# Question of expected number of consecutive coin flip with increasing bias [closed]

This is a question I found on the book and I don't know how to tackle it. Thanks to any help or hint in advance.

I have a coin that, I could get the head 100% at the first flip, $$\frac{1}{3}$$ at the second flip, $$\frac{1}{5}$$ at the third flip ...... $$\frac{1}{2i - 1}$$ at the $$i^{th}$$ flip.

Suppose I would get a point when I get 4 consecutive flips (4 heads or 4 tails). If I flip 100 times, what is the expected number of the points I would get?

Suppose $$n\ge4$$. 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 $$EN_T=\sum_{j=1}^{n-3}E\prod_{k=0}^3 (1-I_{j+k}) \\ =\sum_{j=1}^{n-3}\prod_{k=0}^3 (1-EI_{j+k}) =\sum_{j=1}^{n-3}\prod_{k=0}^3\Big(1-\frac1{2j-1+2k}\Big) =\frac{4 (n-3) \left(n \left(6 n^2+4 n-39\right)+23\right)}{3 (2 n-5) (2n-3) (2 n-1)}+2 H_{n-4}-4 H_{2 n-7},$$ where $$H_p$$ is the $$p$$th harmonic number. The expected number of points you would get is $$EN_H+EN_T$$ for $$n=100$$, which is $${\scriptsize\frac{234670858218227240827580740387990697718075439354345716280274779893948745206696917816972687}{ 2635106162757236442495826303084698495565581115509040892412867358728390 766099042109898375}}\approx89.0556\approx100,$$ which makes sense.