Suppose we flip a coin n-times. By a series of repetitions we mean successive tosses in which the same side of the coin lands (heads or tails each time). For example, in HHHTTHTH ( => 8) there are 5 repetition sequences.
HHH-TT-HTH -> HHH are 3 and TT are 2, summary there is 5 repetitions
What is the expected number of repetition sequences for n-flips ?? How to resolve this, i was trying draw it as probability tree and calculate summary of probability successes but i didnt get anything with any sense.
/// I updated this post
I tried to solve this after your advices, but I think I do something wrong
Step 1 We toss a symetric coin so chances for H or T are equal = 0.5
Let: $ a_n $ - number of repetion sequences for n flips
In first toss we have: $ a_1=1 $
For next tosses $ n+1 $ we have got 50% chances to continue sequence from $n$
As formula it's $ a_{n+1}=\frac{a_n+1}{2} $
or 50% to beginn new series ofrepetitions: $a_{n+1}=\frac{1}{2}*1$
Let's write both together as recurent formula: $a_{n+1}=\frac{1}{2}*\left(a_n+1\right)+\frac{1}{2}*1$
After simplify: $a_{n+1}=\frac{a_n}{2}+\frac{3}{2}$
Try proof that
Calculate this for n flips:
$a_{n+1}=\frac{a_n}{2}+\frac{3}{2}$
$a_n=\frac{a_n}{2}+\frac{3}{2}$
$a_{n-1}=\frac{a_{n-1}}{2}+\frac{3}{2}$
⋮
$a_2=\frac{a_1}{2}+\frac{3}{2}$
$a_1=1$
After substituting and calculating the above equations, we get the following:
$a_n=1+\frac{n-1}{2}\ = \frac{n+1}{2} $
And my question is whether the above provisions are OK and how to proof this? Math iteration, induction? I don't know how to do this proof.
Try induction: if $a_n$ is the expected number of repetition sequences after $n$ coin tosses then it's clear that $a_1 = 1$. Furthermore if we toss another coin, the probability this is part of a 'new' repetition sequence is precisely $1/2$ (since this is just the event we get a different toss to the previous one). Thus $a_{n+1} = a_n + 1/2$ implying $a_n = 1+\frac{n-1}{2}$.