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A biased coin is tossed with probability 0.6 of landing on Heads (H) and probability 0.4 of landing on Tails (T). A group is a maximal sequence of consecutive tosses that land on the same side. For instance, the groups of HTTTHTHTT are (H)(TTT)(H)(T)(H)(TT)

What is the expected number of groups after 10 tosses?

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    Hint: work with indicator variables...Let $X_i$ for $i\in \{1,\cdots 9\}$ indicate whether or not $X_i\neq X_{i+1}$ – 2017-02-02
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    Better yet, let $X_i$ indicate whether the coin is a head, and $Y_i$ indicate whether $X_i\neq X_{i+1}$. – 2017-02-03

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For $i\in\{1,..,10\}$ we let $X_i$ indicate that toss #$i$ is a head.   Then $\mathsf E(X_i)=\mathsf P(X_i=1)=0.6$.

For $i\{1,..,9\}$ we let $Y_i$ indicate that tosses #$i$ and #$(i+1)$ are both the different.   That is that $Y_i=1$ when $X_i\neq X_{i+1}$ and $0$ otherwise.   Also let $Y_{10}=1$.

Thus $\sum_{i=1}^{10} Y_i$ counts the number of groups.   Now find the expectation of this.

$$\mathsf E(\sum_{i=1}^{10} Y_i) ~=~ \ldots$$

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    Thanks. So, I Calculated this expected value to be 1+2*(n-1)*p*(1-p) where p is 0.6 and n is the number of tosses. But now how would i answer the follow up: What is the probability of (strictly) exceeding 6 groups after 10 tosses? – 2017-02-03