Posterior Predictive Distribution for a coin toss 
In this question, i can work out that the posterior is supposed to be a Beta (r+1, n-r+1) distribution.
However, what I am struggling with is how to compute f(X_n+1|theta). Is this the binomial distribution with r+1 replacing r, because it's the probability of achieving r+1 heads in X_n+1 flips?
Or is it simply theta, because it's the probability of seeing an (r+1)th head in 1 extra flip, having observed r heads in n flips?
Obviously depending on which is the correct f, the answers vary significantly.
Any help would be greatly appreciated! Thanks a lot
 A: Taking a Bayesian approach one considers the results on the first
$n$ tosses to provide information about the Heads probability $\theta.$
So if you got $r = 400$ heads in $n = 1000$ previous tosses, your
posterior distribution would be $\theta \sim Beta(401, 601).$
Taking the posterior mean as the success probability for toss $1001$
you would have $E(\theta) = 401/1002 = 0.4002.$ Some would prefer
to use the posterior mode $400/1000 = 0.4000$ or the posterior
median $0.4001,$ but it doesn't much matter. In any case, the
Bayesian approach would predict $P(Heads_{1001}) \approx 0.4.$
The reason that Bayesian modeling is used ever more frequently
in polling is that past experience can inform future predictions.
Often the posterior from the last poll helps to formulate the
prior for the next.
Some would argue that you have to take a realization $\theta$
at random from  $Beta(401, 601)$ and then toss a coin with
that Heads probability. The following R code performs that
experiment a million times, with the expected result. (In more
complicated Bayesian inference models the result is not obvious,
and numerical or analytical integration is necessary.)
 m = 10^6;  x = numeric(m)
 for (i in 1:m) {
   th = rbeta(1, 401, 601)
   x[i] = rbinom(1, 1, th) }
 mean(x == 1)  # P(Heads)
 ## 0.400291

In (b), for the probability of 3 Heads in 5 extra flips the probability
is 0.2306:
 dbinom(3, 5, .4002)
 ## 0.2305920

 m = 10^6;  x = numeric(m);  k = 3;  t = 5  
 for (i in 1:m) {
   th = rbeta(1, 401, 601)
   x[i] = rbinom(1, t, th) }
 mean(x == k)
 ## 0.23003

There is an old joke question about how to distinguish among
a probabilist, a statistician, and a fool: I have just tossed a coin 20 times
and gotten 20 Heads in a row. What are the chances of a Head
on the next toss? The probabilist, persistently faithful to a belief
in fair coins says 1/2. The statistician (especially a Bayesian
one) says he/she is betting it'll be Heads again because the coin
shows signs of bias. The fool says it's about time for Tails to
come up.
Note: Some reputable experimenters claim the all coins are
essentially fair, so let's assume the 'coin' in this Question
is really a die with H on three faces and T on the other three.
It is pretty well established that one can 'load' a die to be
unfair. (Standard dice have a corner where faces 1, 3, and 5 touch;
putting a heavy weight just inside the opposite corner biases
the die toward odd outcomes.)

