For any natural $j$, let $I_j$ be the indicator of heads in the $j$th flip. Then the number of times 4 consecutive heads appear in $n$ flips is 
$$N_H=\sum_{j=1}^{n-3}\prod_{k=0}^3 I_{j+k}.$$
So,
$$EN_H=\sum_{j=1}^{n-3}\prod_{k=0}^3 EI_{j+k}
=\sum_{j=1}^{n-3}\prod_{k=0}^3\frac1{2j-1+2k}=\frac{4 n^3-18 n^2+23 n-15}{45 (2 n-5) (2 n-3) (2 n-1)}. 
$$
Similarly, the expected number of times 4 consecutive tails appear in $n$ flips is 
$$EN_T=\sum_{j=1}^{n-3}E\prod_{k=0}^3 (1-I_{j+k}) \\
=\sum_{j=1}^{n-3}\prod_{k=0}^3 (1-EI_{j+k})
=\sum_{j=1}^{n-3}\prod_{k=0}^3\Big(1-\frac1{2j-1+2k}\Big)
=\frac{80 (n-3)}{77}. 
$$
The expected number of points you would get is $EN_H+EN_T$ for $n=100$, which is 
$$\frac{177985559863}{1765899135}\approx100.790\approx100,$$
which makes sense.