If so, someone, please advise.
Assuming that everything (so far) in my answer is accurate, then the ultimate answer should simply be the limit as $n \to \infty$ of the estimated probability, as a function of $n.$
It wouldn't surprise me if the calculation (as $n \to \infty$) results in the summation of an infinite # of very small ranges being replaced by an integration formula.