Probability of seeing sun rise tomorrow using Bayes theorem When Richard Price's published the Bayes theorem, he gave the example of a man seeing the sun-rise for the first time and wondering if it happened everyday. With each observation thereafter, he updates his belief (i.e. the sun does rise everyday) using Bayesian reasoning. This question tries to analyze the same problem.

It is a stormy night and an alien prisoner is abandoned on the frigid earth during the first ice age. Scared, he finds a cave and only comes out for a few hours at night to find some food. He has no luck though. After spending 6 nights like this, hunger becomes unbearable and so on the 7th night he decides to venture out for a little longer. After a few hours, he sees the skies changing colors. The only time he saw them change color before was during  thunderstorm lightning on the night he was dropped. This time, however, he witnesses a glorious sunrise. He came from the dark side of a tidally-locked planet (i.e.  one side always faces its star and the other  side is forever dark). On his planet, days are not tied to the sun, but time (24 hours is measured as a day). As the warmth soothes him, he wonders if this is an one off event on earth or if the sun will rise again tomorrow. How can he use Bayes theorem to update his beliefs?

Bayes Theorem
Total number of nights observed            = 7
Total number of times sky changed colors   = 2 (once due to lightning and once due to sunrise)
Total number of times the sunrise happened = 1

P(belief | observation) = P(observation | belief) x P(belief) / P(observation)

P(Belief)      = The sun will rise tomorrow
P(Observation) = Sky changed colors
P(Sun will rise | Sky changed color) = P (Sky changed color | Sun will rise) * P (Sun will rise) / P(Sky changed color)

Day 2
P(Sun will rise | Sky changed color) = (1/2) * (1/7) / (2/7)
                                 = 0.25

Day 3
P(Sun will rise | Sky changed color) = (2/3) * 0.25 / (3/8)
                                 = 0.44

Day 4
P(Sun will rise | Sky changed color) = (3/4) * 0.44 / (4/9)
                                 = 0.74

Day 5
P(Sun will rise | Sky changed color) = (4/5) * 0.74 / (5/10)
                                 = 1.184

Why does the posterior probability go beyond 1? What am I doing wrong here?

First, a simplification.  The event "the sky changes color" is actually irrelevant to this problem, since the individual is not concerned with whether or not the sky changes color, either as an outcome or as a conditioning event.  Consequently, we can ignore it; the data then becomes seven observations, assumed independent and identically distributed, with one of those observations being a sunrise and the other six not.  This will give us a $\mathrm{Binomial}(7,p)$ likelihood for the probability parameter $p$ at the end of the seven days.
We have to assume some prior distribution on $p$ at the start of the seven days.  Given the background of the individual, $p=0$ would be not unreasonable, but we will ignore this and select an arbitrary $\mathrm{Beta}(a,b)$ prior, with $a,b > 0$.  Since the prior is conjugate, the posterior, given a single day's observation $x \in \{0,1\}$ where $1$ represents sunrise and $0$ represents no sunrise, will be a $\mathrm{Beta}(a+x, b+1-x)$ distribution.
After day 1, the posterior will be a $\mathrm{Beta}(a,b+1)$ distribution.  This then becomes the prior for day 2.
After day 2, the posterior will be a $\mathrm{Beta}(a,b+2)$ distribution.   We have added one "failure" to the second parameter of the Beta prior, which has parameters $(a,b+1)$. This then becomes the prior for day 3.  Note that this is exactly the same posterior we would have had with our initial prior and two days of no sunrise added in one step instead of sequentially.
...
After day 7, the posterior will be a $\mathrm{Beta}(a+1,b+6)$ distribution.  We will have sequentially added six "failures" to $b$ and $1$ "success" to $a$.
Let us compare with the results of a one-step update of the original prior with all seven  days of information:
$$f'(p) \propto p^{a-1}(1-p)^{b-1} \times p^1(1-p)^6  = p^a(1-p)^{b+5}$$
which is a $\mathrm{Beta}(a+1, b+6)$ distribution, the same functional form and parameterization we see if we add the seven days of information one at a time.
