What is the probability that when flipping a fair coin ten heads will be thrown in a row I just cant figure this out.. Should I be using binomial distribution? Chance of getting ten tail long series in 100 coin throws
 A: At the moment I cannot see a neat way of doing this. So here is an inelegant way. Let $a_n$ be the number of binary sequences length $n$ with at least one run of ten 1s. Consider a sequence length $n+1$. It must begin in one of the following ways: 0,10,110,1110,11110,111110,1111110,11111110,111111110,1111111110 or 1111111111 (to save you counting the last is ten 1s). The number of sequences containing at least one run of ten 1s is $a_n,a_{n-1},\dots,a_{n-9},2^{n-9}$ respectively. That and $a_0=\dots=a_9=0$ gives us a recurrence relation.
We can ask a computer to solve it and give us the answers for sequences length 10-100:
RecurrenceTable[{a[n + 1] == 
   a[n] + a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4] + a[n - 5] + 
    a[n - 6] + a[n - 7] + a[n - 8] + a[n - 9] + 2^(n - 9), a[0] == 0, 
  a[1] == 0, a[2] == 0, a[3] == 0, a[4] == 0, a[5] == 0, a[6] == 0, 
  a[7] == 0, a[8] == 0, a[9] == 0}, a, {n, 10, 100}]
And it obliges by returning: 1,3,8,..., 55 950 584 378 441 149 993 810 452 680
We now compare this with the total number of sequences length 100 which is $2^{100}$ to get as the chance of getting a run of 10 heads: 0.0441372 or $4.4\%$.
A: The probability of getting a 10 heads in 10 throws = $\frac{1}{2^{10}}$
The expected number of ten head sequence in N throws = $(N-9) * \frac{1}{2^{10}}$
A chain of 11 heads is two 10-head sequences. 
To calculate the probability of getting one 10-head sequence in 100 throws, we must remove the 11+ head sequences that we are counting too many times.
$\frac{91}{2^{10}} - \frac{90}{2^{11}}$
