models of coins and uniform distribution Here is the sentence which I'm confused with 

Imagine I give you a coin, and want you to estimate the probability it
  will spin to heads. Given what you know, the most reasonable prior
  belief is to expect any probability of the coin spinning to heads.
  This can be captured in a uniform prior on pp, the probability that a
  coin when spun will land on heads: var p = uniform(0,1).
You conduct an experiment. You spin the coin 20 times. 15 of them,
  they spin to heads. 

I'm confused with the part uniform(0,1). 
why flipping coin prior can be represent with Uniform(0,1)?
I found some related answer here 
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If random variables (r.v.s) $a_1,a_2,\ldots$ are
  i.i.d.
  uniformly distributed on the set $\{0,1\}$, then
  $\alpha=(0.a_1a_2\ldots)_2$ is a r.v. uniformly distributed on the
  real interval $[0,1]$. To see this, note that for any
  $x=(0.x_1x_2\ldots)_2\in[0,1)$ (always taking, WLOG, the unique binary
  representation of $x$ that has infinitely many $0$s), we have the
  following:  $$\begin{align}\{\alpha > x\} = & \{a_1>x_1\}\cup\\
> &\{\{a_1=x_1\}\cap \{a_2>x_2\}\}\cup\\ &\{\{a_1=x_1\}\cap
> \{a_2=x_2\}\cap\{a_3>x_3\} \}\cup\\ &\ldots \end{align}$$
Now, $P(a_i >x_i) = \frac{1}{2}(1-x_i)$, so the probability of the
  above disjoint union is just  $$\begin{align}P(\alpha>x) &=
> \frac{1}{2}(1-x_1) + \frac{1}{2^2}(1-x_2) +
> \frac{1}{2^3}(1-x_3)+\ldots\\ &= \sum_{i=1}^\infty \frac{1}{2^i} -
> \sum_{i=1}^\infty \frac{x_i}{2^i}\\ &= 1 - x\\ \therefore P(\alpha\le
> x) &= x \end{align} $$ therefore $\alpha$ is a r.v. uniformly
  distributed on $[0,1]$.

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But I'm quite confused with  what is 
$$P(\alpha > x )$$
and why does he concluded 
$$P(\alpha \le x)=x$$
is uniform distribution on [0,1] 
 A: I believe this is intended to be an elementary Bayesian inference problem.
It seems you have decided to let $\theta = P(\text{Heads})$ have the "flat" or "noninformative" 
prior distribution $\mathsf{Unif}(0,1) 
\equiv \mathsf{Beta}(\alpha=1,\beta=1).$ So your prior distribution is 
$p(\theta) = 1.$ 

A uniform prior is not the only possible choice: (a) If you had some prior experience with or knowledge of the coin, you might
  choose a prior distribution that reflects your prior opinion. Perhaps you
  look at the coin, play with it a bit, and decide it seems close to fairly balanced. Then you might choose
  the prior distribution $\mathsf{Beta}(5,5)$ which puts roughly half of its
  probability in $(.4, .6)$ [based on a computation in R, where
  qbeta(c(.25,.75), 5, 5) returns 0.3919 and 0.6080.] (b) Another popular
  choice for a noninformative prior is $\mathsf{Beta}(.5, .5).$ 
  [Google Jeffrey's priors.] Your choice of a prior distribution will 
  influence your final conclusions after you have data and get a posterior
  distribution. (I think @Twis7ed's Comment is suggesting that priors other than
  uniform are possible.)

Then from your experiment, you
have $n=20$ Bernoulli trials resulting in $x = 15$ Heads, so your
likelihood function is $p(x|\theta) = \theta^x(1-\theta)^{n-x}
= \theta^{15}(1-\theta)^5.$ 
Then according to Bayes' Theorem, the posterior distribution (using the uniform prior) is 
$$p(\theta|x) \propto p(\theta) \times p(x|\theta) \propto \theta^{15}(1-\theta)^5,$$
which we recognize as the kernel (PDF without constant) of $\mathsf{Beta}(16,6).$
The posterior distribution can be used to get a point estimate, which
might be the posterior mean $\frac{16}{16+6} = 0.7272$, posterior median
0.7343 (from R), or posterior mode $\frac{15}{20} = 0.75.$
qbeta(.5, 16, 6)
[1] 0.7342603

You could also use the posterior distribution to find a Bayesian 95%
probability interval $(0.53, 0.89)$ (from R). 
qbeta(c(.025, .975), 16, 6)
[1] 0.5283402 0.8871906

Note: Maybe you can get the posterior distribution using the prior
$\mathsf{Beta}(5, 5)$ and the same likelihood function, and see what
difference that alternate choice of prior distribution makes in the
point and interval estimates.
A: With a typical coin it is quite unreasonable to assign a uniform distribution on the interval from $0$ to $1$ as the probability distribution of the frequency with which "heads" appear. For example, the uniform distribution would mean that there is a $10\%$ chance that "heads" appear more than $90\%$ of the time and there is a $10\%$ chance that "heads" appear less than $10\%$ of the time. That is not reasonable as applied to any normal coin. So you have to take it as a given in the math problem that is set here.
The physicist Edwin Jaynes has argued that this is a reasonable prior when you know that "heads" and "tails" are both possible outcomes but you know nothing else. That is certainly different from the situation with most coins.
