The idea of this blog was shown up in my mind on my wedding day. In Indonesian tradition, a groom must say a (long) sentence in front of several chosen people in order to make a marriage. The sentence is called "Ijab Qabul". I was really nervous that day, especially my dad said that I have to memorise the sentence and read it in one breath. If I failed, I have to repeat it again until I make it.

One hour before the ceremony, I decided to practice the sentence in a silent room. My target is to say the sentence 20 times without failing. If I made a mistake, I have to repeat my count from 0. After trying many times, I could only manage saying the sentence 18 times in a row. And I was thinking, "if I can say the sentence 18 times in a row, what is the probability I can say it without a mistake next time?"I didn't have time to think about it on my wedding day. But several days later, I managed to solve the problem.

My method is to use Bayesian probability approach. Let's say \( p \) is the successful probability and \( p \) can be between 0 to 1. Also, denote \( k \) as the number of successful events happen in a row. The probability of successful probability has value of \( p\) given there are \( k\) successful events is

$$ P(p|k) = \frac{P(k|p) P(p)}{P(k)}. $$

Let's calculate the variables on the right hand side one by one. The variable \( P(k|p) \) denotes the probability of getting \( k\) successful events in a row given the successful probability is \( p\). By a simple AND rule,

$$ P(k|p) = p^k. $$

Now, for the probability of successful probability having value of \( p\), \( P(p)\), is a bit different. The range of \( p\) is continuous, so instead of calculating the probability of successful probability having the exact value of \( p\), we can calculate the probability having the value between \( p\) and \( p + dp \), which is

$$ P(p) = dp\ \mathrm{for}\ p \in [0,1]. $$

Lastly, we need to compute the probability of getting \( k\) successful events in a row, considering all possible values of \( p\). This can be done by some integration

$$ P(k) = \int_0^1 P(k|p) dp = \int_0^1 p^k dp = \frac{1}{k+1}. $$

Now combining all the variables, we can get the probability of the successful probability having value of \( p\), given there are \( k\) successful events in a row,

$$ P(p|k) = (k+1) p^k dp. $$

We already obtained the probability distribution of \( p\), but this does not answer our original question, "what is the probability of the next event successful?"

To answer that question, we can calculate the expected value of \( p\),

$$ \mathbb{E}(p|k) = \int_0^1 p P(p|k) = (k+1) \int_0^1 p^{k+1} dp = \frac{k+1}{k+2}. $$

So, the probability of having the next event successful, given there are already \(k\) successful events in a row, is \( (k+1)/(k+2) \).

For my case, I can say the sentence 18 times in a row without a mistake. So the probability of saying it correctly next time is 95%.

Side note: it turns out that during Ijab Qabul, the imaam gave me a paper containing the text, so I can just read it. And also, I realised later that this problem is actually the "sunrise problem".